# 3.8: Exercises on Acids and Bases II

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These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

### Q3.1

At room temperature, a solution is made by adding 50mL of 0.325M $$HF$$ $$(Ka = 6.6 \times 10^{-4})$$ with 10mL of 0.50M HCl. Calculate the following at equilibrium:

1. $$[H_{3}O^{+}]$$
2. $$[OH^-]$$
3. $$[HF]$$
4. $$[Cl^-]$$

For a solution that is 0.138 M $$NH_3$$ $$(Ka=1.9\times 10^{-5})$$ and 0.101 M $$KN_{3}$$ (potassium azide) calculate the following:

1. $$[H_{3}O^{+}]$$
2. $$[N_3^-]$$
3. $$[K^+]$$
4. $$[OH^-]$$

### Q3.3

Calculate the change in pH that results from adding

1. 0.150 mol $$LiNO_2$$ to 1.00 L of 0.150 M $$HNO_{2\ (aq)}$$
2. 0.150 mol $$LiNO_3$$ to 1.00 L of 0.150 M $$HNO_{3\ (aq)}$$

### Q3.4

A solution of volume 50.0 mL contains 12.5 mmol $$HClO_2$$ and 7.25 mmol $$KClO_2$$.

1. What is the $$pH$$ of this solution?
2. If 1.00 mmol $$Ca(OH)_2$$ is added to the solution, what will be the new $$pH$$?
3. If 2.00 mL of 6 M $$HCl$$ is added to the original solution, what will be the new $$pH$$?

### Q3.5

There are various procedures for preparing buffer solutions. To obtain a $$pH = 4.60$$, one such way is to mix 18.00 mL of 0.300 M $$C_5H_5N$$ with 72.00 mL of 0.300 M $$C_5H_5NH^+$$.

1. Show by calculation that the $$pH$$ of this solution is 4.60.
2. Would you expect the $$pH$$ of this solution to remain the same if 100.00 mL of buffer solution were diluted to 500.00 mL? To 1.00 L? Explain.
3. What will be the $$pH$$ of the original 100.00 mL of buffer solution if 0.40 mL of 1.00 M $$HBr$$ was added to it?
4. What is the maximum volume of 1.00 M $$HBr$$ that can be added to the original buffer solution so that the $$pH$$ does not drop below 4.4?

### Q3.6

Use the following data table to answer the following questions Identify one indicator (each) that would change color in an acidic solution, a basic solution, and a neutral solution.

### Q3.7

What color will a solution with bromphenol blue change to with the following species in solution

1. $$Mg(OH)_2$$
2. $$HCl$$
3. $$HF$$

### Q3.8

What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the pH value: $$Na(OH)_2$$ for $$pH=10.92$$

### Q3.9

A solution is prepared with 0.3 M $$NH_3$$ and 0.4 M $$NH_4Cl$$.

Show that this is a buffer solution.

Calculate the $$pH$$ of this buffer solution. $$K_b=1.8 \times 10^{-5}$$

What is the final $$pH$$ if 1 L of 0.2 M $$NaOH$$ is added to 1 L of this buffer solution.

### Q3.10

Hydroflouric acid, $$HF$$, is a stronger acid than water and when in an aqueous solution, the following reaction occurs.

$$HF + H_2O \rightleftharpoons H_3O^+ + F^-$$

Three experiments were completed to obtain a $$pKa$$ value for $$HF$$ at $$0^oC$$. A sample of $$HF$$ was shaken and separated into two layers, one with 1-pentanol, and the other with water. Assume no ionization or disassociation occurs in the 1-pentanol layer.

• The first experiment showed that the water layer contained 5 times as much $$HF$$ as the 1-pentanol layer did.
• The 2nd experiment involved adding .5 M $$LiOH_{(aq)}$$ and 1-pentanol. At equilibrium, the concentration of $$HF$$ was 0.005 M in the 1- pentanol layer and 0.35 M in the aqueous layer.
• The third experiment showed that when .25 M $$LiOH_{(aq)}$$ was added, the concentrations in the 1-pentanol layer was .0015 in the 1-pentanol layer and .12 M in the aqueous layer. At $$0^oC$$, $$pK_w= 14.94$$.

Find $$pK_a$$ value of $$HF$$.

### Q3.11

To increase the ionization of $$H_2S$$, which of the following compounds should be added to the solution?

1. $$NaHSO_4$$
2. $$HCl$$
3. Acetic acid ($$CH_3COOH$$)
4. $$NaCN$$

### Q3.12

For the solution that is 0.330 M $$CH_3CH_2COOH$$ $$K_a= 1.3\times 10^{-5}$$ and 0.010M $$HI$$, calculate

(a) $$[H_3O^+$$

(b) $$[OH^-]$$

(c) $$[CH_3CH_2COOH]$$

(d) $$[I^-]$$

### Q3.13

For a solution that is 0.654 M $$NH_3$$ and 0.202 M $$NH_4Cl$$, calculate :

(a) $$[H_3O^+]$$

(b) $$[OH^-]$$

(c) $$[NH_4^+]$$

d) $$[Cl^-]$$

### Q3.14

Calculate the change in $$pH$$ that results from adding (a) 0.300 mol $$NaNO_2$$ to 1.00 L of 0.200 M $$HNO_{2(aq)}$$; (b) 0.300 mol $$NaNO_3$$ to 1.00 L of 0.200M $$HNO_{3(aq)}$$.

### Q3.15

A solution of volume 1L contains 0.500 mol $$HCHO_2$$ and 0.250 mol $$NaHCOO$$.

1. What is the $$pH$$ of this solution?
2. If 0.1 mol $$Ba(OH)_2$$ is added to to the solution, what will be the new $$pH$$?
3. If 10.0 mL of 12 M $$HCl$$ is added to the original solution what will be the new $$pH$$?

### Q3.16

To obtain a $$pH = 8.00$$ you need to mix 35.00 mL of .220 M $$NH_3$$ with 65.00 mL of .250 M $$NH_4Cl$$.

1. Show the calculation that the $$pH$$ of this solution is 8.00.
2. Would you expect the $$pH$$ of this solution to remain at 8.00 if the 100 mL of the buffer were diluted to 1.00 L? 1000L? Explain.
3. What will be the $$pH$$ of the original solution if 0.20 mL of 1.00 M $$HCl$$ is added?

### Q3.17

Use data from table

 Indicator KHIn Acid -> Anion Bromphenol blue 1.4 x 10-4 yellow -> blue Bromcresol green 2.1 x 10-5 yellow -> blue Bromthymol blue 7.9 x 10-8 yellow -> blue 2,4 – Dinitrophenol 1.3 x 10-4 colorless -> yellow Chlorphenol red 1.0 x 10-6 yellow -> red Thymolphthalein 1.0 x 10-10 colorless -> blue

a) Which of these indicators change color in acidic, basic, or neutral point.

b) What is the approximate $$pH$$ of a solution if bromocresol green indicator turns green? If chlorphenol red turns orange?

### Q3.18

Phenol red indicator changes from yellow to red in $$pH$$ ranges from 6.6 to 8.0. Without making detailed calculations, state what color the indicator will assume in each of the following solutions.

1. 0.10 M $$KOH$$
2. 0.10 M $$CH_3COOH$$.
3. 0.10M $$NH_4NO_3$$
4. 0.10M $$HBr$$
5. 0.10 M $$NaCN$$
6. 0.10 M $$CH_3COOH$$ and 0.10 M $$NaCH_3COO$$.

### Q3.19

Two aqueous solutions are mixed. 75 mL of .333 M $$H_2SO_4$$ and 25 mLof 0.05 M $$NaOH$$. What is the pH of the resulting solution?

### Q3.20

Explain why the volume of 0.100 M $$NaOH$$ required to reach the equivalence point in the titration of 30.00 mL of 0.100 M $$HA$$ is the same regardless of whether $$HA$$ is a strong or a weak acid, yet the $$pH$$ at the equivalence point is not the same.

### Q3.21

Sketch a titration curve for each of the following three weak acids when titrated with 0.100 M $$NaOH$$.

1. 10.00 mL of 0.100 M HX; $$Ka=7.0\times 10^{-3}$$
2. 10.00 mL of 0.100 M HY; $$Ka=3.0\times 10^{-4}$$
3. 10.00 mL of 0.100 M HZ; $$Ka=2.0\times 10^{-8}$$

### Q3.22

Is 0.20M $$Na_2S_{(aq)}$$ solution likely to be acidic, basic, or neutral? Explain.

### Q3.23

Sodium phosphate, $$Na_3PO_4$$, is made by neutralizing phosphoric acid with sodium carbonate to obtain $$Na_2HPO_4$$. The $$Na_2HPO_4$$ is further neutralized to $$Na_3PO_4$$ with $$NaOH$$.

1. a) Write net ionic equations for these reactions.
2. b) $$Na_2CO_3$$ is a much cheaper base than $$NaOH$$. Why do you suppose that $$NaOH$$ must be used as well with $$Na_2CO_3$$ to produce $$Na_3PO_4$$?

### Q3.24

The $$pH$$ of a solution of 19.5 g malonic acid in 0.250 L is 1.47. The $$pH$$ of a 0.300 M solution of sodium hydrogen malonate is 4.26. What are the values of $$K_{a1}$$ and $$K_{a2}$$­ for malonic acid?

### Q3.25

Using appropriate equilibrium constants but without doing detailed calculations, determine whether a solution can be simultaneously.

1. 0.10 M $$NH_3$$ and 0.10 M $$NH_4Cl$$, with $$pH = 6.07$$
2. 0.10 M $$NaC_2H_3O_2$$ and 0.058 M $$HI$$
3. 0.10 M $$KNO_2$$ and 0.25 M $$KNO_3$$
4. 0.050 M $$Ba(OH)_2$$ and 0.70 M $$NH_4Cl$$
5. 0.018 M $$C_6H_5COOH$$ and 0.018 M $$NaC_6H_5COO$$ with $$pH = 4.10$$

### Q3.26

A buffer solution can be prepared by by starting with a weak acid, $$HA$$, and converting some of the weak acid to its salt by titration with a strong base. The fraction of the original acid that is converted to the salt is designated $$F$$.

1. a) Derive an equation similar to (16.2) but express in terms of $$F$$ rather than concentrations.
2. b) What is the $$pH$$ at the point in the titration of phenol, $$C_6H_5OH$$, at which $$F = 0.27\ (pK_a\ of\ phenol = 10.00)$$?

### Q3.27

Complete the derivation of equation (16.5). Then derive equation (16.6)

### Q3.28

A solution is prepared that is 0.200 M $$CH_3COOH$$ and 0.250M $$NaHCOO$$.

1. Show that this is a buffer solution
2. Calculate the $$pH$$ of this solution.

### Q3.29

For water at $$0^oC$$, $$pK_w=14.94$$. Find the approximate value of $$K_a$$ for $$H_2O_2$$ at $$0^oC$$.

### Q3.30

To increase the ionization of formic acid, $$HCOOH_{(aq)}$$, which of the following should be added to the solution?

1. $$NaCl$$
2. $$NaHCOO$$
3. $$H_2SO_4$$
4. $$NaHCO_3$$

### Q3.31

For a solution that is 0.345 M hydrosulfuric acid ($$K_a=1.0 \times 10^{-7}$$) and a pH of 3.2. Find [$$H_3O^+$$].

### Q3.32

For a solution that is 0.35 M $$HC_2H_3O_2$$ and $$C_2H_3O_2$$ calculate:

1. $$pH$$
2. $$[H_3O^+]$$

### Q3.33

Calculate the pH of the buffer that results from mixing 100.0 ml of .250 M $$HCHO_2$$ and 10.0 ml of 1 M $$NaCHO_2$$. $$pKa = 3.744$$

### Q3.34

What is the pH of a 0.07M ammonia solution?

### Q3.35

What are the approximate pHs of a solution for the following indicator strips at the specified color change?

1. methyl red indicator strip changes from orange to yellow
2. bromphenol indicator strip changes from yellow to purple
3. thymol blue indicator strip changes from yellow to blue

### Q3.36

Azolitman indicator strips change from red to blue in the pH range from 4.5 to 8.3. What color will the indicator be for the following:

1. 0.000014 M $$HCl$$
2. 0.00014 M $$HCl$$
3. 0.0014 M $$HCl$$
4. 0.0000014 M $$HCl$$

### Q3.37

Why do we usually try to keep the amount of indicator added relatively small when used as acid-base indicators?

### Q3.38

Define the equivalence point in a titration curve.

### Q3.39

What relationship do the molarity of Base added in a titration and molarity of either strong of weak acid share?

### Q3.40

0.205 M $$HClO$$ reacts with water to form .020 $$MClO^-$$. The pH of the solution is 4.04. Determine $$K_a$$ of $$HClO$$.

### Q3.41

Titrate 50mL of 0.200M $$HNO_3$$ with 0.100M $$NaOH$$. Add 150mL $$NaOH$$. 15mmol HNO3. And Find $$pOH$$ and $$pH$$.

### Q3.42

Is a solution of 0.23 M $$NaI$$ likely to be acidic, basic, or $$pH$$ neutral? Explain.

### Q3.43

In a strong acid base titration, at what $$pH$$ does the equivalence point always occur?

### Q3.44

Find the $$pH$$ for $$HCN$$ at $$0^oC$$ if the $$pK_b$$ of $$HCN$$ is 4.8

$$HCN + H_2O \rightleftharpoons H_3O^+ + CN^-$$

### Q3.45

Can an aqueous solution of $$NH_3$$ and $$NH_4Cl$$ form a $$pH$$ of 6 with equal molarity of each added together?

### Q3.46

A buffer solution contains 0.50M acetic acid ($$Ka=1.8\times 10^{-5}$$) and 0.50M sodium acetate. Find the $$pH$$ of the solution.

### Q3.47

You need bring the $$pH$$ of .325 L of .452 M $$NaOH$$ to 7.00. What volume of which of the following solutions would you use: 5.00 M $$HBr$$ or 5.00 M $$Na_2S$$?

### Q3.48

What is the pH of a 0.025 M $$HCN$$ solution? $$K_a=6.2\times 10^{-10}$$

### Q3.49

Define the common Ion Effect and give an example of it.

### Q3.50

If a solution contains 50 ml of 0.25M of $$NH_3$$ and 75 ml $$NH_4Cl$$, calculate the molarity of $$NH_4Cl$$.

### Q3.51

If given two different solutions, one acid and one base, what is the best method to use to calculate the $$pH$$ of the solution?

### Q3.52

If a mixture of .005 mol $$NaOH$$ and .008 mol $$HCl$$ has a volume of 50ml, what is range of $$pH$$ of the solution?

### Q3.53

If an original solution had a $$pH$$ level of 6.05, in what direction would the $$pH$$ shift (towards or away from equilibrium) if 50ml of $$HCl$$ were added

### Q3.54

Does Bromecresol Green ($$2.1\times 10^{-5}$$) change color in an acidic or basic solution?

### Q3.55

If Chlorphenol Red indicator turns red at $$ph= 6.0$$, then will it change colors at 0.10 M $$KOH$$?

### Q3.56

Two aqueous solutions are mixed together, 100 ml of 0.025M $$H_2SO_4$$ and a given amount of 0.03M $$NaOH$$. What is the total volume at equilibrium?

### Q3.57

Why is the volume the same if 0.2 M $$HA$$ in a solution if 0.2 M $$NaOH$$ is either a weak or strong acid?

### Q3.58

Sketch a titration curve of a hypothetical weak acid when .1M of $$NaOH$$ is added. Assume 100ml at 0.1M of $$C_2H_4O_2$$.

### Q3.59

With the titration of 50ml of 0.1M $$NaOH$$ with 0.1M $$HCl$$, solve for the volume of $$HCl$$

### Q3.60

Is the solution 0.1M $$HCl$$ likely to be acidic, basic, or neutral?

### Q3.61

Define the term of stoichiometric concentration.

### Q3.62

A buffer solution can be made by taking a weak acid and converting a portion to its salt. This is done by titration with what?

### Q3.63

If a $$pH$$ of an unknown substance is set at 5, would a strong acid or a strong base solution be needed for the substance to reach equilibrium?

### Q3.64

A solution is made of 0.1M $$CH_3COOH$$ and 0.1M $$NaHCOO$$. Calculate the range of $$pH$$ of this solution.

### Q3.65

In a solution of 0.153 M CH_3COOH ($$K_a=1.8\times 10^{-5}$$). Calculate $$[H_3O^+]$$.

### Q3.66

In a Solution of 0.548M $$HCl$$ ($$K_a=7.2\times 10^{-4}$$) and pure water, calculate $$pH$$.

### Q3.67

In a 300 mL solution of 0.1 M $$HClO_{2(aq)}$$ calculate the $$pH$$ change when adding 10 mL of 0.2 M $$HCl_{(aq)}$$ to the solution ($$K_a=1.1\times 10^{-2}$$).

### Q3.68

A 55 mL solution is made up of 0.45 mmol $$HClO_2$$ and 6.0 mmol $$KClO_2$$. Calculate the $$pH$$.

### Q3.69

In a buffer solution of 50 mL of 0.3 M $$H_3PO_4$$ and 25 mL of .3M $$NaH_2PO_4$$ with a pH of 1.82 what would the $$K_a$$ of this solution be?

### Q3.70

Between an indicator with a $$K_{Hln}=2\times 10^{-4}$$, an indicator with a $$K_{Hln}=2\times 10^{-10}$$, and an indicator with a $$K_{Hln}=2\times 10^{-7}$$, which would be used in a titration of strong acid and strong base to find the endpoint?

### Q3.71

Bromophenol blue changes from yellow to blue-violet in the pH range 3-5. Which of the following titrations would this be best to use in: strong acid-strong base, strong acid-weak base, strong base-weak acid?

### Q3.72

When 20mL of 0.05M KOH of is titrated with 0.05M $$HI$$, an indicator goes from yellow to red when 10mL of the weak acid is added. What is the $$pK_{Hln}$$ of the indicator? Does this give the accurate endpoint of titration?

### Q3.73

What is the $$pOH$$ of a solution with 200mL of 0.007M $$HCl$$ and 300mL of 0.02M $$NaOH$$?

### Q3.74

Which will be greater, the volume of 0.2M $$LiOH$$ is needed to reach the equivalence point with 20mL of 0.2M $$HF$$ or with 20mL of 0.2M $$CH_3COOH$$? What about their $$pH$$ at the equivalence point?

### Q3.75

Based on $$K_b$$, how do titrations curves of strong acids with weak bases differ?

### Q3.76

Calculate the pH of the solution in the titration of 0.05M $$HBr$$ with 40mL of 0.02M $$CsOH$$ when, 1) 5mL 2)10mL 3) 25mL of $$HBr$$ is added.

### Q3.77

Does 0.05M $$NaHSO_4$$ produce an acidic, basic, or neutral solution? (given that $$pK_{a2}=1.92$$)

### Q3.78

How can you obtain $$Na_3SO_4$$ from sulfuric acid and sodium carbonate? Write out equations.

### Q3.79

Show by calculations that the stoichiometric concentration of $$CH_3COOH$$ is 0.1 M is required to obtain an aqueous solution with the $$pH$$ value of about 2.8 .

### Q3.80

Determine the pH values of $$NaH_2PO_{4\ (aq)}$$ and $$Na_2HPO_{4\ (aq)}$$ when $$pK_{a1}= 2.15$$, $$pK_{a2}= 7.20$$, $$pK_{a3}= 12.38$$. Note that $$H_2PO_4^-$$ occurs simultaneously.

### Q3.81

Define what the term simultaneously means in terms of hydrolysis. Determine whether a solution can be simultaneously (these don’t need calculations).

a. 0.20M $$NH_3$$ and 0.20 $$NH_4Cl$$

b. 0.20 M $$NaC_2H_3O_2$$ and 0.116 M $$HI$$

c. 0.036 M $$C_6H_5COOH$$ and 0.036 M $$NaC_6H_5COO$$

### Q3.82

Determine the amount (in mL) of 6.0 M $$NaOH$$ that is required to adjust the pH of the initial buffer solution to a $$pH$$ of 7.4.

Note that a student working with this buffer mixes together 400 mL of 1.80 M $$Na_2HPO_4$$ and 130 g of $$NaH_2PO_4$$.

### Q3.83

What is the buffer range for a benzoic acid/ sodium benzoate buffer? $$[Ka=6.3\times 10^{-5}]$$.

### Q3.84

Define what a common ion effect is when a strong acid supplies the common ion $$H_3O^+$$, the equilibrium shifts which way to form more $$CH_3COOH$$? When a strong basic supplies the common ion $$OH^-$$, the equilibrium shifts which way to form more $$NH_3$$?

### Q3.85

What is the effect on the pH of adding 0.0060 mol $$HCl$$ to 0.3 L of a buffer solution that is 0.250 M $$CH_3COOH$$ and 0.560 M $$NaCH_3COO$$?

Find the pH of $$2.0\times 10^{-8}$$M $$HCl$$. Note that water makes a significant contribution to the $$[H_3O^+]$$. ($$K_w= [H_3O^+][OH^-]$$)

### Q3.87

Define what ionization energy is and determine which of the following would decrease the ionization of acidic acid, $$CH_3COOH_{(aq)}$$: $$NaCl$$, $$NaHCOO$$, $$H_2SO_4$$, and $$NaHCO_3$$.

### Q3.88

0.45M Hypochlorous acid ($$HOCl$$) with a $$K_a=2.9\times 10^{-8}$$ is mixed with 0.001M $$HCl$$.

a) Find $$[H_3O^+]$$

b) Find $$[OH^-]$$

c) Find $$[OCl^-]$$

d) Find $$[Cl^-]$$

### Q3.89

0.3M $$C_5O_5H_5N$$ with a $$K_b= 1.5\times 10^{-9}$$ is mixed with 0.015M $$C_5H_5NHI$$

a) Find $$[OH^-]$$

b) Find $$[C_5H_5NH^+]$$

c) Find $$[I^-]$$

d) Find $$[H_3O^+]$$

### Q3.90

A solution of volume 0.25 L contains 0.2 moles of $$HCN$$ and .004 moles $$CN^-$$.

a) What is the $$pOH$$?

b) If 0.25M $$NaOH$$ is added, what is the $$pH$$?

c) If .076 L of 3M $$HNO_3$$ is added, what is the $$pH$$?

### Q3.91

A manual says that in order to obtain a buffer with a pH of 7.00 you have to mix 32L of 0.644M $$HOCl$$ with .7L of 0.406M $$OCl^-$$.

a) Show by calculation that the $$pH$$ of this solution is 7.00

b) Would the $$pH$$ remain at 7.00 if 1 liter of $$H_2O$$ was added to the solution?

c) What will the $$pH$$ be of the original 1.02L solution if 0.15L of 0.8M $$NaOH$$ is added?

d) What is the maximum volume of 0.8M $$NaOH$$ that can be added to the 1.02L buffer solution so that the $$pH$$ does not change above 7.7?

### Q3.92

Bromocresol green ($$K=2.1\times 10^{-5}$$), as this indicator change from yellow to blue, the $$pH$$ ranges from 4-5. Without making detailed calculations, state what color the indicator will assume in each of the following solutions.

a) 0.3M $$NaOH$$

b) 0.02M $$HClO$$

c) 0.3M $$NH_3$$

d) 0.25M $$HI$$

e) 0.4M $$Na_2CO_2$$

### Q3.94

Explain why the volume of 0.25M $$HCl$$ required to reach the equivalence point in a titration of 50ml of BOH is the same whether it is a strong or a weak base, yet the $$pOH$$ at the equivalence point is not the same.

### Q3.95

Is the solution that is 0.3M $$Na_3PO_4$$ likely to be acidic, basic, or pH neutral? Why?

### Q3.96

For the solution that is 0.330 M $$CH_3CH_2COOH$$ that has $$K_a$$ value to be $$1.3 \times 10^{-5}$$ and 0.01M $$HI$$, calculate

1. (a) $$[H_3O^+]$$
2. (b) $$[OH^-]$$
3. (c) $$[CH_3CH_2COO^-]$$
4. (d) $$[I^-]$$

### Q3.97

For a solution that is 0.654 M $$NH_3$$ and 0.202 M $$NH_4Cl$$, calculate $$[H_3O^+]$$, $$[OH^-]$$, $$[NH_4^+]$$ with $$K_b=1.8\times 10^{-5}$$.

### Q3.98

Calculate the change in $$pH$$ that results from adding (a) 0.300 mol $$NaNO_2$$ to 1.00 L of 0.200 M $$HNO_{2(aq)}$$; (b) 0.300 mol $$NaNO_3$$ to 1.00 L of 0.200M $$HNO_{3(aq)}$$.

### Q3.99

A solution of volume 1L contains 0.500 mol $$HCHO_2$$ and 0.250 mol $$NaHCOO$$.

a) What is the $$pH$$ of this solution?

b) If 0.1 mol $$Ba(OH)_2$$ is added to to the solution, what will be the $$pH$$?

c) If 10.0 mL of 12 M $$HCl$$ is added to the original solution what will be the $$pH$$?

### Q3.100

To obtain a $$pH = 8.00$$ you need to mix 35.00 mL of .220 M $$NH_3$$with 65.00 mL of .250 M $$NH_4Cl$$.

a) Show the calculation that the $$pH$$ of this solution is 8.00.

b) Would you expect the $$pH$$ of this solution to remain at $$pH= 8.00$$ if the 100 mL of the buffer were diluted to 1.00 L? 1000L? Explain.

c) What will be the $$pH$$ of the solution if 0.20 mL of 1.00 M $$HCl$$ is added?

### Q3.101

What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the $$pH$$ value shown?

a) $$Ba(OH)_2$$ for $$pH=10.00$$

b) $$CH_3COOH$$ in 0.300 M $$NaCH_3COO$$ for $$pH=4.50$$?

### Q3.102

You are asked to bring the $$pH$$ of 1.00 L of 0.500 M $$NH_4Cl_{(aq)}$$ to 7.00. How many drops (1 drop = 0.05 mL) of which of the following solutions would you use: 10.0 M $$HCl$$ or 10.0 M $$NH_3$$.

### Q3.103

A handbook lists various procedures for preparing buffer solutions. To obtain a $$pH$$ of 9, the handbook says to mix 36.0 mL of 0.200 M $$NH_3$$ with 64 mL of 0.200 M $$NH_4Cl$$.

a.) Prove that $$pH =9$$ is correct

b.) Would the $$pH$$ remain at 9 if the 100 mL buffer solution was 200 L? Why or why not?

### Q3.104

What are the main principal features of a titration curve for the titration of a strong acid with a strong base?

### Q3.105

The $$pH$$ of a solution of 25 g of malonic acid in 0.300 L is 2.68, with a $$K_{a1}$$ value of $$4.5\times 10^{-12}$$. What is $$[HM^-]$$ in the malonic acid solution?

### Q3.106

Show the derivation of the Henderson-Hasselbalch equation.

## Solutions

### S1

$$Moles\ of\ HCl\ added = Moles\ of\ H_{3}O^{+} = 0.01L(0.005M)=5\times 10^{-4}\ mol$$

$$Moles\ of\ HF\ added = (0.05L)(0.325M)=0.01625\ mol$$

$$Total\ Volume= 50mL + 10mL = 60mL = 0.06L$$

$$M_{H_{3}O^{+}\ initial}=\frac{5\times 10^{-4}\ mol}{0.06L}=0.00833M$$

$$M_{HF}=\frac{0.01625\ mol}{0.06L}= 0.271M$$

$$HF_{(aq}) + H_2O_{(l)} \rightleftharpoons F^-_{(aq)} +H_3O^{+}_{(aq)}$$

 ICE Table $$HF_{(aq)}$$ $$H_2O_{(l)}$$ $$F^-_{(aq)}$$ $$H_3O^+_{(aq)}$$ Initial 0.271 M - 0 0.0083 M Change -x - +x +x Equilibrium (0.271 –x) M - x (0.00833 +x) M

$$K_a= \dfrac{[H_3 O^+][F^-]}{[HF]} = 6.6 \times 10^{-4}= \dfrac{x(0.00833+x)}{0.271-x}$$

Interpret the equation above we have:

$$6.6\times 10^{-4}(0.271-x) = 0.00833x+ x^{2}$$

$$x^{2}+0.00899x-1.789\times10^{-4}=0$$

$$x=0.009615M$$

a) $$[H_{3}O^{+}]=0.00833+x=0.00833+0.009615=0.0179M$$

b) At room temperature $$pH+pOH = 14$$,

$$pH=-log[H_{3}O^{+}] = -log(0.0179)=1.75$$

$$pOH=-log[OH^{-}]=14-1.75=12.25$$

Therefore, $$[OH^{-}]=10^{-12.25}=5.623\times 10^{-13}M$$

c)$$[HF] = 0.271-x=0.271-0.096165=0.2614M$$

d)$$[Cl^{-}]=[HCl]=0.00833M$$

### S2

$$NH_3 +H_2O \rightleftharpoons H_3O^+ + N_3^-$$

 ICE Table $$NH_3$$ $$H_2O$$ $$H_3O^+$$ $$N_3^-$$ Initial 0.138 M - 0 0.101 M Change -x - +x +x Equilibrium (0.138-x) - x 0.101 +x

$$K_a= \frac{[H_3 O^+ ][[N_3^-]}{[NH_3]}=1.9\times 10^{-5}= \frac{x(0.101+x)}{(0.138-x)}\approx \frac{0.101x}{0.138}$$

$$x\approx 2.6 \times 10^{-7}$$

The assumption that x<< 0.101 is correct. $$[H_3O^+] = 2.6 \times 10^{-7}\ M$$

b)

$$[N_3^- ]=(0.101+x)=(0.101+2.6 \times 10^{-7)})\approx 0.101\ M$$

c)

$$[K^+]=[[N_3^-]\approx 0.101\ M$$

d)

$$[OH^-]=\frac{K_w}{[H_3O^+]}=\frac{(1.0 \times 10^{-14})}{(2.6 \times 10^{-7})}=3.8 \times 10^{-8}\ M$$

### S3

a) First, determine the $$pH$$ of 0.150 M $$HNO_2$$.

$$HNO_{2(aq)} +H_2O_{(l)} \rightleftharpoons NO_{2(aq)}^- + H_3O^+_{(aq)}$$

 ICE Table $$HNO_{2 (aq)}$$ $$H_2O_{(l)}$$ $$NO_{2(aq)}^-$$ $$H_3O^+_{(aq)}$$ Initial 0.150 - 0 0 Change -x - +x +x Equilibrium 0.150-x - x x

$$K_a=\frac{[NO_2^-][H_3O^+]}{[HNO_2]}=7.2\times 10^{-4}=\frac{x^2}{0.150-x}$$

Through a successful approximation:

$$x \approx 0.01004\; M=[H_3O^+]$$

Thus $$pH \approx -\log(0.01004) = 2.0$$

When 0.150 mol $$LiNO_2$$ is added to 1.00 L of a 0.150 M $$HNO_2$$, a solution with $$[NO_2^-] = 0.150\ M = [HNO_2]$$ is formed. To find the change in the pH we use the Henderson-Hasselbalch approximation at equilibrium

$$pH=pK_a=-\log(7.2 \times 10^{-4} ) =3.14$$

Thus the addition has caused $$pH$$ to increase by 1.14

b) $$LiNO_3$$ contributes nitrate ions to the solution. Since there is no molecular $$HNO_{3(aq)}$$ in equilibrium with the hydrogen and nitrate ions, there is no equilibrium to be shifted by the addition of the nitrate ions. Thus, the $$[H_3O^+]$$ and the $$pH$$ are not affected by the addition of $$LiNO_3$$ to the solution. The $$pH$$ changes are not the same because unlike the first solution, the second solution does not have an equilibrium to shift, just a total ionic strength.

### S4

a) The $$pH$$ of the solution can be determined by the Henderson-Hasselbalch equation.

$$pH=pK_a+ log\frac{[ClO_2^-]}{[HClO_2]}=-log(1.1\times 10^{-2}) +log\frac{(\frac{7.25\ mmol}{50.0\ mL})}{(\frac{12.5\ mmol}{50.0\ mL})}$$

$$pH=1.96-0.24=1.72$$

b)

$$Ca(OH)_2 \rightarrow Ca^{2+}+2OH^{-}$$

$$Amount\ of\ added\ OH^- = 1.0\ mmol\ Ca(OH)_2 \times \frac{2OH^-}{Ca(OH)_2}=2.0\ mmol$$

The additional $$OH^-$$ reacts with the chlorous acid and produces chlorite ion.

$$HClO_2 +OH^- \rightleftharpoons ClO_2^-+H_2O$$

 ICE Table $$HClO_2$$ $$OH^-$$ $$ClO_2^-$$ $$H_2O$$ Initial 12.5 mmol 2 mmol 7.25 mmol - Change - 2 mmol - 2 mmol + 2 mmol - Equilibrium 10.5 mmol 0 9.25 mmol -

$$pH=pK_a + \log (\frac{[ClO_2^-]}{[HClO_2]})$$

$$=-log(1.1\times 10^{-2})+\log\frac{(\frac{9.25\ mmol}{50.0\ mL})}{(\frac{10.5\ mmol}{50.0\ mL})}$$

$$=1.96-0.055=1.90$$

c)

$$Amount\ of\ added\ H_3O^+= 2.00\ mL ×(6\ M)=12\ mmol\ H_3O^+$$

The additional $$H_3 O^+$$ reacts with the chlorite ion to produce chlorous acid.

$$ClO_2^-+H_3O^+ \rightleftharpoons HClO_2 + H_2O$$

 ICE Table $$ClO_2^-$$ $$H_3O^+$$ $$HClO_2$$ $$H_2O$$ Initial 7.25 mmol 12 mmol 12.5 mmol - Change - 7.25 mmol - 7.25 mmol + 7.25 mmol - Equilibrium 0 4.25 mmol 19.75 mmol -

The buffer’s capacity has been exceeded. The excess strong acid present determines the pH of the solution.

$$[H_3O^+] = \frac{4.75 mmol}{50.0\ mL + 2.0\ mL} = 0.091\ M$$

$$pH=-log\ [H_3O^+]=-log(0.091)= 1.04$$

### S5

a) Use the Henderson-Hasselbalch equation to determine the pH of the solution. The volume of the total solution is

$$18.0\ mL + 72.0\ mL = 100.0\ mL$$

$$pKa=14-pKb=14-8.82=5.18$$

$$[C_5H_5N]=\frac{18.0\ mL\times 0.300M\ C_5H_5N}{100.0\ mL}=\frac{5.4\ mmol}{100.0\ mL}=0.054\ M$$

$$[C_5H_5NH^+]=\frac{72.0\ mL\times 0.300M}{100.0\ mL}=0.216$$

$$pH=pKa+log\frac{[C_5 H_5 N]}{[C_5 H_5 NH^+]}=5.18+log\frac{0.054\ M}{0.216\ M}=4.57\approx 4.60$$

b) The solution has $$[OH^- ]=10^{-9.43}=3.7\times 10^{-10}\ M$$

The Henderson-Hasselbalch equation assumes that:

$$[C_5 H_5 N]\ll1.5 \times 10^{-9}\ll [C_5 H_5 NH^+]$$

If the solution is diluted to 500.00 mL, $$[C_5 H_5 N]=0.0108\ M$$, and $$[C_5 H_5 NH^+]=4.3\times 10^{-3}\ M$$. Only one of the $$[C_5 H_5 N]$$ is consistent with this assumption. So the pH would not remain the same.

If the solution was diluted to 1.00ml, $$[C_5 H_5 N]=0.0054\ M$$, and $$[C_5 H_5 NH^+]=2.2\times 10^{-3}\ M$$. Again, only the $$[C_5 H_5 N]$$ is consistent with the assumption, so in this case as well, the $$pH$$ would not remain the same.

c) The 0.40 mL of added 1.00 $$HBr$$ does not significantly change the volume of the solution. However, it does add $$0.40\ mL\times 1.00\ M\ HBr=0.40\ mmol\ H_3 O^+$$. This added $$H_3O^+$$ reacts with $$C_5H_5N$$, decreasing its amount from $$5.4\ mmol\ C_5 H_5 N$$ to $$5.0\ mmol\ C_5 H_5 N$$, and increasing the amount of $$C_5 H_5 NH^+$$ from $$21.6\ mmol$$ to $$22\ mmol$$.

$$pH=5.18+log\frac{(\frac{5.00\ mmol\ C_5 H_5 N}{100.40\ mL})}{(\frac{22.00\ mmol C_5 H_5 NH^+}{100.40\ mL})}=4.54$$

d) From part c) we see that the total volume of the solution does not affect the $$pOH$$ of the solution as long as the Henderson-Hasselbalch equation is observed. Let $$x$$ represent the number of mmols of $$H_3O^+$$ added, through $$1.00\ M\ HBr$$. This increases $$C_5 H_5 NH^+$$ and decrease $$C_5 H_5 N$$.

$$pH=4.4=5.18+log\frac{5.40-x}{21.6+x}$$

$$log\frac{5.40-x}{21.6+x}=4.4-5.18=-0.78$$

By inverting we get:

$$\frac{21.6+x}{5.40-x}=10^{0.78}=6.03$$

$$21.6+x=6.03(5.4-x)=32.6-6.03x$$

$$x=\frac{32.6-21.6}{1.00+6.03}=1.6\ mmol\ H_3O^+$$

$$Vol\ 1.00\ M\ HBr=1.6\ mmol\ H_3O^+ \times \frac{1\ mmol\ HBr}{1\ mmol\ H_3O^+}\times \frac{1\ mL\ soln}{1.00\ mmol\ HBr}=1.6\ mL\ 1.00\ M\ HBr$$

### S6

Solve for the $$pK_{HIn}$$ using the usual formula; $$pK_{HIn} = log(K_{HIn})$$

The $$pH$$ that gives color change is within 1 pH of this value. For example, Bromphenol Blue is an acidic indicator because, since its $$pK$$ value is 3.9, it has a color change range of between 2.9 and 4.9, all acidic $$pH$$ values. Furthermore, Bromthymol Blue changes in neutral solutions, and Thymolphthalein changes in basic solutions.

What is the approximate $$pH$$ of a solution that turns blue in Bromthymol Blue

As we learned from the first problem, the Brothymol Blue is changes in neutral $$pH$$ range. So, we can safely say that the $$pH$$ is somewhere between 6.1 and 8.1

### S7

1. $$Mg(OH)_2$$ : This solution is basic, so the acidic indicator will show its basic color, which is blue.
2. $$HCl$$: This solution is acidic, it will show its acidic color, yellow
3. $$HF$$: This is a weak, but will still show the yellow acidic color

### S8

$$[Na(OH)_2]\ for\ pH=10.92$$

$$pOH=14-10.92=3.08$$

$$[OH^-]=10^{-3.08}=8.32\times 10^{-4}$$

$$[Na(OH)_2]=\frac{8.32\times 10^{-4}\ mol\ OH^-}{1\ L}\times \frac{1\ mol\ [Na(OH)_2]}{2\ mol\ OH^-}=4.16\times 10^{-4}$$

### S9

A buffer solution will react with both acids and bases to prevent significant changes in $$pH$$. The chlorine is a spectator ion in the ammonium chloride and is therefore irrelevant for the purposes of this question.

$$NH_3+H_3O^+\rightleftharpoons NH_4^+ + H_2O$$

$$NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O$$

This is a buffer solution because based on these reactions that are occur, should either an acid or base be added.

We can use the Henderson-Hasselbach apporximation:

$$pK_a=-log(1.8 \times 10^{-5})= 4.745$$

$$pK_b= 14- pK_a= 9.255$$

$$pH= 9.255 + log (\frac{0.3}{0.4})= 9.130$$

To find the new pH, find the number of moles of each compound using the new volume.

• Moles $$NH_3 = 0.3\ M \times 1\ L= 0.3\ mol$$
• Moles of $$NH_4^+= 0.4\ M \times 1\ L= 0.4\ mol$$
• Moles of $$OH^-= 0.2\ M \times 1\ L= 0.2\ mol$$

New number of moles using ice table and identifying limiting reactant:

• $$0.2\ mol NH_4^+$$
• $$0\ OH^-$$
• $$0.1\ mol\ NH_3$$

$$pH= 9.255 + \log ( \frac{0.1}{0.05})= 9.556$$

### S10

From experiments 1 and 2:

$$[HF]+[F^-]= 0.35M$$

$$5\times (.005\ M) + [F^-] = 0.35M$$

$$[F^-]=0.325M$$

$$[HF]=0.025M$$

$$[H_3O^+]=10-(pK_w-pOH)$$

$$pOH= -log(0.5-0.325)= 0.757$$

$$[H_3O^+]=10-(14.94-.757)= 6.56\times 10^{-15}$$

$$K_a= \frac{[H_3O^+][F^-]}{[HF]}= 8.52989\times 10^{-14}$$

$$pK_a= 13.07$$

### S11

The answer is D. Since $$Na$$ is a spectator ion, $$CN^-$$ is left to react with the $$H_3O^+$$ created from the initial $$H_2S$$ reaction. This causes the reaction to shift to the right, creating more HS- ions.

### S12

 Equation CH3CH2COOH + H2O(l) ↔ CH3CH2COO- + H3O+ Initial 0.330 M -- 0 0.01M Change -x -- +x +x Equilibrium (0.330 M – x) M -- x (0.01 + x) M

$$K_a = \frac{x \times (0.01 + x)}{(0.330\ M – x)} \approx \frac{0.01x}{0.330}$$

$$[CH_3CH_3COO^-] = 4.3 \times 10^{-4}\ M$$

$$[H_3O^+] = [HI^-] = 0.0104\ M$$

$$[OH^-] = \frac{K_w}{[H_3O^+]}$$

$$[OH^-] = 9.6 \times 10^{-13}\ M$$

### S13

$$K_b = 1.8 \times 10^{-5}$$

 Equation NH3 + H2O(l) ↔ NH4+ + OH- Initial 0.654 M --- 0.202 M 0 Change -x --- +x +x Equilibrium (0.654 M – x) M --- (0.202+ x) M x

$$K_b = x\times \frac{(0.202 + x)}{(0.654\ M – x)}\approx \frac{0.202x}{0.330}$$[OH^-] = 2.9\times 10^{-5}\ M[NH_4^+] = [Cl^-] = 0.202\ M[OH^-] = \frac{K_w}{[H_3O^+]}[H_3O^+] = 3.4 \times 10^{-10}\ M$$### S14 a)  Equation HNO2 + H2O(l) ↔ HNO2- + H3O+ Initial 0.200 M -- 0 0 Change -x -- +x +x Equilibrium (0.200 M – x) M -- x x$$K_a = 7.2\times 10^{-4}K_a = \frac{x^2}{(0.200\ M – x)}\approx \frac{x^2}{0.200}x = 0.012\ MpH = -log[H_3O^+]pH = 1.9pH = pK_a – (-log[M])\delta pH = 1.22$$b) Since there is no net change in equilibrium just a shift in ionic strength thus the $$pH$$ of the second solution will be different. ### S15 a)$$pH = pK_a +log \frac{[base]}{[acid]}pH = 3.44$$b)$$OH^- = 0.1\ mol Ba(OH)_2 \times \frac{2\ mol\ OH^-}{ 1\ mol\ Ba(OH)_2} = 0.2\ mol\ OH^-$$ Equation HCHO2 + OH- ↔ CHO2- + H2O(l) Initial 0.5 mols 0.2 mols 0.250 mols 0 Change -0.2 mols -0.2 mols 0.2 mols 0 Equilibrium 0.3 mols 0 0.45 mols 0$$pH = pK_a +log \frac{[base]}{[acid]}pH = 3.92$$c)$$H_3O^+ = 0.01L \times 12M\ HCl = 0.12\ mol\ H_3O^+$$ Equation CHO2- + H3O+ ↔ HCHO2 + H2O(l) Initial 0.25 mol 0.50 mol -- Change -0.25 mol -0.25 mol +0.25 mol -- Equilibrium 0 0.12 mol 0.75 mol --$$H_3O^+ = \frac{0.12\ mol}{(1L + 0.01 L)} = 0.119\ MpH = 0.92$$### S16 a)$$pH = pK_a +log\frac{[base]}{[acid]}[NH_3] = 0.035\ L \times \frac{0.220\ M\ NH_3}{0.1\ L} = 0.077\ M[NH_4^+] = 0.065\times \frac{0.250\ M}{0.1\ L} = 0.163 MpH = 9.00$$b) The $$pH$$ would change since adding more solution will dilute the concentration and thus reduce the $$pH$$ to something near $$pH = 7$$. c) $$pH$$ should still be 9 since the amount of acid is so insignificant ### S17 a)  Indicator Changes color in Bromphenol blue Acidic Bromcresol green Acidic Bromthymol blue Neutral 2,4 – Dinitrophenol Acidic Chlorphenol red Acidic Thymolphthalein Basic b) Bromcresol green would be green at $$pH = 4.7$$ Chlorphenol red would be orange at $$pH = 6.0$$ ### S18 1. a) red 2. b) yellow 3. c) yellow 4. d) yellow 5. e) red 6. f) yellow ### S19$$Amount\ H_3O^+ = 0.075\ L \times 0.333\ M\times \frac{2\ mol H_3O^+}{1\ mol\ H_2SO_4} = 0.05\ molAmount\ OH^- = 0.025\ L\times 0.05\ M \times \frac{1\ mol}{1\ mol} = 0.0013 mol[H_3O^+] = \frac{mole\ H_3O^+}{\ L\ solution} = 0.487\ molepH = -log[H_3O^+]pH = 0.31$$### S20 Because the molarities of both solutions is the same all of the base will react with the acid. Depending on the strength of the acid at the equivalence point the amount of moles of acid equals the amount of moles of base therefore the volume doesn’t change but the $$pH$$ will depending on the type of acid or base. ### S21$$K_a = \frac{x^2}{0.05}x = [H_3O^+]K_b = \frac{K_w}{K_a} =\frac{x^2}{0.05}$$a)$$[H_3O^+] = 0.019\ MpH = 1.7x = [OH^-] = \sqrt{0.05\times \frac{K_w}{K_a}}= 2.7\times 10^{-7}pOH = 6.57pH = 7.42$$b)$$pH = 8.11$$c)$$pH = 10.2$$### S22 The solution should be basic due to the hydrolysis of the sulfide anion of a weak acid. ### S23 a)$$H_3PO_{4(aq)}+CO_3^{2-} \rightleftharpoons H_2PO_{4(aq)}^-+HCO_{3(aq)}^- H_2PO_{4(aq)}^-+ CO_3^{2-} \rightleftharpoons HPO_{4(aq)}^{2-}+ HCO_{3(aq)}^-HPO_{4(aq)}^{2-}+OH^-_{(aq)} \rightleftharpoons PO_{4(aq)}^{3-}+H_2O_{(l)}$$b) The inexpensive base is used in the first 2 ionizations and produces the bulk of the products. In the last ionization $$NaOH$$ must be used because the solution is not strong enough to pull the last hydrogen ion. ### S24$$Malonic\ acid = 104.06 \frac{g}{mol}Moles\ malonic\ acid = 19.5g\times \frac{1\ mol}{104.06\ g} = 0.187\ molesConcentration = \frac{0.187\ moles}{0.250\ L} = 0.784 M$$ Equation H2A + H2O ↔ HA- + H3O+ Initial 0.784 M - 0 0 Change -x - +x +x Equilibrium (0.784 – x) M - x x$$K_a = \frac{x^2}{(0.784 – x)}pH = 1.47[H_3O^+] = 10^{-1.47} = 0.034\ MK_{a1} = 0.0016K_{a2} = \frac{x^2}{0.300 – x}pH = 4.26[H_3O^+] = 10^{-4.26} = 5.5 \times 10^{-5}\ MK_{a2}= 1\times 10^{-8}$$### S25 1. a) No 2. b) No 3. c) Yes 4. d) Yes 5. e) Yes 6. e) No ### S26 a)$$\frac{[conjugate base]}{[weak acid]} = \frac{(equil. Amount conj. Base)}{(equil. Amount weak acid)} = \frac{(F \times initial Amt. weak acid)}{(1–F)\times initial Amount weak acid} = \frac{F}{1 – F}pH = pK_a + log\frac{F}{(1 – F)}$$b)$$pH = 10.00 + log \frac{0.27}{(1 – 0.23)} = 9.56$$### S27 1. $$for\ \ H_2PO_4^-: pH = \frac{1}{2}(pK_{a1} + pK_{a2})$$ 1. $$for\ \ HPO_4^{2-}: pH = \frac{1}{2}(pK_{a2} + pK_{a3})$$ ### S28 a)$$CHO_{2(aq)}^-+H_{3(aq)}O^+ \rightleftharpoons HCHO_{2(aq)}+H_2O_{(l)}HC_2H_3O_{2(aq)} +OH^-_{(aq)} \rightleftharpoons H_2O_{(l)}+ C_2H_3O_{2(aq)}^-$$Since neither species effects the $$pH$$ significantly we call this solution a buffer solution. b)$$HCHO_{2(aq)} + H_2O_{(l)} \rightleftharpoons CHO_{2(aq)}^- + H_3O^+_{(aq)}HC_2H_3O_{2(aq)}+H_2O_{(l)} \rightleftharpoons C_2H_3O_{2(aq)}^- + H_3O^+_{(aq)}[Na^+] = 0.250[OH^-] = 0[H_3O^+] = x[C_2H_3O_2^-] = y[CHO_2^-] = z[HC_2H_3O_2] = 0.200 – y[HCHO_2] = 0.250 – zK_A= 1.8\times 10^{-5} = \frac{(xy)}{0.150 – y}K_F = 1.8\times 10^{-4}= \frac{xz}{0.250 – z}y = \frac{0.150\times K_A}{K_A + x}z = \frac{0.250\times K_F}{K_F + x}[H_3O^+] = 4.0\times 10^{-5}\ M$$### S29$$H_2O_{2(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+HO_{2(aq)}^-[H_2O_2] + [HO_2^-] = 0.259\ M[HO_2^-] = 0.235\ M[H_2O_2] = 0.0242\ M[H_3O^+] = 10^{-(pK_w-pOH)} = 7.7\times 10^{-14}K_a = \frac{(7.7\times 10^{-14})(0.235)}{(0.0242)} = 7.4\times 10^{-13}pK_a = 12.14[H_2O_2] + [HO_2^-] = 0.123\ M[HO_2^-] = 0.1096\ M[H_2O_2] = 0.0134\ M[H_3O^+] = 10^{-(pKw-pOH) }= 7.41\times 10^{-14}K_a = \frac{(7.41\times 10^{-14})(0.1096)}{(0.0134)} = 6.06\times 10^{-13}pK_a = 12.22Average\ pK_a = 12.17$$### S30 d, because it is a weak base and will shift the formic acid ionization equilibrium to the right. ### S31$$pH=-log [H_3O^+]3.2=-log[H_3O^+][H_3O^+]=6.31\times 10^{-4}$$### S32$$HC_2H_3O_{2(aq)} + H_2O_{(l)} \rightleftharpoons C_2H_3O_{2(aq)}^- + H_3O_{(aq)}^+$$ ICE Table $$HC_2H_3O_{2(aq)}$$ $$H_2O_{(l)}$$ $$C_2H_3O_{2(aq)}^-$$ $$H_3O^+_{(aq)}$$ Initial 0.35 M - 0.62 M 0 Change -x - +x +x Equilibrium 0.35-x - 0.62+x x$$K_a= 1.8 \times 10^{-5} = \frac{(0.62+x)(x)}{(0.35-x)}$$Assume: $$x<<0.62$$$$K_a=1.8 \times 10^{-5}=\frac{0.62x}{0.35}x= 1.02 \times 10^{-5}=[H_3O^+]$$Definition of pH$$pH=-log[H_3O^+]pH=-\log(1.02 \times 10^{-5}) = 4.99$$### S33$$pH=pKa + log\frac{[base]}{[acid]}$$Solve for molarities: $$HCHO_2$$ is 0.227 and $$NaCHO2$$ is 0.09 fill in were appropriate in formula and…$$pH = 4.14$$### S34$$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8\times 10^{-5}$$Since $$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$$,$$[NH_4^+] = [OH^-]Let\ x = [OH^-] = [NH_4^+]K_b = \frac{(x)(x)}{(7.00\times 10^{-2} – x)} = 1.8\times 10^{-5}$$Solving for x using the quadratic formula,$$x = 1.11\times 10^{-3}\ M\ OH^-$$Hence:$$ pOH = -log[OH^-] = 2.95pH = 14 - pOH = 11.05$$So the $$pH$$ of a $$7.00\times 10^{-2}$$ M solution of ammonia is $$11.05$$ ### S35 1. $$pH= 6.0$$ 2. $$pH= 3.0$$ 3. $$pH= 8.0$$ ### S36 formula: $$pH = - log [H_3O^+]$$$$HCl \rightarrow H^+ + Cl^-$$a) $$[H^+]=[H_3O^+]=0.000014\ M$$ $$pH= -log(0.000014) = 4.85$$ So the indicator will be blue. b) $$pH = - log (.00014) = 3.85$$ So the indicator will remain red. c) $$pH = - log (.0014) = 2.85$$ So the indicator will remain red. d) $$pH = - log (.0000014) = 5.85$$ So the indicator will turn blue ### S37 An indicator is a weak acid and can therefore alter the $$pH$$ of the mixture if added in excess amounts ### S38 A point at which both acid and base have been consumed and neither in excess ### S39 Whether strong or weak acid, the necessary amount of $$HA$$ is the same regardless of a strong or weak acid, but the $$pH$$ at the equivalence point is not the same ### S40$$4.04 = pK_a + log \frac{0.020}{0.205}pK_a = 5.05K_a = 8.90\times 10^{-6}$$### S41$$(150ml)(0.100M)=15\ mmol\ NaOH20-15= \frac{5}{50+200}=0.02M= [NaOH]pOH=-log (0.02) = 1.70pH=14-1.70= 12.3$$### S42 Since sodium ions do not affect the $$pH$$ of a solution, only the iodine ion needs to be considered. The iodide ion is an anion, meaning in an aqueous solution it will take an $$H^+$$ ion from water to form $$HI$$ and $$OH^-$$. Therefore, the solution will be basic. ### S43 $$pH = 7$$ ### S44$$pK_a + pK_b = 14pK_a +4.8=14pK_a = 9.2pH = pK_a + log\frac{[conjugate\ base]}{[acid]}pH = 9.2 + log(1)pHa = 9.2$$### S45 A solution can be prepared with equal conjugate base and weak acid but would not form the given pH because the buffer ratio would be 1.00 making the $$pH = pK_a = 9.26$$. ### S46$$HC_2H_3O_{2(aq)}+H_2O_{(l)}\rightleftharpoons C_2H_3O_{2(aq)}^- + H_3O^+_{(aq)}$$ ICE Table $$HC_2H_3O_{2(aq)}$$ $$H_2O_{(l)}$$ $$C_2H_3O_{2(aq)}^-$$ $$H_3O^+_{(aq)}$$ Initial 0.50M - 0.50M 0 Change -x - +x +x Equilibrium 0.50-x - 0.50+x x$$1.8 \times 10^{-5}= \frac{(0.50+x)x}{0.50-x}x=1.8 \times 10^{-5}pH=-\log(1.8 \times 10^{-5})= 4.74$$### S47 $$NaOH$$ is a base, so an acid must be used to neutralize the solution, therefore $$HBr$$ is the correct substance to use. $$NaOH$$ and $$HBr$$ are both strong acids/bases, so the stoichiometric ratio between them is 1:1.$$Moles\ NaOH = (0.325 L)(0.452 M) = 0.147\ moles\ NaOHVolume\ HBr = \frac{0.147\ mol}{5.00\ M} = 0.0280\ L = 28.0\ mL$$### S48$$HCN_{(aq)} + H_2O_{(l)}\rightleftharpoons H_3O^+_{(aq)}+ CN^-_{(aq)}6.2\times 10^{-10} = \frac{x^2}{(0.025 – x)}$$assume x<<<0.025, then solve for x$$x = 3.94\times 10^{-6}\ M = [H_3O^+]pH = - log (3.94\times 10^{-6})$$### S49 The effects of a solution that consists the same ions. $$PbCl_{2(s)} \rightleftharpoons Pb_{2(aq)}^+ + 2Cl_{(aq)}^-$$ ### S50 0.375 M ### S51 Using Ice Table ### S52 2< pH < 5 ### S53 Away from equilibrium ### S54 Acidic Solution ### S55 Yes ### S56 12 L ### S57 The molarity is the same therefore the amount does not matter. ### S58 Start from bottom and the direction of the curve is going up. The equivalence point is around a $$pH$$ of 7. ### S59 0.05 L ### S60 Acidic ### S61 The ratio of the concentration of one reaction with another reaction of the same substance. ### S62 Strong Base ### S63 Strong Base ### S64 7.5 < pH < 9.0 ### S65$$CH_3COOH_{(aq)}+H_2O_{(l)}\rightleftharpoons H_3O^+_{(aq)}+CH_3COO^-_{(aq)}$$ CH3COOH H2O H3O+ CH3COO- I 0.153 -- 0 0 C -x -- +x +x E 0.153-x -- x x$$K_a=\frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}1.8\times 10^{-5}=\frac{x^2}{(0.153-x)}1.8\times 10^{-5}\times (0.153-x)=x^20=-x^2 -1.8\times 10^{-5}x+2.754\times 10^{-6}x=[H_3O^+]=1.65\times 10^{-3}$$### S66  HNO2(aq) H2O(l) H3O+(aq) NO2-(aq) I 0.548 -- 0 0 C -x -- +x +x E 0.548-x -- x x$$K_a=\frac{[H_3O^+][Cl^-]}{[HCl]}= 7.2\times 10^{-4}=\frac{x^2}{0.548-x}x=[H_3O^+]=0.0198pH=-log[H_3O^+]pH=-log(0.0198) pH=1.70$$### S67  HClO2(aq) H2O(l) H3O+(aq) ClO2(aq) I 0.10 -- 0 0 C -x -- +x +x E 0.10-x -- x x$$K_a=\frac{[H_3O^+] [ClO_2^-]}{[HClO_2]}1.1\times 10^{-2}=\frac{x^2}{(0.1-x)}x=[H_3O^+] =0.028119pH=-log[H_3O^+]pH=1.55pH=pK_a=-log(1.1\times 10^{-2})=1.95\Delta pH=1.95-1.55=0.4$$### S68$$pH=pK_a+log\frac{[ClO_2^-]}{[HClO_2]}pH=-log (1.1\times 10^{-2})+log\frac{(\frac{6mmol}{55mL})}{(\frac{0.45mmol}{55mL})}pH=1.95+log(13.33)pH=3.08$$### S69$$[H_3PO_4]= \frac{(50\ mL\times 0.3\ M)}{75\ mL}= 0.2\ M[H_2PO_4^-]= \frac{(25mL\times 0.3M)}{(25mL+50mL)}=0.1\ MpH=pK_a+log\frac{[H_2PO_4^-]}{[H_3PO^4]}1.82=pK_a+log(\frac{0.1}{0.2})pK_a=2.12 K_a=10^{-2.12}K_a=7.58\times 10^{-3}$$### S70 $$K_{Hln}$$ of $$2\times 10^{-7}$$. A titration between a strong acid and strong base reach an endpoint around a neutral $$pH$$, therefore the $$K_{Hln}$$ should be similar to the $$K_a$$, $$1\times 10^{-7}$$. An indicator with a larger $$K_{Hln}$$, will be used in acidic solutions. An incidator with a smaller $$K_{Hln}$$ will be used in basic solutions. Remember, a large $$K_{Hln}$$ will correspond with a smaller $$pH$$, which means more acidic. ### S71 Strong acid-weak base. The endpoint is reached in a low $$pH$$, because the acid fully dissociated, but the base does not. ### S72$$KOH: \frac{(0.05moles)}{L}\times (0.02L)=0.001molesHI: \frac{(0.05mole)}{L}\times (0.01L)=0.0005moles0.001-0.0005=0.0005moles\ KOH\ in\ excessStrong\ acid,\ fully\ dissociates,\ so\ 0.0005\ moles\ of\ OH^-:[OH^-]=\frac{0.0005}{(.01L+0.02L)}=0.0166MpOH=-log(0.0166)=1.77pH=14-1.77=12.22pK_{Hln}=pH\ (at\ endpoint)= 12.22$$This does not give the accurate endpoint of titration, because that should occur around a $$pH$$ of 7, since it is strong acid-strong base titration. ### S73$$HCl: \frac{(0.007moles)}{L}\times (0.200L)=0.0014molesNaOH: \frac{(0.02moles)}{L}\times (0.300L)=0.006moles$$This is strong acid, strong base titration, so they both fully dissociate into ions.$$ 0.006-0.0014=0.0046\ moles\ of\ OH^-\ unreacted[OH^-]= \frac{0.0046}{(0.2L+0.3L)}=0.0092MpOH=-log(0.0092)=2.036$$### S74 The volumes will be the same, because they have the same molarity. However, the $$pH$$ will be different because $$HF$$ is a stronger acid than $$CH_3COOH$$, so its $$pH$$ will be around or lower 7, while $$CH_3COOH$$ is a weak acid, so the $$pH$$ will be greater than 7 (since the strong base, LiOH, has fully dissociated, but the $$CH_3COOH$$ has not.) ### S75 The greater the $$K_b$$, the closer the $$pH$$ will be to 7 at the equivalence point (inflection point of graph), and the steeper the slope will be. However, because it is a weak base, the $$pH$$ will be lower than 7 no matter what. The lower the $$K_b$$, the less steep the slope will be and the lower the pH at the equivalence point will be. ### S76$$CsOH: \frac{(0.02\ moles)}{L}\times (0.04L)=0.0008\ moles\ CsOH\frac{0.05mole}{L}\times (0.005L)=0.00025\ moles\ HBrmoles\ OH^- = 0.0008-0.00025=0.00055[OH^-]=\frac{0.00055}{(0.04L+0.005L)}=0.0122pOH=1.91pH=14-1.91=12.09\frac{(0.05mole)}{L}\times (0.01L)=0.0005\ moles\ HBrmoles\ OH^- = 0.0008-0.0005 = 0.0003[OH^-]=\frac{0.0003}{(0.01L+0.04L)}=0.006\ MpOH=2.22pH=14-2.22=11.79\frac{(0.05mole)}{L}\times (0.025L)=0.0013\ moles\ HBr$$Now more moles of acid than base so,$$moles\ H^+= 0.0013-0.0008 = 0.0005[H^+] = \frac{0.0005}{(.025+0.04)}=0.0076pH=-log(0.0076)=2.13$$### S77 It will be basic. The copper is just a spectator ion. The sulfide ion will form a bond with the $$H$$ from the water. However, because $$HS$$ is a weak acid, the concentration of $$OH^-$$ will be greater than $$H^+$$ concentration of the dissociation of $$HS$$. ### S78 Sulfuric acid can be reduced with sodium carbonate, and then further reduced with $$NaOH$$.$$H_2SO_4 + CO_3^{2-} \rightarrow HSO_4^- +HCO_3^-HSO_4^- + OH^- \rightarrow SO_4^{2-} + H_2O$$### S79$$CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-$$ CH3COOH H2O H3O+ CH3COO- I 0.1 -- 0 0 C -x -- +x +x E 0.1-x -- x x$$K_a= \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]} = 1.8\times 10^{-5}Assume\ x \ll 0.1\ \rightarrow 1.3\times 10^{-3} = [H_3O^+]pH= -log [H_3O^+] = -log[1.3\times 10^{e-3}]pH= 2.89$$### S80 $$for\ H_2PO_4^-\ :\ pH = \frac{1}{2}(pK_{a1} + pK_{a2})=\frac{1}{2 }( 2.15+ 7.20) = 4.68$$ $$for HPO_4^{2-}\ :\ pH =\frac{1}{2 }(pK_{a2} + pK_{a3}) = \frac{1}{2 }(7.20 + 12.38) = 9.79$$ ### S81 Simultaneously means ionization as an acid and ionization as a base. 1. yes 2. no 3. yes ### S82 $$K_{a1} = 7.1\times 10^{-3}$$, $$K_{a2} = 6.3\times 10^{-8}$$, $$K_{a3}= 4.2\times 10^{-13}$$$$pH= pK_a + log \frac{[base]}{[acid]}pK_a= -log(6.3\times 10^{-8})x= 0.3852\ moles\ NaOH\ neededmL = \frac{0.3852\ moles\ NaOH}{6.0M\ NaOH} \times 1000 = 64.2\ mL$$### S83 As a result with buffers;$$pH = -log [H_3O^+] =-log[K_a]pH = -log[6.3\times 10^{-5}]= 4.2buffer\ range = 4.2\pm 1= [3.2, 5.2]$$### S84 The common ion effect is the suppression of the ionization of a weak electrolyte caused by adding more of an ion that is a product of this ionization. Add $$H_3O^+: CH_3COOH +H_2O \rightleftharpoons H_3O^+ + CH_3COO^- K_a= 1.8\times 10^{-5}$$ Equilibrium shifts to form more $$CH_3COOH$$ Add $$OH^-: NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- K_b= 1.8\times 10^{-5}$$ Equilibrium shifts to form more $$NH_3$$ ### S85$$(0.3\ L)(0.56\ M) = 0.168\ mol\ CH_3COO^-(0.3\ L)(0.25\ M) = 0.075\ mol\ CH_3COOH$$ CH3COO- H3O+ H2O CH3COOH I 0.168 mol 0.006 mol -- 0.075 mol C -0.006 mol -0.006 mol -- +0.006 mol E 0.162 mol 0 -- 0.018 mol$$[CH_3COO^-] = \frac{0.162\ mol}{0.3\ L} = 0.54\ M[CH_3COOH] = \frac{0.081\ mol}{0.3\ L} = 0.27\ MpH= pK_a + log \frac{[base]}{[acid]}pH= 4.74 + log \frac{[0.54M]}{[0.27M]} = 5.04$$### S86$$HCl_{(aq)} + H_2O_{(l)}\rightleftharpoons H_3O^+_{(aq)} + Cl^-_{(aq)}H_2O_{(l)} + H_2O_{(l)}\rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}$$ H2O H2O H3O+ OH- I -- -- 10-8 0 C -- -- +x +x E -- -- 10-8+x x$$K_w = (1\times 10^{-8}+x)(x) = 1.0\times 10^{-14}$$Use the quadratic equation to solve for x;$$x= 1.1\times 10^{-7}[H_3O^+]= (1\times 10^{-8}+x)= (1\times 10^{-8}+ 1.1 \times 10^{-7}) = 1.2\times 10^{-7}pH = -log[H_3O^+] = -log[1.2\times 10^{-7}] = 6.92$$### S87 Ionization energy is the energy required to completely remove an electron from a gaseous atom or ion. $$NaCl$$, $$NaHCOO$$, and $$NaHCO_3$$ will decrease the ionization of acidic acid. ### S88 a)  HOCl H2O H3O+ OCl- I 0.45 - 0.001 0 C - x - + x + x E 0.45 - x - 0.001 + x x$$2.9\times 10^{-8} = \frac{(0.001+x)(x)}{(0.45-x)}x= 1.28\times 10^{-5}[H_3O^+] = 0.001+ x = 0.0010128M$$b)$$[H_3O^+][OH^-]=K_w[OH^-]= 9.9\times10^{-14}\ M$$c)$$[OCl^-]=x= 1.28\times 10^{-5}\ M$$d)$$[Cl^-]=[H_3O^+]= 0.0010128M$$### S89 a)  CH5N H2O OH- CH5NH+ I 0.3 - 0 0.015 C -x - +x +x E 0.3-x - x 0.015+x$$1.5\times 10^{-9}= \frac{(0.015+x)(x)}{(0.3-x)}x= 3.0\times 10^{-8}[OH^-]= x=3.0 \times 10^{-8}\ M$$b)$$[C_5H_5NH^+]=0.015M$$c)$$[I^-]=0.015M$$d)$$[H_3O^+][OH^-]= K_w[ H_3O^+](1.3\times 10^{-7})=1.0\times 10^{-14}[ H_3O^+]= 7.69\times 10^{-8}$$### S90 a)$$pH=pK_a+log \frac{[B]}{[A]}pH= 9.21+ log\frac{0.004}{0.025}pH= 7.414pOH=14-pH=6.586$$b)$$pH= pK_a+ log \frac{(0.004+0.025)}{(0.25-0.025)}pH= 8.32(3)(0.076)= 0.228\ moles\ HNO_3\frac{(0.228-0.004)}{(0.25+0.076)}=0.687\ MpH= -log(0.687)= 0.163$$### S91 a)$$(0.32L)(0.406M)= 0.13\ moles\ B(0.7L)(0.644M)= 0.451\ moles\ ApH= pK_a+ log\frac{[B]}{[A]}pH= 7.54+log\frac{0.13}{0.451}pH= 7$$b) Yes, even if you add 1L of $$H_2O$$, the mole ratio will not change c)$$(0.15L)(0.8M)= 0.12\ moles\ NaOHpH= 7.54+ log \frac{(0.13+0.12)}{(0.45-0.12)}pH=7.54$$d)$$7.7=7.54+log\frac{(0.13+x)}{(0.451-x)}x=0.2135\ moles\ of\ NaOH(0.8M)(V)=0.2135\ molV= 0.2669L\ of\ 0.8M\ NaOH$$### S92 a) blue b) yellow c) blue d) yellow e) blue f) blue ### S93$$(0.73M)(0.025L)= 0.01825\ moles\ HI2[(0.29M)(0.01L)]= 2(0.0029)=0.0058\ moles\ Ca(OH)_20.01825-0.0058= 0.01245\ mol[H_3O^+]=\frac{0.01245}{0.035}= 0.3557MpH=-log (0.3557)=0.448914-pH=pOH=13.55$$### S94 It doesn't matter because the strength of the base doesn't change the number of moles of that base. At the equivalence point, the amount of moles of acid equals the moles of base regardless of strength. The $$pOH$$ will not be the same however because when titrating a weak base at the equivalence point, there will be a concentration of its conjugate acid which acts acidic in solution. ### S95 It will act basic because $$PO_4^{3-}$$ is a base in solution because it is the conjugate base to $$HPO_4^{2-}$$ ### S96  Equation CH3CH2COOH + H2O(l) ↔ CH3CH2COO- + H3O+ Initial 0.330 M --- 0 0.01M Change -x --- +x +x Equilibrium (0.330 M – x) M --- x (0.01 + x) M$$K_a = \frac{x(0.01 + x)}{(0.330\ M – x)} \approx \frac{0.01x}{0.330}[CH_3CH_2COO^-] = 4.3\times 10^{-4}\ M[H_3O^+] = [I^-] = 0.0104\ M[OH^-] = \frac{K_w}{[H_3O^+]}[OH^-] = 9.6\times 10^{-13}\ M$$### S97  Equation NH3 + H2O(l) ↔ NH4+ + OH- Initial 0.654 M --- 0.202 M 0 Change -x --- +x +x Equilibrium (0.654 M – x) M --- (0.202+ x) M x$$K_b = \frac{x(0.202 + x)}{(0.654\ M – x)} \approx \frac{0.202x}{0.654}[OH^-] = 2.9\times 10^{-5}\ M[NH_4^+] = [Cl^-] = 0.202\ M[OH^-] = \frac{K_w}{[H_3O^+]}[H_3O^+] = 3.4\times 10^{-10}\ M$$### S98  Equation HNO2 + H2O(l) ↔ HNO2- + H3O+ Initial 0.200 M --- 0 0 Change -x --- +x +x Equilibrium (0.200 M – x) --- x x$$K_a = 7.2\times 10^{-4}K_a = \frac{x^2}{0.200M-x}\approx \frac{x^2}{0.200M}x = 0.012\ MpH = -log[H_3O^+]pH = 1.9pH = pK_a – (-log[M])\Delta pH = 1.22$$b) Since there is no net change in equilibrium just a shift in ionic strength thus the $$pH$$ of the second solution will be different. ### S99 a)$$pH = pK_a +log\frac{[base]}{[acid]}pH = 3.44$$b)$$OH^- = 0.1\ mol\ Ba(OH)_2 \times \frac{2\ mol\ OH^-}{1\ mol\ Ba(OH)_2} = 0.2\ mol\ OH^-$$ Equation HCHO2 + OH- ↔ CHO2- + H2O(l) Initial 0.5 mols 0.2 mols 0.250 mols --- Change -.2 mols -0.2 mols +0.2 mols --- Equilibrium 00.3 mols 0 0.45 mols ---$$pH = pK_a +log\frac{[base]}{[acid]}pH = 3.92$$c)$$H_3O^+ = 0.01L \times 12M\ HCl = 0.12\ mol\ H_3O^+$$ Equation CHO2- + H3O+ ↔ HCHO2 + H2O(l) Initial 0.25 mols 0.37 mols 0.50 mols --- Change -0.25 mols -0.25 mols +0.25 mols --- Equilibrium 0 0.12 mols 0.75 mols ---$$[H_3O^+] = \frac{0.12\ mol}{1L + 0.01\ L} = 0.119\ MpH = 0.92$$### S100 a)$$pH = pK_a +log\frac{[base]}{[acid]}[NH_3] = 0.035\ L \times \frac{0.220\ M\ NH_3}{0.1\ L} = 0.077\ M[NH_4^+] = 0.065 \times \frac{0.250\ M}{0.1\ L} = 0.163\ MpH = 9.00$$b) The $$pH$$ would change since adding more solution will dilute the concentration and thus reduce the $$pH$$ to something near $$pH = 7$$. c) $$pH$$ should still be 9 since the amount of acid is so insignificant ### S101 a)$$pOH = 14 – 10 = 4[OH^-] = \frac{10^{-4}\ moles}{1L}\times \frac{1\ mol\ Ba(OH)_2}{2\ mol\ OH^-}=0 .00005\ M$$b)$$pH = pK_a + log\frac{[base]}{[acid]}pK_a = 4.744.5 – 4.74 = log\frac{[base]}{[acid]}10^{-0.24} = \frac{[0.300M]}{[acid]} = 0.52\ M$$### S102$$K_b = \frac{(0.500 M)(1.0\times 10^{-7}\ M)}{x\ M} = 1.8\times 10^{-5}[NH_3] = 2.8\times 10^{-3}\ MV = 1L \times (2.8\times 10^{-3}\ M\ NH_3) \times \frac{1\ L}{10.00\ mol\ NH_3} \times \frac{1\ drop}{0.05\ mL} = 3\ drops$$### S103 a)$$0.2M\ NH_3 = 0.0072\ mol\ NH_30.2M\ NH_4Cl = 0.0128\ mol\ NH_4ClMoles\ of\ base\ in\ excess = 0.0128 - 0.0072 = 0.0056\ molTotal\ volume = 36\ mL + 64\ mL = 100\ mL\ (1L)[OH^-] = \frac{0.0056\ mol}{1\ L} = 0.0056\ MpOH = -log(0.0056) = 2.2512 – 2.25 → pH =9.75$$b.)$$[OH^-] = \frac{0.0056\ mol}{200\ L} = 2.8\times 10^{-5}$$No, the $$pH$$ would be different because you are changing the concentration of the base, and thereby change the $$pH$$ and $$pOH$$ values. chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Buffers ### S104 1.) The $$pH$$ starts low. 2.) The $$pH$$ increases slowly until right before the equivalence point. 3.) At equivalence point, the $$pH$$ spikes by approximately 6 units for only a couple drops of base. 4.) After equivalence point, the $$pH$$ rises slowly again. 5.) Any acid-base indicator with color change in a $$pH$$ range of 4-10 will work. chemwiki.ucdavis.edu/Analytical_Chemistry/Quantitative_Analysis/Titration/Titration_Of_A_Strong_Acid_With_A_Strong_Base ### S105$$25g\ malonic\ acid\ in\ 0.300\ L = 0.801\ M\ malonic\ acidH_2M + H_2O \rightleftharpoons HM^- + H_3O^+K_{a1} = \frac{[HM^-] [H_3O^+]}{[H_2M]}[H_3O^+] = -log(2.68) = .00214.5\times 10^{-12} = \frac{[HM^-](0.0021)}{0.801}[HM^-] = 1.72\times 10^{-9}$$chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Aqueous_Solutions/The_pH_Scale/Determining_and_Calculating_pH ### S106$$K_a = \frac{[H^+][A^-]}{[HA]}log(K_a) = log(\frac{[H^+][A^-]}{[HA]})logK_a = log[H^+] + log(\frac{[A^-]}{[HA]})-pK_a = -pH + log(\frac{[A^-]}{[HA]})pH = pK_a + log(\frac{[A^-]}{[HA]})

chemwiki.ucdavis.edu/Physical_Chemistry/Acids_and_Bases/Buffers/Henderson-Hasselbalch_Approximation

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