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15: Acids and Bases I

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.

Q15.1a

Label each of the following species as an acid or a base according to the Brønsted–Lowry theory.

  1. \(NH_3\)
  2. \(H_2O\)
  3. \(CH_3COOH\)
  4. \(NH_2^-\)

S15.1a

  1. base (proton acceptor)
  2. both an acid (proton donor) and a base (proton acceptor)
  3. base (proton acceptor)
  4. base (proton acceptor)

Q15.1b

According to the Bronsted-Lowry theory, label each of the following as an acid or a base.

  1. \(HNO_3\)
  2. \(CH_3COO^-\)
  3. \(NH_4^+\)
  4. \(H_3O^+\)

S15.1b 

  1. acid
  2. base
  3. acid
  4. acid

Q15.1c

Label each of the following as an acid or a base according to the Bronster-Lowry theory.

  1. \(HBr\)
  2. \(NaOH\)
  3. \(HCl\)
  4. \(HF\)
  5. \(HI\)

S15.1c

  1. Strong acid
  2. Strong base
  3. Strong acid
  4. Weak acid
  5. Strong acid

Q15.1d

 

Label each of the following as an acid or a base, according to the Bronsted-Lowry theory.
  1. \(H_2PO_4^-\) 
  2. \(CH_4\)
  3. \(H_2\) 
  4. \(HSO_4^-\)  
  5. \(CH_3OH\)

S15.1d

  1. Proton acceptor so it is a base.
  2. Proton donor so it is an acid.
  3. Proton donor so it is an acid.
  4. Proton acceptor so it is a base.
  5. Proton donor so it is an acid.

Q15.1e

According to the Brønsted–Lowry theory, label each of the following as an acid or a base.

  1. \(NaOH\)
  2. \(HBr\)
  3. \(LiOH\)
  4. \(HCl\)
  5. \(HF\)

S15.1e

  1. Strong Base
  2. Strong Acid
  3. Strong Base
  4. Strong Acid
  5. Weak Acid

Q15.2a

Find the empirical formula and the strength of conjugate acid in water for each of the following. If these is no conjugate acid, answer none.

  1. \(CO_3^{2-}\)
  2. \(HS^-\)
  3. \(NaOH\)
  4. \(NH_3\)

S15.2a

   a. \(CO_3^{-2} + H_2O  \rightarrow OH^- + HCO_3^- \)  

      Conjugate acid:  \(HCO_3^- \) Strength: Weak (relatively low Ka value)

   b. \(HS_- + H_2O \rightarrow OH^- +H_2S \)

       Conjugate acid:\(H_2S\) Strength: Weak (relatively low Ka value)

   c. \(NaOH + H_2O \rightarrow H_2^O+ OH^- + Na^+\)

       Conjugate acid: \(Na^+\) Strength: Weak (NaOH is a strong conjugate acid)

   d. \(NH_3 + H_2O \rightarrow OH^- + NH_4^+\) 

       Conjugate acid: \(NH_4^+\) Strength: weak (low Ka value)

Q15.2b

Write the formula of the conjugate base in the reaction of each acid with water.

  1. \(H_2O\)
  2. \(H_2CO_3\) 
  3. \(H_2SO_4\)
  4. \(HCO_3^-\) 
  5. \(HSO_4^-\)

 

S15.2b

  1. \(OH^-\)
  2. \(HCO_3^-\)
  3. \(HSO_4^-\)
  4. \(CO_3^{2-}\)
  5. \(SO_4^{2-}\)

 

Q15.2c

Write the conjugate base in the reaction of each acid with water.
  1. \(HNO_3\)
  2. \(H_2SO_4\)
  3. \(CH_4\)
  4. \(H_2O\)

S15.2c

  1. \(HNO_2^-\)
  2. \(HSO_4^-\)
  3. \(CH_3^-\)
  4. \(OH^-\)

Q15.2d

From these formulas, what is the conjugate base?

  1. \(HCl + H_2O \rightarrow H_3O^+ + Cl^- \)
  2. \(H_2PO_4^- + HF \rightarrow H_3PO_4 + F^- \)
  3. \(CH_3CO_2H + NaOH \rightarrow CH_3CO_2Na + H_2O \)

S15.2d

  1. \(CA=H_3O^+\)  \(CB = Cl^- \)
  2. \(CA = H_3PO_4 \) \(CB = F^-\) 
  3. \(CA = H_2O \) \(CB=CH_3CO_2Na\)

Q15.3a

For each of the following, identify the acids and bases involved in both the forward and reverse directions.

  1. \(HClO_2 + H_2O \rightleftharpoons ClO^- + H_3O^+ \)
  2. \(HCl + OH^- \rightleftharpoons Cl^- + H_2O \)
  3. \(HS^- + H_2O \rightleftharpoons H_2S + OH^- \)
  4. \(HCl + H_2PO_4^- \rightleftharpoons Cl^- + H_3PO_4 \)

S15.3a

  1. \(\underbrace{HClO_2}_ {acid} + \underbrace{ H_2O}_{base} \rightleftharpoons \underbrace{ ClO^-} _ {base} + \underbrace{ H_3O^+}_{acid} \)
  2. \(\underbrace{HCl} _ {acid} + \underbrace{ OH^-} _{base} \rightleftharpoons \underbrace{ Cl^-} _ {base} + \underbrace{ H_2O} _{acid} \)
  3. \(\underbrace{ HS^-}_{base} + \underbrace{ H_2O}_{acid} \rightleftharpoons \underbrace{ H_2S}_ {acid} + \underbrace{ OH^-}_{base}\)
  4. \(\underbrace{ HCl } _ {acid} + \underbrace{ H_2PO_4^-}_{base} \rightleftharpoons  \underbrace{Cl^-}_{base} + \underbrace{ H_3PO_4}_{acid}\)

Q15.25a

A 0.5 M solution of aqueous hydrofluoric acid is found to have a pH of 3.0. Calculate Ka for the hydrofluoric acid.

\[HF + H_2O \rightleftharpoons F^- + H_3O^+\]

S15.25a

Given the pH of 3.0 , the \([H_3O^+]\) can be obtained.
\[pH= 10^{-3} = 0.001= [H_3O^+]\]
 
Equation \[HF + H_2O \rightleftharpoons F^- + H_3O^+\]
ICE Table \(HF\) \(F^-\) \(H_3O^+\)
Initial \(0.5\) \(0\) \(0\)
Change \(-0.001\) \(+0.001\) \(+0.001\)
Equilibrium \(0.449\) \(0.001\) \(0.001\)



\[K_a=\dfrac{(0.001)^2}{0.499} =  2.0 \times 10^{-6}\]

Q15.25b

A 0.311 M solution of fluoroacetic acid is found to have a pH = 1.43. Calculate the Ka of fluoroacetic acid.

\[CH_2FCOOH_{(aq)} + H_2O \rightleftharpoons H_3O^+ + CH_2FCOO^-\]

S15.S25b

\[[H_3O^+] = 10^{-1.43} = 0.0372\; M\]

\[CH_2FCOOH(aq) + H_2O \rightleftharpoons H_3O^+ (aq) + CH_2FCOO^- \]
ICE Table \(CH_2FCOOH\) \(H_3O^+\) \(CH_2FCOO^-\)
Initial \(0.311M\) \(0\) \(0\)
Change \(-0.0372M\) \(+0.0372M\) \(+0.0372M\)
Equilibrium \(0.2738M\) \(0.0372M\) \(0.0372M\)

 

\[K_a = \dfrac{[H_3O^+][CH_2FCOO^-]}{[CH_2FCOOH]}\]

\[K_a = \dfrac{[0.0372M][0.0372M]}{[0.2738M]}\]

\[K_a = 0.00505\]

Q15.25c

Consider the following reaction: 

\[H_3BO_3(aq) + H_2O(l) \rightarrow H_2BO_{3}^-(aq) + H_3O^+(aq)\]

Initially, 12.1 mol of boric acid are placed in a 1.43 L flask. The Ka for boric acid is \(5.8 \times 10^{-10}\). Calculate the pH of the resulting solution.

S15.25c

\[[H_3BO_3] = \dfrac{12.1mol}{1.43L} = 8.46M\]

\[5.8 \times 10^{-10} = \dfrac{[H_2BO_3^-][H_3O^+]}{[H_3BO_3]} = \dfrac{x^2}{8.46-x}\]

\[ x= 7.0\times 10^{-5} M\]

\[ [H3O^+] = x = 7.0\times 10^{-5} M\]

\[  pH = -log (7.0\times 10^{-5}) = 4.15\]

Q15.25d

Fluoroacetic acid occurs in a gifblaar, one of the most poisonous of all plants.  A 0.227 M solution of the acid is found to have a pH=2.03. Calculate Ka of fluoroacetic acid when the equilibrium concentration of \(CH_2FCOO^-(aq)\) is 0.123M.

S15.25d

Given: \(CH_2FCOOH(aq)+H_2O  \rightleftharpoons  H_3O^+(aq)+CH_2FCOO^-(aq) \)

\(Ka=?\)

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

\[2.03=-\log([H_3O^+])\]

\[[H_3O^+]=0.0093M\]

\[CH_2FCOOH(aq)+H_2O  \rightleftharpoons  H_3O^+(aq)+CH_2FCOO^-(aq) \]

\[K_a = \dfrac{(0.123M)(0.0093)}{0.227M}\]

\[K_a = 0.0050 \]

Q15.27a

What mass of benzoic acid,\(C_6H_5COOH\), would you dissolve in 50 mL of water to produce a solution with a pH = 5.85?

Given \(K_a = 6.5 \times 10^{-5} \) 

S15.27a

\[C_6H_5COOH + H_2O \rightleftharpoons H_3O^+ + C_6H_5COO^-\]

\[K_a = 6.5 \times 10^{-5} \]

\[[H_3O^+] = 10^{-5.85} = 1.41 \times 10^{-6} M \]

 

\[C_6H_5COOH + H_2O \rightleftharpoons H_3O^+ + C_6H_5COO^-\]
ICE Table \(C_6H_5COOH\) \(H_3O^+\) \(C_6H_5COO^-\)
Initial \(S\) \(0\) \(0\)
Change \(- 1.41 \times 10^{-6} M\) \(+1.41 \times 10^{-6} M\) \(+1.41 \times 10^{-6} M\)
Equilibrium \(S - 1.41 \times 10^{-6} M\) \(1.41 \times 10^{-6} M\) \(1.41 \times 10^{-6} M\)

 

\[K_b = \dfrac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}\]

\[\dfrac{[1.41 \times 10^{-6}][1.41 \times 10^{-6} M]}{[S-1.41 \times 10^{-6} M]} = 6.3 \times 10^{-5}\]

\[S = 1.44 \times 10^{-6} M\]

\[50 mL(\dfrac{1L}{1000mL})(1.44 \times 10^{-6}M) = 7.2 \times 10^{-8} \text{moles } C_6H_5COOH\] 

\[Mass = 7.2 \times 10^{-8} moles (\dfrac{122.1 g C_6H_5COOH}{1 mol})\]

\[Mass = 8.79 \times 10^{-6} g\]

Q15.27b

What mass of Hydrazidic acid, \(HN_3\), would you use dissolve in 500 ml of water to produce a solution with pH= 2.00?

\[HCN_{(aq)} + H_2O \rightleftharpoons CN^- + H_3O^+\]

Given \(K_a= 1.9 \times 10^{-5}\)

S15.27b

\[[H_3O^+] = 10^{-2.00} = 0.01M \]
 
\(HN_{3(aq)} + H_2O_{(l)} \rightleftharpoons N_3^- +H_3O^+_{(aq)} \)
ICE \(HN_{3(aq)}\) \(N_3^-\) \(H_3O^+_{(aq)}\)
Initial \(x\) \(0\) \(0\)
Change \(-0.01\) \(+0.01\) \(+0.01\)
Equilibrium \(x-0.01\) \(0.01\) \(0.01\)

\[1.9 \times 10^{-5} = \dfrac{(0.01)^2}{x-0.01}\]

\[x= 0.5273\]

\[[HN_3]= 0.5273M\]

  • \(\text{Moles of H} N_3 = 0.5273M (0.5L)= 0.2637 moles\)
  • \(Mass = (0.2637mol)(\dfrac{43.03g}{mol}) = 11.35g\)

Q15.27c

What mass of \(HCl\) would you dissolve in 300.0 ml of water to produce a solution with a pH = 1.0?

S15.27c

\[[H^+] = 10^{-1} = 0.1M\]

Assuming HCl completely dissociate.

\[Mass = 0.1M(0.300L)(\dfrac{36.46g}{mol}) = 1.09g\]

Q15.27d

What would the pOH be if you had a 1.5 M sample of Chlorous acid, with a \(Ka = 1.1 \times 10^{-2}\)?

S15.27d

ICE Table \[HClO_2 + H_2O \rightleftharpoons H_3O^+ + ClO_2\]
Initial \(1.5M\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(1.5-x\) \(x\) \(x\)

 

\[K_a= \dfrac{x^2}{1.5-x}=1.1 \ time 10^{-2}\]

 

\[x=0.1231 = [H_3O^+]\]

\[pOH= 14- pH\]

\[pOH= 14- (-log(0.1231))\]

\[pOH=13.09\]

Q15.31a

The solubility of ammonia, NH3, is given as 51.8g per 1000 g H2O. What is the approximate pH of a saturated aqueous solution of ammonia? Assume water density is 1.0g per mL.
\(K_b = 1.8 \times 10^{-5} \)

S15.31a

$$NH_3+ H_2 O \rightleftharpoons NH_4^++OH^-$$
$$K_b=1.8\times 10^{-5}$$
 
$$NH_3 =(\dfrac{51.8 gNH_3}{1000 g H_2O})\dfrac{1.00g H_2O}{1 mL}\dfrac{1000 mL}{1 L}\dfrac{1 mol NH_3}{17.031 gNH_3}=3.04 M NH_3$$
 
$$NH_3 (aq)+H_2O(l) \rightleftharpoons NH_4^+ (aq) + (OH)^- (aq)$$
ICE Table \(NH_3\) \(NH_4^+\) \(OH^-\)
Initial \(3.04\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(3.04-x\) \(x\) \(x\)
 
$$K_b=\dfrac{[OH^- ][NH_4^+]}{[NH_3]}=\dfrac{x^2}{3.04-x}=1.8\times 10^{-5}=\dfrac{x^2}{3.04}$$    assuming x≪3.04
 
$$x=\sqrt{3.04(1.8\times 10^{-5})}=0.0074$$ 
The quadratic equation roots formula provides the same answer.
 
\[ \dfrac{0.0074}{3.04} \text{100%} = \text{0.243% < 5%} \]
$$pOH= -log⁡[OH^- ]=-log⁡(0.0074)=2.13$$  
$$pH=14.00-pOH=11.87$$
 

Q15.31b

The solubility of 1- naphthylamine, \(C_{10}H_7NH_2\), a substance used in the manufacture of dyes, is given in a handbook as 1 g per 590 mL \(H_2O\). What is the approximate \(pOH\) of a saturated aqueous solution of 1-naphthylamine?                

\[C_{10}H_7NH_2 + H_2O \rightleftharpoons C_{10}H_7NH_3^+ + OH^- \]

with \(pK_b = 3.92\)

S15.31b

\[1 g( \dfrac{1 mole}{143.19 g})= 0.0070 mol\]

\[Molarity = \dfrac{1 g}{590 mL H_2O} = \dfrac{0.0070 mol}{0.590 L} = 0.0118 M\]

ICE Table \(C_{10}H_7NH_2\) \(C_{10}H_7NH_3^+\) \(OH^-\)
Initial \(0.0118\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(0.0118-x\) \(x\) \(x\)

\[K_b = \dfrac{x^2}{0.0118-x} = 10^{-3.92} \]

\[x = [OH^-] = 0.00113M \]

\[pOH = -log[OH^-] = 2.95 \]

\[pH = 14-pOH = 11.05\]

Acids and Bases

Q15.31c

The solubility of  \(C_{10}H_7NH_2 \) is 1 g per 450 g of water. (Assume density 1g/1mL) What is the pH of the saturated aqueous solution?

\(pKb = 3.92 \)

S15.31c

\[C_{10}H_7NH_2 + H_2O \rightleftharpoons C_10H_7NH_3^+ + OH^- \]

\[pK_b = 3.92\]

\[K_b = 10^{-3.92} = 1.2 \times 10^{-4}\]

\[Molarity = \dfrac{1g C_{10}H_7NH_2}{450g H_2O}(\dfrac{1g H_2O}{1 mL})(\dfrac{1000 mL}{1L})(\dfrac{1 moles C_{10}H_7NH_2}{143.2g C_{10}H_7NH_2}) = 0.0155M \]

\[\dfrac{x^2}{0.0155-x} = 1.2 \times 10^{-4}\]

\[x = 0.0013M = [OH^-]\]

\[pOH = -log[OH^-]\]

\[pOH = 2.9\]

\[pH = 14 - pOH\]

\[pH = 11.1\]

Q15.31d

What is the process for finding the pH of a acid or base in water?

S15.31d

Figure out if the acid base is monoprotic, diprotic etc. pH is calculated by taking the -log of \(H^+\) concentration.

Q15.33a

A particular mustard is found to contain 4.1% acetic acid, \(CH_3COOH\) by mass. What mass of this mustard should be diluted with water to produce 0.675 L of a solution with pH= 4.86?
Given \(K_a = 1.8 \times 10^-5 \)
 

S15.33a

The \([H3O^+]\), due to the stoichiometry of the reaction, is equal to \([CH_3COO^-]\). 
$$[H3O^+]=10^{-pH}=10^{-4.86}=1.38\times 10^{-5}=[C_2 H_3 O_2^-]$$
Solve for S, the concentration of \(HC_2H_3O_2\) in the 0.675 L solution before it dissociates. 
 
\(HC_2H_3O_2(aq)+H_2O(l) \rightleftharpoons C_2H_3O_2^-(aq) + H_3O^+(aq)\)
ICE \(HC_2H_3O_2\) \(C_2H_3O_2^-\) \(H_3O^+\)
Initial \(S\) \(0\) \(0\)
Change \(-1.38 \times 10^{-5}\) \(1.38 \times 10^{-5}\) \(1.38 \times 10^{-5}\)
Equilibrium \(S - 1.38 \times 10^{-5}\) \(1.38 \times 10^{-5}\) \(1.38 \times 10^{-5}\)
 
$$K_a=\dfrac{[H_3 O^+ ][C_2H_3O_2^-]}{[HC_2H_3O_2]}=1.8\times 10^{-5}=\dfrac{(1.38\times 10^{-5})^2}{S-1.38 \times 10^{-5}}$$
 
 
$$S= 2.438 \times 10^{-5} M$$ 
Now we determine the mass of mustard needed.
 
\[\text{Mustard Mass} = 0.675L \times \dfrac{2.438 \times 10^{-5} HC_2H_3O_2}{1L} \times \dfrac{60.05gHC_2H_3O_2}{1 mol} \times \dfrac{100.0g mustard}{4.1g HC_2H_3O_2} = 0.0241g\]

Q15.33b

A solutions is 10% Hydrochloric acid by mass. What mass of the solution is needed to produce 2 L with pH = 6.02?

S15.33b 

$$Molarity = [H^+] = 10^{-pH} = 10^{-6.02} = 9.55 \times 10^{-7} M$$

$$\text{mole of solute} = 2L(10^{-6.02}) = 1.91 \times 10^{-6} moles $$


$$\text{MM of HCl} =1(1.008) + 1(35.45) = 36.458$$

$$\text {HCl mass} = \dfrac{1.91 \times 10^{-6} (36.458[MM])}{(0.1)[Percent by mass]} = 6.96 \times 10^{-4} g$$

Q15.33c

A vinegar is 4.3 % acetic acid, \(CH_3COOH\) by mass. What mass of vinegar should be diluted with water to produce 0.750 L of a solution with pH = 5.85?

Given \(K_a = 1.8 \times 10^{-5} \)

S15.33c

\[CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-\]

\[K_a = 1.8 \times 10^{-5}\]

\[10^{-5.85} = 1.4 \times 10^{-6} M = [CH_3COO^-] = [H_3O^+]\]

\[CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-\]
ICE Table \(CH_3COOH\) \(H_3O^+\) \(CH_3COO^-\)
Initial \(S\) \(0\) \(0\)
Change \( - 1.41 \times 10^{-6} M\) \(+1.41 \times 10^{-6} M\) \(+1.41 \times 10^{-6} M\)
Equilibrium \(S - 1.41 \times 10^{-6} M\) \(1.41 \times 10^{-6} M\) \(1.41 \times 10^{-6} M\)

 

\[K_a = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}\]

\[\dfrac{(1.4 \times 10^{-6} M)^2}{S - 1.41 \times 10^{-6}M} = 1.8 \times 10^{-5}\]

\[S = 1.52 \times 10^{-6} M CH_3COOH\]

\[Mass =  0.750L (1.52 \times 10^{-6}M CH_3COOH)(\dfrac{60.05g CH_3COOH}{1 mol CH_3COOH})(\dfrac{100g vinegar}{4.3g CH_3COOH}) = 0.0016 g \]

Q15.33d

A particular vinegar is found to contain 7.5% acetic acid, \(CH_3COOH\), by mass. What mass of this vinegar should be diluted with water to produce 1.25L of a solution with pH=5.42?

Given \(K_a = 1.8 \times 10^{-5} \)

S15.33d

\[CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-\]

\[K_a = 1.8 \times 10^{-5}\]

\[10^{-5.42} = 3.8 \times 10^{-6} M = [CH_3COO^-] = [H_3O^+]\]

\[CH_3COOH + H_2O \rightleftharpoons H_3O^+ + CH_3COO^-\]
ICE Table \(CH_3COOH\) \(H_3O^+\) \(CH_3COO^-\)
Initial \(S\) \(0\) \(0\)
Change \( - 3.8 \times 10^{-6} M\) \(+3.8 \times 10^{-6} M\) \(+3.8 \times 10^{-6} M\)
Equilibrium \(S - 3.8 \times 10^{-6} M\) \(3.8 \times 10^{-6} M\) \(3.8 \times 10^{-6} M\)

 

\[K_a = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}\]

\[\dfrac{(3.8 \times 10^{-6} M)^2}{S - 3.8 \times 10^{-6}M} = 1.8 \times 10^{-5}\]

\[S = 4.6 \times 10^{-6} M CH_3COOH\]

\[Mass =  1.25L (4.6 \times 10^{-6}M CH_3COOH)(\dfrac{60.05g CH_3COOH}{1 mol CH_3COOH})(\dfrac{100g vinegar}{7.5g CH_3COOH}) =  0.0046g \]

Q15.39a

What is the a) degree of ionization and b) the ionization of benzoic acid in a solution that is 0.24 M \(C_6H_5COOH\)?
$$C_6 H_5 COOH+ H_2 O\rightleftharpoons H_3 O^++C_6 H_5 COO^-$$
\(pK_a = 4.20 \)

S15.39a

First compute the \([H_3O^+ ]\) in the solution:
 
\(C_6H_5COOH(aq)+H_2 O(l)\rightleftharpoons H_3O^+ (aq)+C_6 H_5COO^- (aq)\)
ICE Table \(C_6H_5COOH\) \(H_3O^+\) \(C_6 H_5COO^-\) 
Initial \(0.24\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \((0.24-x)M\) \(xM\) \(xM\)
 
$$K_a=\dfrac{[H_3O^+ ][C_6H_5 COO^-]}{[C_6H_5COOH]}=\dfrac{x^2}{0.24-x}=10^{-4.20}=6.31\times 10^{-5}\approx \dfrac{x^2}{0.24}$$
$$x=3.9 \times 10^{-3} M$$
We have assumed that x≪0.24 M,an assumption that is correct.
 
a)
$$\alpha=\dfrac{[H_3O^+ ]_{equil}}{[C_6H_5COOH]_{equil}}=\dfrac{3.9\times 10^{-3}M}{0.24M}=0.016=\text{degree of ionization}$$
b)
\[\text{% ionization} = \alpha \text {100% = (0.016)(100%) = 1.6%} \]

Q15.39b

What is the (a) degree of ionization and (b) percent ionization of propionic acid in a solution that is 0.32 M.

S15.39b

\[CH_3CH_2CO_2H + H_2O \rightleftharpoons H_3O^+ + CH_3CH_2CO_2\]

\[pK_a = 5.01\]

\[pK_a = -log[K_a]\]

\[K_a = 10^{-5.01}\]

\[CH_3CH_2CO_2H + H_2O \rightleftharpoons H_3O^+ + CH_3CH_2CO_2^-\]

\[CH_3CH_2CO_2H + H_2O \rightleftharpoons H_3O^+ + CH_3CH_2CO_2^-\]
ICE Table \(CH_3CH_2CO_2H\) \(H_3O^+\) \(CH_3CH_2CO_2\)
Initial \(0.32M\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \((0.32-x)M\) \(x\) \(x\)

 

\[K_a = \dfrac{x^2}{0.32-x} = 9.8 \times 10^{-6}\]

\[x = 1.77 \times 10^{-3} M\]

\(\alpha = \dfrac{1.77 \times 10^{-3}M}{0.32M} = 0.0055 \text{degree of ionization}\)

\(\alpha \times 100 = \text{0.55% ionization}\)

Q15.39c

The pH of a 0.100 M solution for chlorous acid (\(HClO_2\)) is 1.57 (10 points).

  1. Calculate \(K_a\) for this acid
  2. Calculate the % ionization of this acid.

S15.39c

1. Ka of Acid

First you will need know what's \(H^+\) concentration. You will discover that by calculating the anti-log of the pH.

1. Find H+ concentration by calculating the anti-log of pH.

\[[H^+] = 10^{-1.57} = 0.02691M\]

\[K_a= \dfrac{[H^+][ClO_2^-]}{[HClO_2]}\]

 

We know the concentration of \(H^+\), which will coincide with the concentration of \(ClO_2^-\). So:

\[K_a= \dfrac{(0.02691)(0.02691)}{0.100 - 0.02691}\]

\[K_a=9.9 \times 10^{-3}\]

 

2. Percent ionization

\[[H^+] = 10^{-1.57} = 0.02691M\]

\[\dfrac{0.02691}{0.100} = 0.2691 \]

\[0.2691 (100) = \text{% ionization = 26.91%}\]

Q15.41a

What is the initial molarity of an aqueous solution of \(C_6H_5NH_2\) if it is 5.0% ionized?
Given \(K_b = 7.4 \times 10^{-10} \)

S15.41a

Let x be the initial concentration of \(C_6H_5NH_2\),, therefore, the amount dissociated is 0.050x
 
\[C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5(NH)_3^+ + OH^-\]
ICE Table \(C_6H_5NH_2\) \(OH^-\) \(C_6H_5(NH)_3^+\)
Initial \(xM\) \(0\) \(0\)
Change \(-0.050xM\) \(+0.050xM\) \(+0.050xM\)
Equilibrium \(x-0.050xM\) \(0.050xM\) \(0.050xM\)
 
$$K_b=7.4\times 10^{-10}=\dfrac{[C_6H_5(NH)_3)^+][OH^-]}{[C_6 H_5(NH)_3]} =\dfrac{(0.050x)^2}{0.095x}=0.0263x$$
$$[C_6H_5(NH)_3]=x=\dfrac{7.4\times 10^{-10}}{0.0263}=2.81\times 10^{-8}M$$

Q15.41b

What is the initial molarity of an aqueous solution of \(NH_3\) if it is 1.2% ionized?

Given \(K_b = 1.8 \times 10^{-5} \)

S15.41b

\[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\]

\[K_b = 1.8 \times 10^{-5}\]

ICE Table \[NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\]
Initial \(x\) \(0\) \(0\)
Change \(-0.012x\) \(+0.012x\) \(+0.012x\)
Equilibrium \(x-0.012x\) \(0.012x\) \(0.012x\)

 

 

\[K_b = 1.8 \times 10^{-5} = \dfrac{[0.012x]^2}{0.988x}\]

\[x = 0.1235M\]

Q15.43a

What is the concentration of \(H^+\) of 0.001 M solution of \(CH_3COOH\) with a \(K_a = 6.5 \times 10^{-6}\)?

S15.43a

$$Ka = \dfrac{[H^+][A^-]}{[HA]}$$

We can assume that \([H^+] = [A^-] = x\) since this is a buffer solution. 

\[6.5 \times 10^{-6} = \dfrac{x^2}{0.001 - x}\]

\[x = [H^+] = 7.74 \times10^{-5} M\]

\[ \text{% ionization} = \dfrac{8.06\times 10^{-5}}{0.001} \text{(100%) = 8.06%}\]

Ionization Constants

Q15.43b

What is the percent ionization of 2.5 M \(HNO_2\) in water?

Given \(K_a = 4.5 \times 10^{-4} \)

S15.43b

First we identify the balanced reaction

\[HNO_3 + H_2O \rightleftharpoons H_3O^+ + NO_3^-\]

Now we can use the ICE table to address the question, which is asking information about the equilibrium properties of the acid.

ICE Table \( HNO_3 \) \( H_2O \) \( H_3O^+ \) \( NO_3^- \)
Initial 2.5 M - 0 0
Change -x - +x +x
Equilibrium 2.5-x - x x


\[X = \dfrac{moles}{\text{liter of ionize }HNO_3}\]

\[K_a = \dfrac{[H^+][NO_2^-]}{[HNO_2]}\]

$$[H^+] = [NO_2^-] = x$$

$$[HNO_2] = 2.5 -x$$

Assume \(x \ll  2.5\)

$$[HNO_2] = 2.5$$

$$4.5\times 10^{-4} = \dfrac{x^2}{2.5}$$

$$x = 0.0335$$

\(x \ll  2.5\) so no need for quadratic formula

Q15.43c

What is the percent ionization of Benzoic Acid, in 0.1M \(C_6H_5COOH\)?
Given \(K_a = 6.4 \times 10^{-5} \)

S15.43c

You can use an ice table to describe the 0.1 M \(C_6H_5COOH\)
\[C_6H_5COOH + H_2O \rightleftharpoons H_3O^+ + C_6H_5COO^-\]
ICE Table \(C_6H_5COOH\) \(H_3O^+\) \(C_6H_5COO^-\)
Initial \(0.1M\) \(0\) \(0\)
Change \(-xM\) \(+xM\) \(+xM\)
Equilibrium \(0.1-xM\) \(xM\) \(xM\)
 
 
We need to calculate \(x = [H_3O^+]\) 
$$K_a=6.4\times 10^{-5}=\dfrac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]} = \dfrac{x^2}{0.1M-x}\approx \dfrac{x^2}{0.1M}$$
$$x=0.00253 M$$
 
$$\text{Percent Ionization} =  \dfrac{[H_3 O^+]}{[C_6 H_5COOH]}100=\dfrac{0.00253 M}{0.1M}100 =\text{2.53% ionization}$$

Q15.43c

What is the percent ionization of carbonic acid in 2.0M \(H_2CO_3\)?

Given \(K_a = 4.4 \times 10^{-7} \)

S15.43c

ICE Table \[H_2CO_3 + H_2O \rightarrow H_3O^+ + HCO_3^-\]
Initial \(2.0M\) \(0\) \(0\)
Change \(-x\) \(+x\) \(x\)
Equilibrium \(2-x\) \(x\) \(x\)

 

Assume: \(x \ll  2.0\)

\[K_a = \dfrac{x^2}{2-x} = \dfrac{x^2}{2} = 4.4 \times 10^{-7}\]

\[x = [HCO_3^-] = 9.38 \times 10^{-4}M \]

\[\dfrac{9.38\times 10^{-4}}{2.0} \times \text{100% = 0.047%} \]

Q15.43d

A solution containing hydrochloric acid, \(HCl\) is 11.3 percent \(HCl\) by mass. What mass of this solution should be diluted with water to produce 2 L of a solution with pH of 3.89?

S15.43d

Knowns/Givens:

  • \(pH = -log[H^+]\)
  • \([H^+] = 10^{-pH}\)
  • \(pH = 3.89\)
  • \(HCl = 36.5 \dfrac{g}{mol}\)

 

\[[H^+]=10^{-3.89} = 1.288 \times 10^{-4}M \]

Therefore, the moles of \(HCl\) solute in 2 Liters of solution would be:

\[1.228 \times 10^{-4}M (2L) = 2.576\times 10^{-4} moles\] 

That many moles would have a mass \(2.576 \times 10^{-4} mol(\dfrac{36.5g}{mol}) = 0.0094 g\)

\[ \text{Mass of solution} = 0.0094g \dfrac{100}{11.3} = 0.083g \]

Q15.45a 

Explain why \([CO_3^{2-}]\) in 2.0 M \([H_2CO_3]\) is not simply 1/2 \([H_3O^+]\), but much much less than 1/2 \([H_3O^+]\) 

S15.45a

Carbonic acid, \([H_2CO_3]\), is a weak acid, so there is much less \(CO_3^{2-}\)being produced after the second ionization. Comparatively, there is very little \(H_3O^+\) being produced. 

Q15.45b

A solution is 5.3 % \(HBr\) by mass. What is the mass needed to produce 1.3 L of product with a pH = 4.35?

S15.45b

$$ [H^+] = 10^{-pH} = 10^{-4.35}M$$

$$\ 1.3 L(10^{-4.35}) = 5.8068 \times 10^{-5} [Moles]$$

$$\text{MM of HBr} =1(1.008) + 1(79.90) = 80.908$$ 

$$\text{grams HBr}=  \dfrac{5.8068 \times 10^{-5}moles(80.908g.mol^{-1})}{0.053} = 0.08864 grams$$

Q15.45c

Explain why\([PO_4^{3-}]\) in 1.00 M \(H_3PO_4\) is not simply \( \dfrac{1}{3} [H_3O^+]\), but much less than( \dfrac{1}{3} [H_3O^+]\).

S15.45c

The 1st ionization produces the highest \([H_3O^+]\) and in any subsequent ionizations the \([PO_4^{3-}]\) decreases in every ionization which would be 3 because there are three hydrogen atoms attached.

Q15.45d

Explain why \([SO_4^{2-}]\) in 1.00 M \(H_2SO_4\) is not simply \(0.5[H_3O^+]\), but much, much less than \(0.5[H_3O^+]\).

S15.45d

The decomposition of sulfuric acid to the sulfate ion is a two-step process. First, the sulfuric acid undergoes a reaction that reduces a portion of the \(H_2SO_4\) to \(HSO_4^-\) and \(H_3O^+\). Second, the \(HSO_4^-\) goes through its own reaction, reducing a portion of it to \(SO_4^{2-}\), the sulfate ion, and additional \(H_3O^+\). The reason \(0.5[H_3O^+]\) is not an accurate calculation for the concentration of the sulfate ion is the fact that \(H_3O^+\) is produced in both reactions, while the sulfate ion is only produced in the second reaction. Additionally, each reaction has its own corresponding \(K_a\) value which determines the concentrations of the \(H_3O^+ \)and the other products. The correct calculation of \([SO_4^{2-}]\), thus, takes both these equations into and their \(K_a\) values into consideration.

Q15.47a

For the dissociation reaction of the polyprotic acid \(H_2SO_4\), determine \(H3O^+\), \(HSO_4^-\) and \(SO_4^{2-}\) for both a concentration of 0.05 M and 0.1 M.
Given \(K_{a1} = 1.0 \times 10^{3} \) and \( K_{a2} = 1.2 \times 10^{-2} \)

S15.47a

\(H_2SO_4 \) is a strong acid and it completely dissociates in water. 

a. \([H_2SO_4] = 0.05M \)

$$K_{a2}=  \dfrac{[SO_4^{2-} ][H_3 O^+ ]}{[HSO_4^-]}=1.2 \times 10^{-2} = \dfrac{x(x+0.05)}{0.05-x}$$

$$x=0.0085M  = [SO_4^{2-} ] $$ 

\[ [H_3O^+] = 0.0085 + 0.05 = 0.0585M \]

\[[HSO_4^-] = 0.05 - 0.0085 = 0.0415M \]

b. \([H_2SO_4] = 0.1M \)

$$K_{a2}=  \dfrac{[SO_4^{2-} ][H_3 O^+ ]}{[HSO_4^-]}=1.2 \times 10^{-2} = \dfrac{x(x+0.1)}{0.1-x}$$

$$x=0.0098M  = [SO_4^{2-} ] $$ 

\[ [H_3O^+] = 0.0098 + 0.1 = 0.1098M \]

\[[HSO_4^-] = 0.1 - 0.0098 = 0.0902M \]

Q15.47b

Determine the \([H_3O^+]\), \([HS^-]\), \([S^{2-}]\) for the following \(H_2S(aq)\)solutions. a) 1.0 M \(H_2S\) b) 0.5M \(H_2S\) c) 0.0001 M \(H_2S\).

Given \(K_{a1} = 1.0 \times 10^{-7}\) and \(K_{a2} = 1.0 \times 10^{-19}\)

S15.47b

Using a) as example for calculation.

\[H_2S(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HS^-(aq)\]

\[K_{a1} = 1.0 \times 10^{-7}\]

\[K_{a2} = 1.0 \times 10^{-19}\]

ICE Table \[H_2S(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HS^-(aq)\]
Initial \(1.0M\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \((1.0 - x)M\) \(x\) \(x\)

 

 

\[K_{a1} = \dfrac{x^2}{1.0-x} \approx \dfrac{x^2}{1.0} = 1.0 \times 10^{-7} \]

\[x = [HS^-] = [H_3O^+] = 3.2 \times 10^{-4} M \]

ICE Table \[HS^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + S^{2-}(aq) \]
Initial \(3.2 \times 10^{-4} \) \(3.2 \times 10^{-4} M\) \(0\)
Change \(-y\) \(+y\) \(+y\)
Equilibrium \(3.2 \times 10^{-4} - y\) \(3.2 \times 10^{-4} + y\) \(y\)

 

\[K_{a2} = 1.0 \times 10^{-19} = \dfrac{(y)(3.2 \times 10^{-4} + y)}{3.2 \times 10^{-4} - y} \]

\[y = [S^{2-}] = 1.0 \times 10^{-19} \]

a) \(x = 3.2 \times 10^{-4} M [HS^-], 1.0 \times 10^{-19} M [S^{2-}]\)

b) \(x = 2.2 \times 10^{-4} M [HS^-], 1.0 \times 10^{-19} M [S^{2-}]\)

c) \(x = 3.1 \times 10^{-6} M [HS^-], 1.0 \times 10^{-19} M [S^{2-}]\)

Q15.47c

Determine concentration of \(H^+\) for the following \(HS\) solution. \([HS] = 1.0 \times 10^{-10} M\).

\(K_{a1} = 1.0 \times 10^{-7} \)

S15.47c

\(HS\) is a very weak acid and \(1.0 \times 10^{-10} M\) is a tiny concentration, the concentration of \(H^+\) will be tiny.

\[ 1.0 \times 10^{-7} = \dfrac{x^2}{1.0 \times 10^{-10} - x}\]

\[ x = [H^+] = 9.99 \times 10^{-11} \]

Q15.47d

Find the  \([H_3O^+]\), \([HC_2O_4^-]\) and \([C_2O_4^{2-}]\) for the following aqueous oxalic acid concentrations given \(K_{a1} = 5.6 \times 10^{-2}\) and \(K_{a2} = 5.4 \times 10^{-5}\) :

  1. 30M
  2. 75M
  3. 4M.

S15.47d

 

n = initial concentration of \(H_2C_2O_4\)

\[H_2C_2O_4(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HC_2O_4^-(aq)\]  
ICE Table \(H_2C_2O_4\) \(H_3O^+\) \(HC_2O_4^-\)
Initial \(n\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(-x\) \(x\) \(x\)

 

 

\[K_{a1} = \dfrac {[H_3O^+][HC_2O_4^-]}{[H_2C_2O_4]}=\dfrac {x^2}{n-x} = 5.6\times 10^{-2}\]

 

\[ HC_2O_4^-(aq) + H_2O(l) \rightleftharpoons H_3O^+ + C_2O_4^{2-} \]
ICE Table \( HC_2O_4^- \) \(H_3O^+ \) \(C_2O_4^{2-} \)
Initial \(x\) \(x\) \(0\)
Change \(-y\) \(+y\) \(+y\)
Equilibrium \(x-y\) \(x+y\) \(y\)

 

\[K_{a2} = \dfrac {[H_3O^+][C_2O_4^-]}{[HC_2O_4^-]} = \dfrac{(x+y)(y)}{x-y} =  5.4 \times 10^{-5}\]

a)   

\[x = 1.3M = [HC_2O_4^-]\]

\[y = 5.4 \times 10^{-5}M=[C_2O_4^{2-}]\]

\[y+x = 1.3M = [H_3O^+] \]

b)  

\[x = 2.02M = [HC_2O_4^-]\]

\[5.4 \times 10^{-5}M=[C_2O_4^{2-}]\]

\[y+x = 1.3M = [H_3O^+] \]

c)  

\[x = 0.45M = [HC_2O_4^-]\]

\[y = 5.4 \times 10^{-5}M=[C_2O_4^{2-}]\]

\[y+x = 1.3M = [H_3O^+] \]

Q15.49a

Find the pH of \(HCl\) and \(NaOH\) for the following solutions:

  1. 1.5M \(HCl\)
  2. 0.30M \(NaOH\)
  3. 0.064M \(HCl\)

S15.49a

a. \[pH = -log(1.5) = 0.176 \]

b. \[pH = 14 + log(0.30) = 13.48 \]

c. \[pH = -log(0.064) = 1.19\]

Q15.49b

Find the concentration of \(H^+ \) with 0.05 M of carbonic acid.
\(Ka_1 = 4.5 \times 10^{-7} \)
\(Ka_2 = 4.7 \times 10^{-11} \)

S15.49b

\(Ka_1\) and \(Ka_2\) are given in a table as \(4.5 \times 10^{-7}\) and \(4.7 \times 10^{-11}\) respectively. 
 
\[H_2CO_3 + H_2O \rightleftharpoons HCO_3^- + H_3O^+\]
ICE Table \(H_2CO3\) \(H_3O^+\) \(HCO_3^-\)
Initial \(0.05M\) \(0\) \(0\)
Change \(-xM\) \(+xM\) \(+xM\)
Equilibrium \(0.05-xM\) \(xM\) \(xM\)
 
 
$$K_{a1}=\dfrac{[HCO_3^- ][H_3 O^+ ]}{[H_2CO_3]}=4.5\times 10^{-7}= \dfrac{x^2}{0.05-x}  \approx \dfrac{x^2}{0.05}$$
$$x=0.000149M$$
$$K_{a2} = \dfrac{y(x+y)}{x-y}=4.5\times 10^{-11}= \dfrac{(0.000149 +y)}{0.000149 - y} $$
$$y = 4.7 \times 10^{-11}M$$
\[ [H_3O^+] = x + y = 0.000149 + 4.7 \times 10^{-11} = 0.000149M \]

Q15.49c

Calculate the \([H_3O^+]\), \([HSO_4^-]\), \([SO_4^{2-}]\) in (a) 1.75 M \(H_2SO_4\); (b) 1.5 M \(H_2SO_4\); (c) 1.0 M \(H_2SO_4\)

\(K_{a2} = 0.011\)

S15.49c

\(H_2SO_4 is a strong acid. Completely dissociates. Below is the example calculation for part a. Same procedures are done for b and c

\[H_2SO_4(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HS^-(aq) \]

\[K_{a2} = 0.011\]

Equation \[HSO_4^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + SO_4^{2-}(aq) \]
ICE Table \(HSO_4^-\) \(H_3O^+\) \(SO_4^{2-}\)
Initial \(1.75M\) \(1.75M\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(1.75-x\) \(1.75+x\) \(x\)

 

 

\[K_{a2} = \dfrac{x(1.75+x)}{1.75-x} \approx \dfrac{1.75x}{1.75}\]

\[x = 0.011M [SO_4^{2-}]\]

  1. \([H_3O^+]\) =1.76 M , \([SO_4^{2-}]\) = 0.011 M, \([HSO_4^-]\) = 1.74 M
  2. \([H_3O^+]\) =1.51 M , \([SO_4^{2-}]\) = 0.011 M, \([HSO_4^-]\) = 1.49 M
  3. \([H_3O^+]\) =1.01 M , \([SO_4^{2-}]\) = 0.011 M, \([HSO_4^-]\) = 0.99 M

Q15.49d

Assuming Kc= 0.54, find the Molarity of \(H_3O^+\) and \(ClO_2^-\) for

  1. 1.2 M \(HClO_2\)
  2. 0.86 M \(HClO_2\)

S15.49d

\[HClO_2(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + ClO_2^- (aq)\]

a) 

\[0.54 = \dfrac{x^2}{1.2-x}\]

\[x = 0.58\]

\[[H_3O^+] = [ClO_2^-] = 0.58 M\]

 

b) )

\[0.54 = \dfrac{x^2}{0.86-x}\]

\[x= 0.46\]

\[[H_3O^+] = [ClO_2^-] = 0.46 M\]

Q15.53

Given pKa=4.72 for hydrazonic acid, write the reaction for \(HN_3\) and calculate the pKb for hydrazonic acid.

S15.53

\[HN_3+H_2O \rightleftharpoons H_3O^+ +N3^-\]

 

\[pK_w = pK_a + pK_b \]

\[pK_b = 14 - 4.72 \]

\[pK_b = 9.28 \]

Q15.55a

Fill in the reactions and identify which of the following 3 reactions is a hydrolysis reaction.
 
$$NH_3+H_2 O \rightarrow ?$$
 
$$NH_4 Cl+H_2 O \rightarrow ?$$
 
$$NaOH+HCl \rightarrow ?$$

S15.55a

$$NH_3+H_2 O \rightarrow ?$$
$$NH_3+H_2 O \rightarrow NH_4^++ OH^- \text{ Hydrolysis reaction}$$ 
 
$$NH_4 Cl+H_2 O \rightarrow ?$$
$$NH_4 Cl+H_2 O \rightarrow NH_4OH+HCl \text{ Hydrolysis reaction}$$ 
 
$$NaOH+HCl \rightarrow ?$$
$$NaOH+HCl \rightarrow NaCl+H_2O \text{ Not Hydrolysis reaction}$$ 
 

Q15.55b

Complete the following equations in those instances in which a reaction will occur. If no reaction occurs, so state.

  1. \(Na^+ + Cl^- + H_2O \rightarrow\)
  2. \(NO_2^- + H_2O \rightarrow\)
  3. \(CN^- + H_2O \rightarrow\)

S15.55b

  1. \(Na^+ + Cl^- + H_2O \rightarrow \text{no reaction}\)
  2. \(NO_2^- + H_2O \rightarrow HNO_2 + OH^-\)
  3. \(CN^- + H_2O \rightarrow HCN + OH^-\)

Q15.63a

For each of the following ions, write two equations – one showing its ionization as an acid and the other as a base:

  1. \(H_2PO_4^-\)
  2. \(HF\)
  3. \(HSO_2^-\)

S15.63a

a.)   

$$H_2PO_4^- + H_2O\rightarrow HPO_4^{2-} + H_3O^+ (acid)$$

\[H_{2}PO_{4}^{-} + H_{2}O \rightarrow H_{3}PO_{4} + OH^{-} (base)\]

b.)   

$$HF + H_2O \rightarrow F^- + H_3O^+ (acid)$$

$$HF + H_2O \rightarrow H_2F^+ + OH^- (base)$$

c.)    

$$HSO_2^- +H_2O \rightarrow SO_2^{2-} + H_3O^+ (acid)$$

$$HSO_2^- + H_2O \rightarrow H_2SO +  OH^- (base)$$

Lewis Acid and Base

Q15.63b

 

Write the equation for the ionization as and acid and one for a base
  1. \(H_2PO_4^-\)
  2. \(HCO_3^-\)
  3. \(HSO_4^-\)

S15.63b

a) $$\text{Acid: } H_2PO_4^- (aq)+H_2O(l ) \rightleftharpoons H_3O^+ (aq)+HPO_4^{2-} (aq)$$
$$\text{Base: } H_2 PO_4^- (aq)+H_2 O(l) \rightleftharpoons H_3PO_4 (aq)+OH^- (aq)$$
 
b) $$\text{Acid: } HCO_3^- (aq)+  H_2 O(l) \rightleftharpoons H_3 O^+ (aq)+CO_3^{2-} (aq)$$
$$\text{Base: } HCO_3^- (aq)+H_2 O(l) \rightleftharpoons H_2CO_3 (aq)+OH^- (aq)$$
 
c) $$\text{Acid: }HSO_4^- (aq)+H_2 O(l) \rightleftharpoons H_3 O^+ (aq)+SO_4^{2-} (aq)$$
$$\text{Base: } HSO_4^- (aq)+H_2 O(l) \rightleftharpoons  H_2SO_4 (aq)+OH^- (aq)$$

Q15.65a

Which is a stronger acid:

  1. \(HNO_3\) or \(H_2SO_4\)?
  2. \(HI\) or \(HF\)?
  3. \(HBr\) or \(HNO_2\)?

S15.65a

  1. \(H_2SO_4\)
  2. \(HI\)
  3. \(HBr\)

Q15.65b

Predict which is the stronger acid

  1. \(HCl\) or \(HF\)
  2. \(HClO\) or \(HClO_3\)
  3. \(HCl\) or \(HBr\)

S15.65b

  1. \(HCl\) 
  2. \(HClO_3\) 
  3. \(HBr\).

S15.65c

Predict which is the stronger base:

  1. \(NaOH\) or \(NH_3\)
  2. \(SO_4^{2-}\) or \(CN^-\)
  3. \(LiOH\) or \(Sr(OH)_2\)

S15.65c

  1. \(NaOH\)
  2. \(CN^{-}\)
  3. \(Sr(OH)_2\)

Q15.67

For the following pairs of acids, indicate which is the stronger acid and why.

  1. \(HF\) vs.\(HCl\)
  2. \(CH_3CF_2COOH\) vs. \(CH_3CHFCOOH\)

S15.67

  1. \(HCl\) has a larger ion, it is the stronger acid.
  2. \(CH_3F_2COOH\) is stronger because it has more \(F\) atoms.

Q15.69

Select the base one with the smallest Kb and the one with the largest Kb and give the reasons for your choice.

  1. \(CH_3CH_2CH_2CH_3\) 
  2. \(NH_3\)                   
  3. \(C_6H_5NH_2\) 

S15.69

The largest \(K_b\) (most basic) is the hydrocarbon chain (a) which has the lowest electronegativity. The smallest Kb(least basic) would be (c) because the NH2 pair is delocalized and thus is less likely to accept a proton.

Q15.73

Which of the following is a weak or strong acid?

  1. \(CCl_3COOH\)
  2. \(HCN\)
  3. \(H_2SO_4\)
  4. \(HI\)

S15.73

  1. Weak acid
  2. Weak acid
  3. Strong acid
  4. Strong acid

Q15.77b

Write an equation for the formation of the \(I_3^-\) anion from molecular iodine, and indicate the Lewis acid and Lewis base.

15.S77b

\[I_2(aq) + I^-(aq) \rightleftharpoons I_3^-(aq)\]

\(I_2\)(aq) = Lewis acid, \(I^-\)(aq) = Lewis base

Q15.77c

Identify the Lewis acid and Lewis Base of the reaction: 

\[6H_2O(l) + Cu^{2+}(aq) \rightarrow [Cu(H_2O)_6]^{2+}(aq)\]

S15.77c

Lewis Acids accept pairs of electrons, therefore the \(Cu^{2+}\) is the Lewis Acid and Lewis Bases donate pairs of electrons (there is a lone pair of electrons on the O atom that are donated to the cupric ion), therefore the water is the Lewis Base.

Q15.79

\(Br_2\) doesn’t dissolve in water well but does dissolve in \(KBr\) because a \(Br_3^-\) anion forms. Write an equation for this formation.

S15.79

\[Br2(aq) + Br^-(aq) \rightleftharpoons Br^{3-}(aq)\]

Q15.86

How does the ion product of water change with temperature? What is the correlation?

S15.86

Yes, they are directly related, as temperature increases so does the ion product and as temperature decreases so does ion product.

Q15.90a

A sample Acetic Acid (found in many Vinegers and Cleaning Supplies) has a \(K_a = 1.8 \times 10^{-5} \) with 0.100 M. Calculate pH of the solution.

S15.90a

 

ICE Table $$HC_2H_3O_2 (aq) + H_2O (l) \rightarrow H_3O^+ (aq) + C_2H_3O_2^- (aq)$$
  \(HC_2H_3O_2 \) \(H_3O^+\) \(C_2H_3O_2^-\)
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1-x +x +x


$$Ka = \dfrac{[H_3O^+] [C_2H_3O_2-]}{ [HC_2H_3O_2]}$$

$$1.8 \times 10^{-5} = \dfrac{x^2}{0.1 - x}$$

$$x = 0.00133 = [H_3O^+]$$

$$pH = -log [H_3O^+] = -log (0.00133) = 2.88$$

Q15.90b

It is possible to write simple equations to relate pH, pK, and molarities (M) of various solutions. Three such equations are shown here.

\( \text{Weak acid: pH = } \frac{1}{2}pK_a - \frac{1}{2}logM \)

\( \text{Weak base: pH = } 14 - \frac{1}{2}pK_b + \frac{1}{2}logM \)

\( \text{Salt of weak acid (pKa) and strong base: pH =} \dfrac{1}{2} pK_w + \dfrac{1}{2} pK_a + \dfrac{1}{2} log(M) \)

  1. What is the name of the famous equation used to derive these three? What is the equation?
  2. Use these equations to determine the pH of a 0.2 M \(CH_3COOH\)(aq), 0.2 M \(NH_3\)(aq), and 0.2 M \(NaCH_3COO\).

\(K_a \text{ of C}H_3COOH = 1.8 \times 10^{-5} \)

\(K_b \text{ of N}H_3 = 1.8 \times 10^{-5} \)

S15.90b

a.)    The Henderson-Hasselbalch Equation.  \[pH = pKa + log(\dfrac{A^-}{HA})\]

b.)    $$ \text{pKa of }CH_3COOH = 4.74.$$ 

$$pH = 0.5(4.74) – 0.5log(0.2)  =2.72$$

$$\text{pKb of N}H_3 = 4.74 $$

$$pH = 14 - 0.5(4.74) + 0.5log(0.2) = 11.3$$

$$pKw = 14 $$

$$pH = 14 - 0.5(14) + 0.5(4.74) + 0.5log(0.2)  = 9.02$$

Henderson-Hasselbalch Approximation

Q15.90c

M = concentration of the given solution.

\( \text{Weak acid: pH = } \frac{1}{2}pK_a - \frac{1}{2}logM \)

\( \text{Weak base: pH = } 14 - \frac{1}{2}pK_b + \frac{1}{2}logM \)

\( \text{Salt of weak acid (pKa) and strong base: pH =} \dfrac{1}{2} pK_w + \dfrac{1}{2} pK_a + \dfrac{1}{2} log(M) \)

a) Derive these equations and point out the assumptions involved in the derivatives.

b) Use these equations to determine the pH of 0.22 M \(CH_3COOH\)(aq), 0.14M \(NH_3\) (aq), 0.32M \(NaCH_3COO\).

\(K_a \text{ of C}H_3COOH = 1.8 \times 10^{-5} \)

\(K_b \text{ of N}H_3 = 1.8 \times 10^{-5} \)

S15.90c

a)

Weak Acid

$$HA(aq) + H_2O(l) \rightleftharpoons A^-(aq) + H_3O^+(aq)$$
ICE \(HA\) \(H_3O^+\) \(A^-\)
Initial \(M\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(M-x\) \(x\) \(x\)

 

Assuming x is a lot less than M.

\[K_a = \dfrac{x^2}{M-x} \approx  \dfrac{x^2}{M} \]

\[x = \sqrt{K_a M}\]

\[pH = -log(x) = -log(K_aM)^{ \dfrac{1}{2}} \]

\[ pH = - \dfrac{1}{2} log(K_a) - \dfrac{1}{2} log(M) \]

\[pH = \dfrac{1}{2} pK_a - \dfrac{1}{2} log(M) \]

 

Same kind of process is done for weak base

$$A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^- (aq)$$
ICE \(A^-\) \(OH^-\) \(HA\)
Initial \(M\) \(0\) \(0\)
Change \(-x\) \(+x\) \(+x\)
Equilibrium \(M-x\) \(x\) \(x\)

 

\[K_b = \dfrac{x^2}{M-x} \approx  \dfrac{x^2}{M} \]

\[x = \sqrt{K_b M}\]

\[pOH = -log(x) = -log(K_bM)^{ \dfrac{1}{2}} \]

\[pOH = - \dfrac{1}{2} log(K_b) - \dfrac{1}{2} log(M) \]

\[pOH = \dfrac{1}{2} pK_b - \dfrac{1}{2} log(M) \]

\[pH = pKw - pOH = 14 - pOH \]

\[pH = 14 - \dfrac{1}{2} pK_b + \dfrac{1}{2} log(M) \]

 

Salt of weak acid \(pK_a\) and strong base: pH

Assuming molar ratio of base and acid are equal. 

\[K_b = \dfrac{K_w}{K_a} \]

\[K_b = \dfrac{[OH^-]^2}{M} \]

\[ \dfrac{K_w}{K_a} = \dfrac{[OH^-]^2}{M} \]

\[ [OH^-] = \sqrt{ \dfrac{MK_w}{K_a} } \]

\[ pOH = log(\sqrt{ \dfrac{MK_w}{K_a} }) \]

\[ pOH = \dfrac{1}{2} log( \dfrac{MK_w}{K_a}) \]

\[ pOH =  \dfrac{1}{2} (log(MK_w) + log(K_a) ) \]

\[ pH = pKw - pOH \]

\[ pH = pKw - \dfrac{1}{2} (log(MK_w) + log(K_a) ) \]

\[ pH = pKw - \dfrac{1}{2} pK_w + \dfrac{1}{2} pK_a + \dfrac{1}{2} log(M) \]

\[ pH = \dfrac{1}{2} pK_w + \dfrac{1}{2} pK_a + \dfrac{1}{2} log(M) \]

 

 

b)

\[pK_a \text{ } CH_3COOH(aq) = -log(1.8 \times 10^{-5})\]

\[pK_a = 4.74\]

\[pH = 0.5pK_a - 0.5log(M)\]

\[pH = 2.70\]

 

\[pK_b \text{ } NH_3(aq) = -log(1.8 \times 10^{-5})\]

\[pK_b = 4.74\]

\[pH = 14.00 - 0.5pK_b + 0.5 log(M)\]

\[pH = 11.2\]

 

 

\[pK_a \text{ } NaCH_3COO(aq) = -log(1.8 \times 10^{-5})\]

\[pK_a = 4.74\]

\[pK_w = 14\]

\[pH = 14 - 0.5log(K_w) + 0.5log(M) + 0.5log(K_a)\]

\[pH = 9.1\]

Q15.90d

Calculate pH of 0.100 M of \(HCl\) solution.

S15.90d

\[pH = -log(0.100M) = 1 \]

Q15.93a

Determine the pH of 0.0100 M succinic acid.

\(K_{a1} = 6.2 \times 10^{-5}\)

\(K_{a2} = 2.3 \times 10^{-6}\)

S15.93a

\[H_2C_4H_4O_4 + H_2O \rightleftharpoons H_3O^+ + HC_4H_4O_4^- \]

\[K_{a1} = 6.2 \times 10^{-5}\]

\[HC_4H_4O_4^- +H_2O \rightleftharpoons H_3O^+ + C_4H_4O_4^{2-}\]

\[K_{a2} = 2.3 \times 10^{-6}\]

ICE Table \[H_2C4H4O4 + H_2O \rightleftharpoons H_3O^+ + HC_4H_4O_4^- \]
Components \(H_2C_4H_4O_4\) \(H_3O^+\) \(HC_4H_4O_4^-\)
Initial \(0.0100\) \(0\) \(0\)
Change \(-x\) \(x\) \(x\)
Equilibrium \(0.01-x\) \(x\) \(x\)

 

\[K_{a1} = \dfrac{x^2}{0.01 - x}\]

\[[H_3O^+] = 7.57 \times 10^{-4}M\]

\[pH = -log[H_3O^+]\]

\[pH = 3.1\]

 

ICE Table \[HC_4H_4O_4^- +H_2O \rightleftharpoons H_3O^+ + C_4H_4O_4^{2-}\]
Components \(HC_4H_4O_4^{-}\) \(H_3O^+\) \(C_4H_4O_4^{2-}\)
Initial \(7.9 \times 10^{-4}M\) \(7.9 \times 10^{-4}M\) \(0\)
Change \(-Y\) \(+Y\) \(+Y\)
Equilibrium \((7.9 \times 10^{-4} - Y)M\) \((7.9 \times 10^{-4} + Y)M\) \(Y\)
 

\[K_{a2} = \dfrac{Y(7.57 \times 10^{-4} + Y)M}{(7.57 \times 10^{-4} - Y)M}\]

\[Y = 2.3 \times 10^{-6}\]

\[[H_3O^+] = 7.9 \times 10^{-4} M\]

\[pH = -log[H_3O^+]\]

\[pH = 3.1\]

Q15.93b   

Determine the pH of 0.204 M sulfuric acid.

\(K_{a1} = 1.0\times 10^3\)

\(K_{a2} = 1.0 \times 10^{-2}\) .

S15.93b

Sulfuric acid is a strong acid. 

pH can be found as such:

\[pH = -log(0.204) = 0.69 \]

\[H_2SO_4(aq) + H_2O (l) \rightleftharpoons HSO_4^-(aq) + H_3O^+(aq) \]

\[1.0 \times 10^3 = \dfrac{x^2}{0.204 - x}\]

\[x = 0.204M\]

\[[H_3O^+]_1 = 0.204M\]

  \[HSO_{4}^- (aq) + H_2O(l) \rightleftharpoons SO_4^{2-}(aq) + H_3O^+(aq)\]

   \[1.0\times 10^{-2} = \dfrac{(0.204+y)(y)}{0.204-y}\]

\[y = 0.00914M\]

\[[H_3O^+] = 0.204 + 0.00914 = 0.213M\]

\[pH = -log(0.213) = 0.671\]

Q15.103a

In your own words, define or explain the following terms or symbols:
  1. \(K_c\)
  2. \(pOH\) 
  3. \(pK_b\) 
  4. Lewis base

S15.103a

\(K_c\) is known as an expression for equilibrium constants. It is expressed in terms of concentrations. If you allow a reaction to reach equilibrium and then measure the equilibrium concentrations of everything, you can combine all of these concentrations into an expression known as \(K_c\).

\(pOH\) is a measure of the concentration of hydroxide ions in a solution. It also measures the alkalinity of a solution. These values are derived from \(pH\) values 
\(K_b\) is the base dissociation constant of a solution. It is closely related to \(K_a\), which is the acid dissociation constant. \(pKb\) is -log of \(K_b\). The lower the \(pK_b\) value, the stronger the base.
A lewis base is a compound that is capable of donating an electron or pair of electrons to an electron accepter compound, also known as a Lewis acid. 

Q15.103b

In your own words define these terms

(a) Kw

(b) pH

(c) pKa

(d) Hydrolysis

(e) Lewis Acid

S15.103b

  1. The equilibrium constant of water also known as 1.0 x 10-14 at 25 Celsius.
  2. A number used to express acidity of a solution base on logarithmic scale. –log[H+]
  3. The Ka is the dissociation constant of an acid. pKa is simply -log of Ka
  4. A reaction involving the breaking of a bond in a molecule using water.
  5. A species which acts as an electron acceptor.