17: Aqueous Equilibria

Common Ion Effect

 

Exercise \(\PageIndex{1.a}\)

Consider a solution of 0.20 M HF, K_a= 6.8x10^-4.

\(HF\rightleftharpoons H^+ + F^-\)

1. What is the % Ionization of the pure HF solution?
2. Would you expect that adding NaF would increase or decrease the % ionization
3. What is the % Ionization if the solution was also 0.20 M NaF?

Answer a.

For pure 0.20 M HF, [HA]_i <100K_a so:

 \[ [H^+]=[F^-]=\sqrt{K_a[HA]_i} =\sqrt{(6.8x10^{-4})(0.20M)}= 0.012M \\ so \\ \%  I=\frac{[A^-]}{[HA]_i}(100)=\frac{0.012}{.2}(100)=5.8 \text{% ionized} \nonumber\] 

Answer b.

Adding fluoride will push the reaction to the left and suppress ionization

Answer c.

R	HF(aq)	⇌	H^+(aq)     +	F^-(aq)
I	0.20 M		0	0.20 M
C	-x		+x	+x
E	0.20-x		x	0.20+x

\[K_{a}=\frac{x\left ( 0.20+x \right )}{0.20-x}=\approxeq \frac{x(0.2)}{0.2}=6.8*10^{-4}\nonumber\]

\[x=\left [ H^{+} \right ]=6.8*10^{-4}\nonumber\]

\[\text{% ionization}=\frac{6.8x10^{-4}}{.2}=0.34 \text{% ionized} \nonumber \]

So the common ion reduced the % ionized from 5.8% to 0.34%

 

Buffer Solutions

Exercise \(\PageIndex{2.a}\)

The Henderson-Hasselbalch equation is

1. \(\left [ H^{+} \right ]=K_{a}+\frac{\left [salt \right ]}{\left [ acid \right ]}\)
2. \(pH=pK_{a}-log\frac{\left [salt \right ]}{\left [ acid \right ]}\)
3. \(pH=pK_{a}+log\frac{\left [ salt \right ]}{\left [ acid \right ]}\)
4. \(pH=pK_{a}+log\frac{\left [ acid \right ]}{\left [ salt \right ]}\)
5. \(pH=log\frac{\left [ acid \right ]}{\left [ salt\right ]}\)

Answer

c. \(pH=pK_{a}+log\frac{\left [ salt \right ]}{\left [ acid \right ]}\)

 

Exercise \(\PageIndex{2.b}\)

Consider a solution of 0.20 M HF and 0.20 M NaF, K_a= 6.8x10^-4 in a 1 liter container. 

\(HF\rightleftharpoons H^+ + F^-\)

1. What is the pH of the solution?
2. What is the change in pH if you add 0.02mol NaOH to one liter of water neutral?
3. What is the change in pH if you add 0.02mol NaOH to one liter of the above solution that is 0.20M HF and 0.20M NaF?

Answer a.

R	HF(aq)	⇌	H^+(aq)     +	F^-(aq)
I	0.20 M		0	0.20 M
C	-x		+x	+x
E	0.20-x		x	0.20+x

\[K_{a}=\frac{x\left ( 0.20+x \right )}{0.20-x}=\approxeq \frac{x(0.2)}{0.2}=6.8*10^{-4}\nonumber\]

\[x=\left [ H^{+} \right ]=6.8*10^{-4}\nonumber\]

\[pH=-log6.8x10^{-4}=3.17 \nonumber \]

Answer b.

pOH=-log0.02=1.70, so pH = 12.3.

so the change in pH is 12.3=7=5.3

Answer c.

The 0.02mol NaOH converts 0.02mol of HF to A^-, so after reaction

[HF]= 0.20-0.02 = 0.18M 

  and

[A^-]= 0.20+0.02M=2.02M 

\[ph=pK_a+log{F^-}{HA}=-log(6.8x10^{-4})+log\frac{2.02}{0.18} = 3.26 \nonumber \]

so the change in pH is 3.26-3.17 = 0.09

 

 

 

 

Exercise \(\PageIndex{2.c}\)

What is the ratio of HCO₃^- to H₂CO₃ in our blood that has a pH=7.4? Ka₁=4.3x10^-7

Answer

\[pH=PK_a+ log \frac{[A^-]}{[HA]} \\ \; \\ pH-pK_a=log \frac{[A^-]}{[HA]} \;\;\therefore \;\;\frac{[A^-]}{[HA]}=10^{pH-pK_a} \nonumber\]

\[pK_a=-log4.3x10^{-7}=6.37 \nonumber\]

 

\[\frac{\left [ HCO_{3}^{-} \right ]}{\left [ H_{2}CO_{3} \right ]}=10^{7.4-6.367} = 10.8 \nonumber\]

 

Exercise \(\PageIndex{2.d}\)

How many grams of NH₄Cl must be added to 1.00L of 0.10M NH₃ to form a buffer that has a pH=9.0?  Kb=1.8x10^-5 (Regardless the volume change.)

Answer

\[pOH=pKb+log\frac{salt}{base} \;\; \therefore \;\; \frac{salt}{base}=10^{pOH-pK_b} \\ \; \\ \; \; and \;[salt]=[base]10^{pOH-pK_b} \\ now \; pOH=14-pH=14-9=5 \;\; and \;\;pK_b=-log(1.8x10^{-5})= 4.74 \\ \; \\ \text{ammonium chloride is the salt and so }[NH_4Cl]= [0.1M]10^{5-4.74}=0.18M\] so 1 liter needs 0.18mol, 

\[0.18mol*53.45gNH_4Cl/mol=9.62g \nonumber \]

 

Exercise \(\PageIndex{2.e}\)

Which one of the following pairs cannot be mixed to form a buffer solution?

1. H₃PO₄, KH₂PO₄
2. KOH, HF
3. NaC₂H₃O₂, HCl
4. NH₃, NH₄Cl
5. RbOH, HBr

Answer

e. RbOH, HBr

 

Exercise \(\PageIndex{2.f}\)

Consider a solution containing 0.100M fluoride ion and 0.126M hydrogen fluoride. The concentration of hydrogen fluoride after addition of 5.00mL if 0.0100M HCl to 25.0mL of this solution is ______.

Answer

First calculate the moles of each species

\[mol_{F^{-}}= 0.100M*0.0250L=2.50*10^{-3}mol \nonumber \]

\[mol_{HF}=0.126M*0.0250L=3.15*10^{-3}mol \nonumber \]

\[mol_{HCl}=mol_{H^{+}}=0.0100M*0.00500L=5.00*10^{-5}mol \nonumber \]

Now calculate how much of each species is left over, noting all the H⁺ from the HCl reacted with the fluoride

\[x=mol_{H^{+}}=5.00*10^{-5}mol \nonumber \]

R	F-(aq)     +	H+(aq)	⇌	HF(aq)
I	2.50*10^-3 mol	5.00*10^-5 mol		3.15*10^-3 mol
C	-x	-x		+x
E	2.50*10^-3 - x	0		3.15*10^-3 + x

\[F^{-}=\left ( 2.50*10^{-3}mol \right )-\left ( 5.00*10^{-5}mol \right )=2.45*10^{-3}mol \nonumber \]

\[HF=\left ( 3.15*10^{-3}mol \right )+\left ( 5.00*10^{-5}mol \right )=3.20*10^{-3}mol \nonumber \]

To get the HF concentration you need to divide the moles by the total volume, and take into account the dilution factor.

\[\left [ HF \right ]=\frac{3.20*10^{-3}mol}{0.0250L+0.00500L}=0.107M \nonumber \]

 

Exercise \(\PageIndex{2.g}\)

Of the following, which solution has the greatest buffering capacity?

1. 0.821 M HF and 0.217 M NaF
2. 0.821 M HF and 0.909 M NaF
3. 0.100 M HF and 0.217 M NaF
4. 0.121 M HF and 0.667 M NaF
5. They are all buffer solutions and would all have the same capacity.

Answer

b. 0.821 M HF and 0.909 M NaF

 

Exercise \(\PageIndex{2.h}\)

The K_a of acetic acid is 1.7*10^-5. What is the pH of a buffer prepared by combining 50.0mL of 1.00M potassium acetate and 50.0mL of 1.00M acetic acid?

Answer

First note you by adding equal volumes that you diluted each solution in half

\[\left [K CH_{3}COO \right ]=\frac{1.00M*50.0mL}{50.0mL+50.0mL}=0.500M \nonumber \]

\[\left [ CH_{3}COOH \right ]=\frac{1.00M*50.0mL}{50.0mL+50.0mL}=0.500M \nonumber \]

Second, note how by adding equal volumes of equal concentrations of the acid and salt, that pH=pK_a.

\[pH=pK_{a}+log\frac{\left [ CH_{3}COOK \right ]}{\left [ CH_{3}COOH \right ]}=-log\left ( 1.7*10^{-5} \right )+log\frac{\cancel{0.500M}}{\cancel{0.500M}}=4.77 \nonumber \]

 

Exercise \(\PageIndex{2.i}\)

The K_b of ammonia is 1.8*10^-5. What is the pH of a buffer prepared by combining 50.0mL of 1.00M ammonia and 50.0mL of 1.00M ammonium nitrate?

Answer

Note, as you have equal mole of the base and its salt, and so pOH=pK_b=-log(1.8x10^{-5}=4.74, so pH=14-4.74=9.26

Taking into account dilution:

\[\left [ NH_{4}NO_{3} \right ]=\frac{1.00M*50mL}{50.0mL+50.0mL}=0.500M \nonumber \]

\[\left [ NH_{3} \right ]=\frac{1.00M*50mL}{50.0mL+50.0mL}=0.500M \nonumber \]

\[pOH=pK_{b}+log\frac{\left [ NH_jj{4}NO_{3} \right ]}{\left [ NH_{3} \right ]}=-log\left ( 1.8*10^{-5} \right )+log\frac{\cancel{0.500M}}{\cancel{0.500M}}=4.77 \nonumber \]

f

Exercise \(\PageIndex{2.j}\)

Calculate the pH of a solution prepared by dissolving 0.25 mol of benzoic acid (C₇H₅O₂H) and 0.15 mol of sodium benzoate (NaC₇H₅O₂) in 1.00L of solution. K_a = 6.5*10^-5 for benzoic acid.

Answer

\[pH=pK_{a}+log\frac{salt}{acid}=-log(6.5*10^{-5})+log\frac{0.150}{0.250}=4.19+(-0.22)=3.97 \nonumber \]

Exercise \(\PageIndex{2.k}\)

Determine the pH of a solution prepared by adding 0.45 mol of solid KOAc to 1.00L of 2.00M HOAc. K_a = 1.8*10^-5 of HOAc.

Answer

\[pH=pK_{a}+log\frac{\left [ CH_{3}COO^{-} \right ]}{\left [ CH_{3}COOH \right ]}=-log\left ( 1.8*10^{-5} \right )+log\frac{0.45mol}{2.00mol}=4.10 \nonumber \]

 

Exercise \(\PageIndex{2.l}\)

The pH of a solution is prepared by dissolving 0.35 mol of solid CH₃NH₃Cl (methylamine hydrochloride) in 1.00L of 1.10 M CH₃NH₂ (methylamine) is _____. The K_b for methylamine is 4.4*10^-4.

Answer

\[pOH=pK_{b}+log\frac{\left [ CH_{3}NH_{3}^{+} \right ]}{\left [ CH_{3}NH_{2} \right ]}=-log\left ( 4.4*10^{-4} \right )+log\frac{0.35mol}{1.10mol}=2.86 \nonumber \]

\[pH=14-pOH=14-2.86=11.14 \nonumber \]

 

Exercise \(\PageIndex{2.m}\)

Which of the following substances, when added to a solution of hydrofluoric acid, could be used to prepare a buffer solution?

1. HCl
2. NaNO₃
3. NaF
4. NaCl
5. NaBr

 

Answer

c. NaF

 

Exercise \(\PageIndex{2.n}\)

Which of the following could not be added to a solution of sodium acetate to prepare a buffer?

1. sodium acetate
2. acetic acid
3. hydrochloric acid
4. nitric acid
5. more than one of these answers is correct

Answer

a. sodium acetate

 

Exercise \(\PageIndex{2.o}\)

Calculate the fluoride ion concentration (in M) in a 1.0L aqueous solution containing 0.40 mol of HF and 0.10 mol of HCl. (Ka for HF = 6.8*10^-4).

Answer

R	HF(aq)	⇌	H^+(aq)     +	F^-(aq)
I	0.40 mol		0.10 mol	0
C	-x		+x	+x
E	0.40-x		0.10+x	x

\[K_{a}=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ HF \right ]}=\frac{\left [ 0.10+x \right ]\left [ x \right ]}{\left [ 0.40-x \right ]}=6.8*10^{-4}\nonumber\]

\[\frac{0.10(x)}{0.40}=6.8*10^{-4}\nonumber\]

\[x=\frac{0.40}{0.10}*6.8*10^{-4}=2.7*10^{-3}M\nonumber\]

 

Exercise \(\PageIndex{2.p}\)

Calculate the pH of 1.0L aqueous solution containing 0.30mol of HF and 0.10mol of HCl. (K_a for HF = 6.8*10^-4)

Answer

R	HF(aq)	⇌	H^+(aq)     +	F^-(aq)
I	0.30 mol		0.10 mol	0
C	-x		+x	+x
E	0.30-x		0.10+x	x

\[K_{a}=\frac{\left [ H^{+} \right ]\left [ F^{-} \right ]}{\left [ HF \right ]}=\frac{\left [ 0.10+x \right ]\left [ x \right ]}{\left [ 0.30-x \right ]}=6.8*10^{-4}\nonumber\]

\[\frac{0.10(x)}{0.43}=6.8*10^{-4}\nonumber\]

\[x=\frac{0.40}{0.10}*6.8*10^{-4}=2.7*10^{-3}M\nonumber\]

\[pH=-log\left [ H^{+} \right ]=-log\left [ 0.10+2.7*10^{-3}M \right ]=0.99 \nonumber \]

 

Acid-Base Titrations

Textbook: Section 17.3

Exercise \(\PageIndex{3.a}\)

How many milliliters of 0.0750M NaOH are required to titrate 50.0ml of 0.025M HCl?

Answer

\[OH^{-}(aq)+H^{+}\rightleftharpoons H_{2}O(l) \nonumber \]

\[\frac{50.0mL*0.025M}{0.075M}=16.7mL \nonumber \]

 

Exercise \(\PageIndex{3.b}\)

How many milliliters of the same NaOH are needed to titrate 25.0ml of HCl solution that has 1.85g of HCl per liter?

Answer

\[OH^{-}(aq)+H^{+}\rightleftharpoons H_{2}O(l) \nonumber \]

\[\frac{\frac{1.85g/L}{36.45g/mol}*25.0mL}{0.075M}=16.9mL \nonumber \]

 

Exercise \(\PageIndex{3.c}\)

How many milliliters of the same NaOH are needed to titrate 20.0ml of 0.050M HBr?

Answer

\[OH^{-}(aq)+HBr\rightleftharpoons H_{2}O(l)+Br^{-}(aq) \nonumber \]

\[\frac{0.050M*20.0mL}{0.075M}=13.3mL \nonumber \]

 

Exercise \(\PageIndex{3.d}\)

Calculate the pH of the solution formed when 20.0ml of 0.100M NaOH is added to 40.0ml of 0.100M of HC₂H₃O_2. Ka=1.8x10^-5

Answer

You are neutralizing a weak acid (acetic acid) with a strong base (NaOH). You first need to find the equivalence point and determine if they are in stoichiometric proportions, or if one is in excess.  This in effect tells you which region of the titration curve you are in. The vol. base required to neutralize all the acid is

\[M_aV_a=M_bV_b \; \therefore \; V_b=V_a(\frac{M_a}{M_b})= 40.0mL(\frac{1.0M}{1.0M})= 40.0mL\]

This makes total sense as you have equal concentrations, so it will take equal volumes to neutralize each other.

You have added half the amount of base required to neutralize the weak acid, and so you are at a half titer, and the Henderson Hasselbach equation becomes
pH = pK_a, as half of the weak acid has been neutralized, so

\[pH=-log\left ( 1.8*10^{-5}M \right )=4.74 \nonumber \]

Of course you could have used eq. 17.3.8, noting that the ration of the salt to the weak acid is the ratio of the volume of strong base added to the volume still needed to get to the equivalence point:

\[ pH = pK_a + log{V_b}{V_{eq}-V_b}= pK_a+ log\frac{20}{40-20}= pK_a+ log\frac{20}{20}=pK_a=-log\left ( 1.8*10^{-5}M \right )=4.74 \nonumber \]

or if you want to spend a real lot of time:

\[ [HA] = \frac{n_{HA_{initial}}-n_{HA_{consumed}}}{V_{total}} =\frac{M_aV_a(initial)-M_bV_b}{V_a+V_b}  \; \;  and [A^-]=\frac{M_bV_b}{V_a+V_b} \nonumber \] 

Then the Henderson Hasselbach eq becomes

\[ pH = pK_a + log\frac{\frac{M_bV_b}{V_a+V_b}}{\frac{M_aV_a(initial)-M_bV_b}{V_a+V_b} } = -log\left ( 1.8*10^{-5}M \right )+ log\frac {\frac{0.1M*0.400l-0.1M*0.20l}{0.40l+0.20l}}{\frac{0.1M*0.20l}{0.40l+0.20l}} \\ \; \\  -log\left ( 1.8*10^{-5}M \right )+ log\frac {\frac{0.04molel-0.02mol}{0.60l}}{\frac{0.02mol}{0.60l}}  \\ \; \\   = -log\left ( 1.8*10^{-5}M \right )+ log\frac {\frac{0.02mol}{0.60l}}{\frac{0.02mol}{0.60l}}=  -log\left ( 1.8*10^{-5}M \right ) =4.74\]

 

 

 

Exercise \(\PageIndex{3.e}\)

Calculate the pH at the equivalence point in the titration of 50.0ml of 0.20M HC₂H₃O₂ with 0.05M NaOH. Ka=1.8x10^-5

Answer

All of the weak acid has been neutralized to its salt, and so you need to calculate it's concentration after dilution.

\[\frac{50.0mL*0.20M}{0.05}M=200.0mL \nonumber \]

\[\left [ C_{2}H_{3}O_{2}^{-} \right ]=\frac{0.20M*0.050L}{\left ( 0.050+0.200 \right )L}=0.040M \nonumber \]

\[C_{2}H_{3}O_{2}^{-}(aq)+H_{2}O(l)\rightleftharpoons HC_{2}H_{3}O_{2}(aq)+OH^{-}(aq) \nonumber \]

Now calculate K_b^' (of the conjugate base, the salt)

\[K_{b}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{1.8*10^{-5}}=5.56x10^{-10}  \]

Since \([C_{2}H_{3}O_{2}^{-}] > 100K_b, \; \; pOH=-log(\sqrt{K_b'[A^-]})\)

 

\[pOH=-log(\sqrt{5.56x10^{-10}(0.040M)}= -log(4.716x10^{-6})=5.33 \nonumber \]

\[pH=14-5.33=8.67\]

 

To solve in two steps:

\[pOH = -log\sqrt{K_b'[A^-]}=-log\sqrt{\frac{K_w}{K_a}(\frac{M_aV_a}{V_a+V_b})} \\ \; \\ pOH=-log\sqrt{(\frac{1x10^{-14}}{1.8x10^{-5}})( \frac{(0.20M)0.050l}{0.050l+0.200l})}=5.33 \\ \; \\ pH=14-pOH=14-5.33 = 8.67\]

 

 

 

Exercise \(\PageIndex{3.f}\)

Determine the pH when 50.0ml of 0.010M NaOH is added to 40.0ml of 0.010M HC₂H₃O₂.

Answer

First, calculate the volume of base required to neutralize the acid, the equivalence point.

\[M_bV_b=M_aV_a \therefore V_B=V_a(\frac{M_a}{M_b}) = 40mL(\frac{0.10M}{0.10M}) = 40 mL\]

Since the volume added is greater than the volume needed to neutralize the acid we are in the excess base region.

\[ [OH^-]_{excess} = \frac {M_b(V_b(total)-V_b(equiv))}{V_b+V_a}=\frac{0.010M(50.0ml-40.0ml)}{50.0+40.0}=0.0011M\]

pOH=-log0.0011=2.95

pH=14-pOH=14-2.95-11.05

 

 

Exercise \(\PageIndex{3.g}\)

Consider the titration of 25.0mL of 0.723M HClO₄ with 0.273M KOH.

1. What is the H₃O⁺ concentration before any KOH is added?
2. What is the H₃O⁺ concentration after addition of 10.0mL of KOH?
3. What is the H₃O⁺ concentration after addition of 66.2mL of KOH?
4. What is the H₃O⁺ concentration after addition of 80.0mL of KOH?

Answer a.

\[\left [ H_{3}O^{+} \right ]=\left [ HClO_{4} \right ]=0.723mol \nonumber \]

Answer b.

\[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \]

\[moles\,KOH=0.273M*0.0100L=0.00273mol \nonumber \]

\[moles\,H_{3}O^{+}=moles\,HClO_{4}-moles\,KOH=0.018075mol-0.00273mol=0.015345mol \nonumber \]

\[\left [ H_{3}O^{+} \right ]=\frac{0.015345mol}{0.0250L+0.0100L}=0.438M \nonumber \]

Answer c.

\[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \]

\[moles\,KOH=0.273M*0.0662L=0.0180725mol \nonumber \]

The solution is neutral because the acid and base have the same amount of moles. So the concentration of H₃O⁺ is 1.00*10^-7.

Answer d.

\[moles\,HClO_{4}=0.723M*0.0250L=0.018075mol \nonumber \]

\[moles\,KOH=0.273M*0.0800L=0.02184mol \nonumber \]

\[excess\,moles=moles\,KOH-moles\,HClO_{4}=0.02184mol-0.018075mol=0.003765mol \nonumber \]

\[\left [ OH^{-} \right ]=\frac{0.003765mol}{0.0800L+0.0250L}=0.0359M \nonumber \]

\[\left [ H_{3}O^{+} \right ]=\frac{10^{-14}}{\left [ OH^{-} \right ]}=\frac{10^{-14}}{0.0359M}=2.79*10^{-13}M \nonumber \]

 

Exercise \(\PageIndex{3.h}\)

A _____ yields a titration curve with an initial pH of 1.00, an equivalence point at pH 7.00, and a relatively long, nearly vertical middle section.

1. strong acid titrated by a strong base
2. strong base titrated by a strong acid
3. weak acid titrated by a strong base
4. weak base titrated by a strong acid
5. weak base titrated by a weak acid

Answer

a. strong acid titrated by a strong base

 

Exercise \(\PageIndex{3.i}\)

An initial pH of 13.00, an equivalence point at pH 7.00 and a relatively long, nearly vertical middle section corresponds to a titration curve for _____.

1. strong acid titrated by a strong base
2. strong base titrated by strong acid
3. weak acid titrated by strong acid
4. weak base titrated by strong acid
5. weak base titrated by weak acid

Answer

b. strong base titrated by strong acid

 

Exercise \(\PageIndex{3.j}\)

The pH of a solution prepared by mixing 45mL of 0.183M KOH with a 65mL of 0 .145M HCl is_____?

Answer

These are both strong

\[moles\,KOH=0.183*0.045L=0.008235mol \nonumber \]

\[moles\,HCl=0.145*0.065L=0.009425mol \nonumber \]

There is more acid than base, so pH is determined by the excess acid

\[moles\,H^{+}_{excess}=moles\,HCl-moles\,KOH=0.009425mol-0.008235mol=0.00119mol \nonumber \]

\[\left [ H^{+} \right ]=\frac{0.00119mol}{0.045+0.065}=0.01082M \nonumber \]

\[pH=-log\left [ H^{+} \right ]=-log(0.01082M)=1.97 \nonumber \]

 

Titration of Weak Acid by Strong Base

Exercise \(\PageIndex{3.1.a}\)

25mL of 0.15 M CH₃COOH        K_a = 1.8*10^-5

titrated with 0.15M NaOH

What is the pH at V_B (volume of base) = 0?

Answer

\[pH=-log\left [ H^{+} \right ]=-log\sqrt{K_{a}\left [ HA \right ]_{i}}\nonumber \]

\[pH=-log\sqrt{(1.8*10^{-5})*0.15M}=-log(0.0016431677)=2.78\nonumber \]

Exercise \(\PageIndex{3.1.b}\)

What is the pH if 25mL of 0.15 M CH₃COOH (K_a = 1.8*10^-5) is titrated with 10.0mL of 0.15M NaOH

Answer

First calculate the volume to reach the equivlalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\]

We are in the buffer region, and have neutralized 10/25ths, and have 15/25ths to go.so 

 

\[pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}= -log1.8*10^{-5} + log{10}{15}=4.47 \nonumber \]

 

Exercise \(\PageIndex{3.1.c}\)

25mL of 0.15 M CH₃COOH        K_a = 1.8*10^-5

titrated with 0.15M NaOH

What is the pH at V_B at half equivalence (12.5mL)?

Answer

\[pH=pKa\nonumber \]

\[pH=-log\left ( 1.8*10^{-5} \right )=4.74\nonumber \]

Exercise \(\PageIndex{3.1.d}\)

What is the pH if 25mL of 0.15 M CH₃COOH (K_a = 1.8*10^-5) titrated with 15 mL of 0.15M NaOH

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\]

We are in the buffer region, and have neutralized 15/25ths, and have 10/25ths to go.so 

 

\[pH=pK_{a}+log\frac{\left [ A^{-} \right ]}{\left [ HA \right ]}= -log1.8*10^{-5} + log{15}{10}=4.92 \nonumber \]

Exercise \(\PageIndex{3.1.e}\)

What is the pH at the equivalence point if 25mL of 0.15 M CH₃COOH (K_a = 1.8*10^-5) is titrated with 0.15M NaOH

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\]

\[\left [ OH^{-} \right ]=\sqrt{K_{b}'\left [ A^{-} \right ]} = \sqrt{\frac{K_w}{K_a}[A^-]} \\ \; \\ [OH^-]=\sqrt{\frac{K_w}{K_a}[\frac{M_bV_b}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{1.8x10^{-5}}[\frac{(0.15M)(25ml}{25ml+25ml}]} =6.45x10^{-6} \nonumber \]

\[pOH=-log[OH^-]=-log6.45x10^{-6} =5.19 \\ \; \\ pH=14-pOH=14-5.19=8.81 \]

Exercise \(\PageIndex{3.1.f}\)

What is the pH when 25mL of 0.15 M CH₃COOH (K_a = 1.8*10^-5) is titrated with 30.0 mL of0.15M NaOH

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 25mL\frac{0.15M}{0.15M}=25mL\]

So there are 5 mL of excess base

\[[OH^-]=\frac{(0.15M)(5mL)}{25 mL + 30 mL} = 0.0136\]

\[pH=14-pOH=14-(-log(0.0136))=12.13

 

Exercise \(\PageIndex{3.1.g}\)

What is the molarity of an HOAc solution if 25.5mL of this solution required 37.5mL of 0.175M NaOH to reach the equivalence point?

Answer

\[M_{a}V_{a}=M_{b}V_{b}\nonumber \]

\[M_{a}=\frac{M_{b}V_{b}}{V_{a}}=\frac{0.175M*0.0375L}{0.0255L}=0.257M\nonumber \]

Exercise \(\PageIndex{3.1.h}\)

Consider an experiment where 35.0mL of 0.175M HOAc is titrated when 0.25M in NaOH. What is the pH at equivalence point for this titration? The Ka for HOAc is 1.8 * 10^-5.

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 35mL\frac{0.175M}{0.25M}=24.5mL\]

\[\left [ OH^{-} \right ]=\sqrt{K_{b}'\left [ A^{-} \right ]} = \sqrt{\frac{K_w}{K_a}[A^-]} \\ \; \\ [OH^-]=\sqrt{\frac{K_w}{K_a}[\frac{M_bV_b}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{1.8x10^{-5}}[\frac{(0.175M)(35ml)}{24.5ml+35ml}]} =7.56x10^{-6} \nonumber \]

\[pOH=-log[OH^-]=-log7.56x10^{-6} =5.12 \\ \; \\ pH=14-pOH=14-5.12=8.88

 

 

Exercise \(\PageIndex{3.1.i}\)

50.50mL of 0.116 M HF is titrated with 0.1200 M NaOH. (Ka for HF is 6.8 * 10^-4)

1. How many mL of the base are required to reach the equivalence point?
2. What is the pH after 50.50mL of base has been added?

Answer a.

\[M_{1}V_{1}=M_{2}V_{2}\nonumber \]

\[V_{2}=\frac{M_{1}V_{1}}{M_{2}}=\frac{0.116M*0.0505L}{0.1200M}=48.82mL\nonumber \]

Answer b.

The amount of excess base is 50.50mL-48.82mL=1.68mL

\[ [OH^-] = \frac{M_bV_b(excess)}{V_T}= \frac{0.1200M(1.68mL)}{50.50mL+50.50mL}=0.0019M \nonumber\]

\[pOH=-log0.0019=2.70 \;\; pH=14-pOH=14-2.7=11.300\]

Exercise \(\PageIndex{3.1.j}\)

65.50mL of 0.1610 M HF is titrated where 0.1200 M NaOH. What is the pH at the equivalence point? (Ka for HF is 6.8 * 10^-4)

Answer

First calculate the volume to reach the equivalence point.

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 65.50mL\frac{0.1610M}{0.1200M}=87.88mL\]

\[\left [ OH^{-} \right ]=\sqrt{K_{b}'\left [ A^{-} \right ]} = \sqrt{\frac{K_w}{K_a}[A^-]} \\ \; \\ [OH^-]=\sqrt{\frac{K_w}{K_a}[\frac{M_bV_b}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{6.8x10^{-4}}[\frac{(0.1200M)(87.88ml)}{65.50ml+87.88ml}]} =1.0x10^{-6} \nonumber \]

\[pOH=-log[OH^-]=-log1.00x10^{-6} =6.00 \\ \; \\ pH=14-pOH=14-6=8.00

 

Titration of Weak Base with Strong Acid

Exercise \(\PageIndex{3.2.a}\)

What is the pH if 25mL of 0.15M NH₃ (K_b = 1.8*10^-5) is titrated with 0.00 mL of 0.15M HCl

What is the pH at V_A (volume of acid) = 0?

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 65.50mL\frac{0.1610M}{0.1200M}=87.88mL\]

\[\left [ OH^{-} \right ]=\sqrt{K_{b}*\left [ NH_{3} \right ]}=\sqrt{1.8*10^{-5}*\left [ 0.15 \right ]}\nonumber\]

\[pOH=-log\left ( \sqrt{1.8*10^{-5}*\left [ 0.15 \right ]} \right )=2.78\nonumber\]

\[pH=14-pOH=14-2.78=11.22\nonumber\]

Exercise \(\PageIndex{3.s.b}\)

What is the pH if 25mL of 0.15M NH₃ (K_b = 1.8*10^-5) is titrated with 10 mL of 0.15M HCl?

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\]

You are in the buffer region, so use the Henderson Hasselbach eq. of a base titrated with an acid. You have titrated 10/25ths of the base and have 15/25ths to go, os the ratio of salt to base is 10 to 15 or 2 to 3

\[pOH=pK_{b}+log\frac{base}{salt}\nonumber\]

\[pOH=-log(1.8*10^{-5})+log\frac{2}{3} = 4.74-0.176=4.57\nonumber\]

\[pH=14-pOH=14-4.57=9.43\nonumber\]

Exercise \(\PageIndex{3.2.c}\)

What is the pH when 25mL of 0.15M NH₃ (K_b = 1.8*10^-5) titrated with 12.5 mL of 0.15M HCl

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\]

You are at half equivalence, so [B]=[BH⁺] (you have turned have the base to its conjugate acid) and pOH=pK_b.

\[pOH=pK_{b}\nonumber\]

\[pOH=-log(1.8*10^{-5})=4.74\nonumber\]

\[pH=14-pOH=14-4.74=9.26\nonumber\]

Exercise \(\PageIndex{3.2.d}\)

What is the pH if 25mL of 0.15M NH₃ (K_b = 1.8*10^-5) is titrated with 15 mL of 0.15M HCl

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\]

You are in the buffer region, so use the Henderson Hasselbach eq. of a base titrated with an acid. You have titrated 15/25ths of the base and have 10/25ths to go, os the ratio of salt to base is 15 to 10 or 3 to 2

\[pOH=pK_{b}+log\frac{base}{salt}\nonumber\]

\[pOH=-log(1.8*10^{-5})+log\frac{3}{2} = 4.74+0.176=4.92\nonumber\]

\[pH=14-pOH=14-4.92=9.08\nonumber\]

Exercise \(\PageIndex{3.2.e}\)

What is the pH when 25mL of 0.15M NH₃ (K_b = 1.8*10^-5) titrated with 25 mL of 0.15M HCl?

Answer

You have neutralized all the base and converted it to its salt

 

\[ [H^+]=\sqrt{K_{a}'\left [ B \right ]} = \sqrt{\frac{K_w}{K_b}[B]} \\ \; \\ [H^+]=\sqrt{\frac{K_w}{K_b}[\frac{M_aV_a}{V_a+V_b}]} =\sqrt{\frac{10^{-14}}{1.8x10^{-5}}[\frac{(0.15M)(25ml)}{25ml+25ml}]} =6.45x10^{-6} \nonumber \]

pH=-log[H⁺]=-log(6.45x10^-6)=5.19

Exercise \(\PageIndex{3.2.f}\)

What is the pH when 25m.0L of 0.15M NH₃ (K_b = 1.8*10^-5) is titrated with 30.0 mL of0.15M HCl?

Answer

First calculate the volume to reach the equivalence point and then determine which region of the titration curve you are in

\[M_aV_a=M_bV_b  \therefore V_b(equiv)= V_a\frac{M_a}{M_b}= 250mL\frac{0.15M}{0.15M}=25mL\]

You are in the excess acid range and have added 5 mL more than is needed to neutralize the base

\[ [H^+]=\frac{\text{moles acid excess}}{V_T} \nonumber \]

\[pH=-log\frac{0.0050(0.15)}{0.025+0.03}=1.87\nonumber\]

 

Solubility of Salts

Textbook: Section 17.4

Exercise \(\PageIndex{4.a}\)

In which of the following aqueous solutions would you expect AgCl to have the lowest solubility?

1. pure water
2. 0.020 M BaCl₂
3. 0.015 M NaCl
4. 0.020 M AgNO₃
5. 0.020 M KCl

Answer

b. 0.020 M BaCl₂, note the chloride salts have a common ion with the silver chloride equilibrium (AgCl(s) <-> Ag^+(aq) + Cl^-(aq)), and so push it to the left (towards the precipitate).  This is the common ion effect and it is inhibiting the ionization.  Since BaCl₂ provides the most chloride, it pushes it to the left the most

Exercise \(\PageIndex{4.b}\)

Given the following table K_sp values, determine which compound listed has the greatest solubility.

Compound	Ksp
CdCO3	5.2*10^-12
Cd(OH)2	2.5*10^-14
AgI	8.3*10^-17
Fe(OH)3	4.0*10^-38
ZnCO3	1.4*10^-11

1. CdCO₃
2. Cd(OH)₂
3. AgI
4. Fe(OH)₃
5. ZnCO₃

Answer

b. Cd(OH)₂

note: options (a), (c) and (e) are the same formula (x²=K_sp) and (e) would be the most soluble because it has the largest K, so you only need to check (b), (d) and (e)

e) \[x^2=K_{sp} \therefore x=\sqrt{K_{sp}}=\sqrt{1.4x10^{-11}}=3.74x10^{-6} \nonumber\]

b) \[x(2x)^2=4X^3=K_{sp} \therefore x= \left ( \frac{K_{sp}}{4} \right )^\frac{1}{3} = \left ( \frac{2.5x10^{-14}}{4} \right )^\frac{1}{3} = 1.84x10^{-5} \nonumber\]

d) \[x(3x)^3=27x^4=K_{sp} \therefore x= \left ( \frac{K_{sp}}{27} \right )^\frac{1}{4} = \left ( \frac{4.0x10^{-38}}{27} \right )^\frac{1}{4} =1.96x10^{-10}\]

Exercise \(\PageIndex{4.c}\)

The solubility of which one of the following will not be affected by the pH of the solution?

1. Na₃PO₄
2. NaF
3. KNO₃
4. AlCl₃
5. MnS

Answer

c. KNO₃ , it is the salt of a strong acid (HNO₃) and a strong base (KOH). Phosphate, fluoride and hydrogen sulfide are all basic salts (look at their K_a's), Al is a Lewis acid.

Exercise \(\PageIndex{4.d}\)

Write the expression relating solubility (x) to K_sp for silver sulfide.

Answer

\[Ag_{2}S(s)\rightleftharpoons 2Ag^{+}(aq)+S^{2-}(aq)\nonumber\]

\[K_{sp}=\left [ Ag^{+} \right ]^{2}\left [ S^{2-} \right ]=(2x)^{2}(x)=4x^{3}\nonumber\]

\[x=(\frac{K_{sp}}{4})^{\frac{1}{3}}\nonumber\]

Exercise \(\PageIndex{4.e}\)

In which aqueous system is PbI₂ least soluble?

1. H₂O
2. 0.5 M HI
3. 0.2 M HI
4. 1.0 M HNO₃
5. 0.8 M KI

Answer

e. 0.8 M KI, this is the common ion effect, you are looking at the equilibrium of PbI₂ and adding a soluble iodide salt or acid (anything that adds iodide) will inhibit ionization.  The more iodide you add the greater the effect, and (e) adds the most

Exercise \(\PageIndex{4.f}\)

Of the substances below, _____ will decrease the solubility of Pb(OH)₂ in a saturated solution.

1. NaNO₃
2. H₂O₂
3. HNO₃
4. NaOH
5. NaCl

Answer

d. NaOH, the common hydroxide inhibits the ionization and causes the lead(II)hydroxide to precipitate out

 

Exercise \(\PageIndex{4.g}\)

What is the Pb⁺² concentration for a saturated solution of PbSO₄?  K_sp=6.3x10^-7

Answer

\[K_{sp}=\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=\left [ x \right ]\left [ x \right ]=x^{2}\nonumber\]

\[x=\sqrt{K_{sp}}=\sqrt{6.3*10^{-7}}=7.94*10^{-4}M\nonumber\]

 

Exercise \(\PageIndex{4.h}\)

What is the Pb⁺² concentration if 0.10mol of Na₂SO₄ is added to 1L of a saturated solution of PbSO₄?

Answer

\[K_{sp}=\left [ Pb^{2+} \right ]\left [ SO_{4}^{2-} \right ]=\left [ x \right ]\left [ 0.10M \right ]\nonumber\]

\[x=\frac{K_{sp}}{\left [ 0.10M \right ]}=\frac{6.3*10^{-7}}{\left [ 0.10M \right ]}=6.3*10^{-6}M\nonumber\]

Exercise \(\PageIndex{4.i}\)

What is the solubility of Ca₃(PO₄)₂?  K_sp=2.0x10^-29

Answer

\[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\]

\[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\]

\[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\]

\[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\]

 

Exercise \(\PageIndex{4.j}\)

What is the Ca²⁺ concentration for a saturated solution of Ca₃(PO₄)₂?

Answer

\[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\]

\[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\]

\[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\]

\[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\]

\[\left [ Ca^{2+} \right ]=3*(7.1*10^{-7}M)=2.1*10^{-6}M\nonumber\]

 

Exercise \(\PageIndex{4.k}\)

What is the PO₄^3- concentration for a saturated solution of Ca₃(PO₄)₂?

Answer

\[Ca_{3}\left ( PO_{4}\right )_{2}(aq)\rightarrow 3Ca^{2+}(aq)+2PO_{4}^{3-}(aq)\nonumber\]

\[K_{sp}=\left [ Ca^{2+} \right ]^{3}\left [ PO_{4}^{3-} \right ]^{2}=\left [ 3x \right ]^{3}\left [ 2x \right ]^{2}=x^{5}(27)(4)\nonumber\]

\[x^{5}=\frac{K_{sp}}{108} x=\left ( \frac{K_{sp}}{108} \right )^{\frac{1}{5}}\nonumber\]

\[x=\left ( \frac{2.0*10^{-29}}{108} \right )^{\frac{1}{5}}=7.1*10^{-7}M\nonumber\]

\[\left [ PO_{4}^{3-} \right ]=2*(7.1*10^{-7}M)=1.4*10^{-6}M\nonumber\]

 

Exercise \(\PageIndex{4.l}\)

What is the K_sp for MgF₂ if F^-=0.00236M?

Answer

\[MgF_{2}(aq)\rightarrow Mg^{+2}(aq)+2F^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Mg^{+2} \right ]\left [ F^{-} \right ]^{2}=\left [ x \right ]\left [ 2x \right ]^{2}\nonumber\]

\[[F^-]=2x=0.00236M\nonumber\]

\[[Mg+2]=x=\frac{0.00236M}{2}=0.00118M\nonumber\]

\[K_{sp}=\left [ 0.00118M \right ]\left [ 0.00236 \right ]^{2}=6.6*10^{-9}\nonumber\]

Exercise \(\PageIndex{4.m}\)

If the molar solubility of CaF₂ at 25°C is 1.25x10^-3mol/L, what is the K_sp at this temperature?

Answer

\[CaF_{2}(s)\rightleftharpoons Ca^{2+}(aq)+2F^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Ca^{2+} \right ]\left [ F^{-} \right ]^{2}=\left [ X \right ]\left [ 2X \right ]^{2}\nonumber\]

\[K_{sp}=\left [ X \right ]4\left [ X \right ]^{2}=4\left [ X \right ]^{3}=4\left [ 1.25*10^{-5} \right ]^{3}=7.81*10^{-9}\nonumber\]

 

Exercise \(\PageIndex{4.n}\)

A saturated solution of NaF contains 4.0 g of salt in 100.0ml of water at 15°C, what is the solubility product for NaF?

Answer

\[\frac{\frac{4.0g}{42g/mol}}{0.10L}=0.952M\nonumber\]

\[NaF(s)\rightleftharpoons Na^{+}(aq)+F^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Na^{+} \right ]\left [ F^{-} \right ]=0.952^{2}=0.91\nonumber\]

Note: sodium fluoride is a soluble salt and we usually do not use solubilty constants for soluble salts.

 

Exercise \(\PageIndex{4.o}\)

The K_sp for AgBr is 5.0x10^-13 at 25°C, what is the molar solubility of AgBr?

Answer

\[AgBr(s)\rightleftharpoons Ag^{+}(aq)+Br^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Ag^{+} \right ]\left [ Br^{-} \right ]=5.0*10^{-13}\nonumber\]

\[x^{2}=5.0*10^{-13}\nonumber\]

\[x=7.1*10^{-7}M\nonumber\]

 

Exercise \(\PageIndex{4.p}\)

Calculate the molar solubility of AgBr in 0.10M NaBr solution?

Answer

R	AgBr(s)	⇌	Ag^+(aq)     +	Br^-(aq)
I	---		0	0.10M
C	---		x	x
E	---		x	.10+x

\[K_{sp}=\left [ Ag^{+} \right ]\left [ Br^{-} \right ]=5.0*10^{-13}=\left(0.10+x\right)x \approxeq (0.10)x\nonumber\]

\[x=5.0*10^{-12}\nonumber\]

 

Exercise \(\PageIndex{4.q}\)

Calculate the solubility of Mn(OH)₂ in grams per liter when buffered at pH = 9.50? K_sp= 5.0x10^-13

Answer

\[\left [ OH^{-} \right ]=10^{-\left ( 14-9.5 \right )}=3.16*10^{-5}\nonumber\]

\[Mn(OH)_{2}(s)\rightleftharpoons Mn^{2+}(aq)+2OH^{-}(aq)\nonumber = x(2x)^2\]

\[Mn^{+2}=x \; , \, OH^-=2x\]

\[K_{sp}=x[OH^-]^2 \; \therefore \;  x=\frac{K_{sp}}{[OH^-]^2}=\frac{5.0x10^{-13}}{[3.16x10^{-5}]^2}=5.07x10^{-4}M \nonumber \]

\[ S=5.07x10^{-4}\frac{mol}{l} \left(  \frac{88.95g}{mol}\right) = 0.045g/l \nonumber \]

Exercise \(\PageIndex{4.r}\)

What is the molar solubility of MgC₂O₄? (K_sp for MgC₂O₄ is 8.6*10^-5)

Answer

\[MgC_{2}O_{4}\rightleftharpoons Mg^{2+}+C_{2}O_{4}^{2-}\nonumber\]

\[K_{sp}=\left [ Mg^{2+} \right ]\left [ C_{2}O_{4}^{2-} \right ]\nonumber\]

\[K_{sp}=(s)*(s)=s^2\nonumber\]

\[s^{2}=K_{sp}\nonumber\]

\[s=\sqrt{K_{sp}}=\sqrt{8.6*10^{-5}}=9.27*10^{-3}mol/L\nonumber\]

 

Exercise \(\PageIndex{4.s}\)

Calculate the concentration (in M) of iodine ions in a saturated solution of lead(II) iodine (K_sp = 1.39*10^-8)

Answer

\[PbI_{2}(s)\rightleftharpoons Pb^{2+}(aq)+2I^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Pb^{2+} \right ]\left [ I^{-} \right ]^{2}\nonumber\]

\[1.39*10^{-8}=x*(2x)^{2}\nonumber\]

\[1.39*10^{-8}=4x^{3}\nonumber\]

\[x^{3}=3.475*10^{-9}\nonumber\]

\[x=(3.475*10^{-9})^{1/3}=1.515*10^{-3}\nonumber\]

\[\left [ I^{-} \right ]=2x=2*1.515*10^{-3}=3.03*10^{-3}\nonumber\]

 

Exercise \(\PageIndex{4.t}\)

Calculate the molar solubility of silver carbonate (K_sp = 6.15*10^-12)

Answer

\[Ag_{2}CO_{3}(s)\rightleftharpoons 2Ag^{+}(aq)+CO_{3}^{2-}(aq)\nonumber\]

\[K_{sp}=\left [ Ag^{+} \right ]^{2}\left [ CO_{3}^{2-} \right ]=(2x)^{2}*x\nonumber\]

\[K_{sp}=(2x)^{2}*x\nonumber\]

\[K_{sp}=4x^{3}\nonumber\]

\[x^3=\frac{K_{sp}}{4} \nonumber \]

\[x=\left( \frac{K_{sp}}{4} \right)^{1/3}  =\left( \frac{6.15*10^{-12}}{4} \right)^{1/3} = 1.15*10^{-4}\nonumber \]

 

Exercise \(\PageIndex{4.u}\)

What is the bromide concentration (in M) in a saturated solutions of mercury(II) bromide. K_sp = 8.0*10^-20.

Answer

\[HgBr_{2}(s)\rightleftharpoons Hg^{2+}(aq)+2Br^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Hg^{2+} \right ]\left [ Br^{-} \right ]^{2}\nonumber\]

\[ [Hc^{+2}]=x, \; \; [Br^-]=2x \nonumber \]

\[K_{sp}=x*(2x)^{2}\nonumber\]

\[K_{sp}=4x^{3}\nonumber\]

\[x^3=\frac{K_{sp}}{4} \nonumber \]

\[x=\left( \frac{K_{sp}}{4} \right)^{1/3}  =\left( \frac{8.0*10^{-20}}{4} \right)^{1/3} = 2.71*10^{-7}\nonumber \]

\[\left [ Br^{-} \right ]=2s=2*(2.71*10^{-7})=5.4*10^{-7}M\nonumber\]

 

Exercise \(\PageIndex{4.v}\)

The solubility of AuCl₃ (as Au³⁺ and Cl^-) in water at 298K is 3.3*10^-7 M. K_sp for AuCl₃ is _____.

Answer

\[AuCl_{3}(s)\rightleftharpoons Au^{3+}(aq)+3Cl^{-}(aq)\nonumber\]

\[K_{sp}=\left [ Au^{3+} \right ]\left [ Cl^{-} \right ]^{3}\nonumber\]

\[K_{sp}=x(3x)^3=27x^4, \;\;\; and \;\;\; x=3.37x10^{-7} \nonumber \]

\[K_{sp}=27\left( 3.37x10^{-7} \right)^4=3.2*10^{-25} \nonumber \]

 

Exercise \(\PageIndex{4.w}\)

What is the molar solubility of PbS? (K_sp (PbS) = 8.0*10^-28)

Answer

\[PbS(s)\rightleftharpoons Pb^{+}(aq)+S^{-}(aq) \nonumber\]

\[K_{sp}=\left [ Pb^{+} \right ]\left [ S^{-} \right ] \nonumber\]

\[K_{sp}=x^2 \nonumber\]

\[x=\sqrt{K_{sp}}=\sqrt{8x10^{-28}}= 2.8*10^{-14}\nonumber\]

 

Exercise \(\PageIndex{4.x}\)

Calculate the maximum concentration (in M) of lead ions in a solution containing 0.181M of sulfide. The K_sp for lead sulfide is 3.4*10^-28.

Answer

\[PbS(s)\rightleftharpoons Pb^{+2}(aq)+S^{-2}(aq) \nonumber\]

\[K_{sp}=\left [ Pb^{+2} \right ]\left [ S^{-2} \right ]  =x^2\nonumber\]

\[ Pb^{+2}]=[S^{-2}]=x \nonumber \]

\[x=\frac{K_{sp}}{[S^{-2}]}=\frac{3.4*10^{-28}}{0.181}=1.9*10^{-27}M\nonumber\]

 

Exercise \(\PageIndex{4.y}\)

The last two problems dealt with the concentration of lead in a saturated solution of lead sulfide, and in the presence of 0.181M of sulfide. (K_sp for lead sulfide is 3.4*10^-28).  Explain how the common ion effect affects the solubility of lead in water, and how this could be used to remove lead, and why it may not be a good idea.

Answer

lead(II)sulfide is an extremely insoluble salt, having a solubility of 2.8 x 10^-14M or 2.8 fM, while in the presence of 0.181M sulfide, it drops to 1.9 rontoM (1.9x10^-27 , which is virtually nonexistent.  The issue is when you use a counter ion to suppress the ionization of a toxic compound like lead, you need to be sure the counter ion is safe.  Sulfate is commonly used, as are phosphates.

 

Precipitation Reactions

Complex Ion Equilibria

17.6

Exercise \(\PageIndex{4.r}\)

Calculate the concentration of Cu²⁺ at equilibrium when NH₃ is added to a 0.010M CuCl₂ solution to produce an equilibrium concentration of [NH₃] = 0.02M.  Neglect the volume change when the ammonia is added. K_f=5x10¹²

Answer

K_f=5x10¹², which is very very large, so assume almost all the copper is converted to the complex, also note, 

\[K_{f}=\frac{\left [ Cu(NH_{3})_{4}^{2+} \right ]}{\left [ Cu^{2+} \right ]\left [ NH_{3} \right ]^{4}}\nonumber\]

\[\left [ Cu^{2+} \right ]=\frac{\left [ Cu(NH_{3})_{4}^{2+} \right ]}{K_{f}\left [ NH_{3} \right ]^{4}}\nonumber\]

\[\left [ Cu^{2+} \right ]=\frac{\left [ 0.01 \right ]}{5*10^{12}\left [ .02 \right ]^{4}}=1.2*10^{-8}\nonumber\]

 

17.6

Solubility and Complex Ions

 

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [e-mail]. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

• Liliane Poirot