# 17: Solubility and Complex ion Equilibria


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## Q17.1a

Write $$K_{sp}$$ expression for the following equilibria.

1. $$PbI_2 (s) \rightleftharpoons Pb^{2+}(aq)+2I^-_{(aq)}$$
2. $$Cu(IO_3)_{2\; ­(s)} \rightleftharpoons Cu^{2+}­(aq)+2IO^-_{3\;_(aq)}$$
3. $$Ag(OH)_{3\; (s)} \rightleftharpoons Al^{3+}(aq)+3OH^-_{(aq)}$$
4. $$CaF_{2\; (s)} \rightleftharpoons Ca^{2+}_{(aq)}+2F^-_{(aq)}$$

## S17.1a

1. $$K_{sp} = [Pb^{2+}][I^-]^2$$
2. $$K_{sp}= [Cu^{2+}][IO_3^-]^2$$
3. $$K_{sp}= [Al^{3+}][OH^-]^3$$
4. $$K_{sp}= [Ca^{2+}][F^-]^2$$

## Q17.1b

Write $$K_{sp}$$ expression for the following equilibria. For example $$MX_{2 (s)} \rightleftharpoons M^{2+}_{ (aq)} + 2 X^-_{ (aq)}$$, $$K_{sp}= [M^{2+}][X^-]^2$$

• $$Al(OH)_{3 (s)} \rightleftharpoons Al^{3+}_{ (aq)} + 3 OH^-_{ (aq)}$$
• $$CaF_{2 (s)} \rightleftharpoons Ca^{2+}_{ (aq)} + 2F^-_{ (aq)}$$
• $$AgI _{(s)} \rightleftharpoons Ag^+_{ (aq)} + I^-_{ (aq)}$$
• $$BaSO_{4 (s)} \rightleftharpoons Ba^{2+}_{ (aq)} + SO^{2-}_{4(aq)}$$

## S17.1b

1. $$K_{sp}= [Al^{3+}][OH^-]^3$$
2. $$K_{sp}=[Ca^{2+}][F^-]^2$$
3. $$K_{sp}=[Ag^+][I^-]$$
4. $$K_{sp}=[Ba^{2+}][SO_4^{2+}]$$

## Q17.1c

Write $$K_{sp}$$ expressions for the following equilibrium reactions

1. $$AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)}+ Br^-_{(aq)}$$
2. $$Al(OH)_{3(s)} \rightleftharpoons Al^{3+}_{(aq)}+3OH^-_{(aq)}$$
3. $$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)}+2Cl^-_{(aq)}$$

## S17.1c

1. $$K_{sp}=[Ag^+ ][Br^-]$$
2. $$K_{sp}=[Al^{3+} ][OH^-]^3$$
3. $$K_{sp}=[Pb^{2+}][Cl^-]^2$$

## Q17.1d

What would the $$K_{sp}$$ expression be for the following reactions?

1. $$NH_4Br_{(s)} \rightleftharpoons NH^+_{4(aq)} + Br^-_{(aq)}$$
2. $$BaCrO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + CrO^{2-}_{4(aq)}$$
3. $$Na_2S_2O_{3(s)}\rightleftharpoons 2Na^+_{(aq)} + S_2O_{3(aq)}^{2-}$$
4. $$Fe_2(Cr_2O_7)_{3(s)} \rightleftharpoons 2Fe^{3+}_{(aq)}+ 3Cr_2O_{7(aq)}^{2-}$$

## S17.1d

• $$[NH_4^+][Br^-]$$
• $$[Ba^{2+}][CrO_4^{2-}]$$
• $$[Na^+]^2[S_2O_3^{2-}]$$
• $$[Fe^{3+}]^2[Cr_2O_7^{2-}]^3$$

## Q17.2

What is the solubility equilibrium and solubility-product constant expression for $$CaF_2$$?

## S17.2

1. $$CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^-_{(aq)}$$

2. $$K_{sp} = [Ca^{2+}][F^-]^2$$

## Q17.2

Write the solubility equilibrium equations that are described by the following $$K_{sp}$$ expressions. For example, $$K_{sp}= [Ra^{2+}][IO_3^-]^2$$ represents $$Ra(IO_ 3)_2 \rightleftharpoons Ra^{2+}_{(aq)} + 2IO^-_{3(aq)}$$

1. $$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2 Cl^-_{(aq)}$$
2. $$Mg_3(PO_4)_{2(s)} \rightleftharpoons 3 Mg^{2+}_{(aq)} + 2 PO_{4(aq)}^{3-}$$
3. $$SrCO_{3(s)} \rightleftharpoons Sr^{2+}_{(aq)} + CO^{2-}_{(aq)}$$
4. $$Hg_2Cl_{2(s)} \rightleftharpoons Hg^{2+}_{(aq)} + 2 Cl^-_{(aq) }$$

## S17.2

1. $$K_{sp}= [Pb^{2+}][Cl^-]^2$$
2. $$K_{sp}= [Mg^{2+}]^3[PO_4^{3-}]^2$$
3. $$K_{sp}= [Sr^{2+}][CO_3^{2-}]$$
4. $$K_{sp}= [Hg_2^{2+}][Cl^-]^2$$

## Q17.2

Write the equilibrium reactions for these $$K_{sp}$$ expressions

1. $$K_{sp}=[Ag^+][CrO_4^{2-}]$$
2. $$K_{sp}= [Ba^{2+} ][SO_4^{2-} ]$$
3. $$K_{sp}=[Pb^{2+} ][CrO_4^{2-}]$$

## S17.2

1. $$Ag_2 CrO_{4(s)} \rightleftharpoons Ag^+_{(aq)}+ CrO^{2-}_{4(aq)}$$
2. $$BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)}+ SO^{2-}_{4(aq)}$$
3. $$PbCrO_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)}+ CrO_{4(aq)}^{2-}$$

## Q17.2

Write the solubility equilibrium equations according to the $$K_{sp}$$ expression provided.

1. $$AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$$
2. $$Ra(IO_3)_{2(s)} \rightleftharpoons Ra^{2+}_{(aq)}+2IO^-_{3(aq)}$$
3. $$Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4(aq)}$$
4. $$MgF_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2F^-_{(aq)}$$

## S17.2

1. $$K_{sp} = [Ag^+ ][Cl^-]$$
2. $$K_{sp} = [Ra^{2+}] [IO­_3^-]^2$$
3. $$K_{sp} = [Ag^+]^2[CrO_4^{2-}]$$
4. $$K_{sp} = [Mg^{2+}][F^-]^2$$

## Q17.3a

Write the solubility product expression to which each one applies:

1. $$K_{sp} (Hg_2Cl_2) = 1.3 \times 10^{-19}$$
2. $$K_{sp} (AgI) = 8.5 \times 10^{-17}$$
3. $$K_{sp} (SrSO_4) = 3.2 \times 10^{-7}$$
4. $$K_{sp} (Fe(OH)_3) = 4 \times 10^{-38}$$

## S17.3a

1. $$K_{sp} = [Hg^{2+}][Cl^-]^2=1.3 \times 10^{-19}$$
2. $$K_{sp} = [Ag^+][I^-]= 8.5 \times 10^{-17}$$
3. $$K_{sp} = [Sr^{2+}][SO_4^{2-}]=3.2 \times 10^{-7}$$
4. $$K_{sp} = [Fe^{3+}][OH^-]^3 = 4 \times 10^{-38}$$

## Q17.3b

The following Ksp values are found in a table. Write the solubility product expression to which each one applies. For example, $$K_{sp} (BaCO_3) = [Ba^{2+}][CO_3^-]= 5.1\times 10^{-9}$$

1. $$K_{sp}(CaSO_4) = 9.1\times 10^{-6}$$
2. $$K_{sp}(PbI_2)=7.1\times 10^{-9}$$
3. $$K_{sp}(Mg(OH)_2)=1.8\times 10^{-11}$$
4. $$K_{sp}(Ag_2CrO_2)=1.1\times 10^{-12}$$

## S17.3b

1. $$[Ca^{2+}][SO_4^{2-}]=9.1\times 10^{-6}$$
2. $$[Pb^{2+}][I^-]^2=7.1\times 10^{-9}$$
3. $$[Mg^{2+}][OH^-]^2=1.8\times 10^{-11}$$
4. $$[Ag^+]^2[CrO_4^{2-}]=1.1\times 10^{-12}$$

## Q17.3c

Write the solubility product expression for each $$K_{sp}$$ value

1. $$K_{sp} (BaF_2 )=3.15 \times 10^{-3}$$
2. $$K_{sp} (Mg(OH)_2 )= 1.71 \times10^{-4}$$
3. $$K_{sp} (BaCO_3 ) = 5.1 \times 10^{-9}$$

## S17.3c

1. $$K_{sp}=[Ba^+] [F^-]^2=3.15 \times 10^{-3}$$
2. $$K_{sp}=[Mg^{2+}] [OH^- ]^2= 1.71 \times10^{-4}$$
3. $$K_{sp}=[Ba^{2+}][CO_3^{2-}]= 5.1 \times 10^{-9}$$

## Q17.3d

Write the solubility product expression for the following.

1. $$Ag_2CrO_{4(s)}$$
2. $$Mg_3(PO_4)_{2 (s)}$$
3. $$PbCl_{2(s)}$$
4. $$Cr(OH)_3$$

## S17.3d

1. $$[CrO_4^{2-}][Ag^+]^2$$
2. $$[Mg^{2+}]^3[PO_4^{3-}]^2$$
3. $$[Pb^{2+}][Cl^-]^2$$
4. $$[Cr^{3+}][OH^-]^3$$

## Q17.5

Which of the following would be more soluble in an acidic solution: $$Zn(OH)_2$$, $$KBr$$, $$AlCl_3$$, $$PbS$$?

## S17.5

$$Zn(OH)_2$$ and $$PbS$$ would be more soluble in an acidic solution.

## Q17.25

A particular water sample that is saturated in $$CaF_2$$ has a $$Ca^{2+}$$ content of 130 ppm (i.e., 130 g $$Ca^{2+}$$ per $$10^6$$ grams of water sample). What is the $$F^-$$ ion content of the water in ppm?

## S17.25

$$CaF_2$$ dissociates into $$Ca^{2+}$$ and $$2F^-$$. $$(CaF_2\rightleftharpoons Ca^{2+} + 2F^-)$$ So the ion content of the $$F^-$$ ion would be double that of the $$Ca^{2+}$$ ion content per $$10^6$$ grams of the water sample. $$\frac{260\ g\ F^-}{10^6\ g} = 260 ppm$$

## Q17.25

How much $$SO_4^{2-}$$ ion content (in ppm) that a sample of water contains when it’s saturated in $$CaSO_4$$ and it has a $$Ca^{2+}$$ content of 40 ppm?

## S17.25

$$[SO_4^{2-}] = 2.76\; M = 2.65 \times 10^{-7}\; ppm$$

Video Solution

## Q17.25a

A sample of water is saturated with $$Mg(OH)_2$$. The sample has a $$Mg^{2+}$$ content of 45 ppm (that is, 45 g $$Mg^{2+}$$ per 106 g of water sample. What is the $$OH^-$$ ion content of the water in ppm?

## S17.25a

$$[Mg^{2+}]= \frac{45\ g\ Mg^{2+}}{10^{6}\ g\ soln}\times \frac{1\ mol\ Mg^{2+}}{24.31\ g\ Mg^{2+}}\times \frac{1000g\ soln}{1\ L\ soln} = 1.85\times 10^{-3}$$

$$[Mg^{2+}][OH^-]^2 = K_{sp} = 1.8\times 10^{-11} = (1.85\times 10^{-3})[OH^-]^2$$

$$[OH^-]= 9.86\times 10^{-5}$$

$$ppm\ OH^- = \frac{9.86\times 10^{-5}}{1\ L\ soln}\times \frac{17\ g\ OH^-}{1\ mol\ OH^-}\times \frac{1\ L\ soln}{1000\ g}\times 10^6\ soln = 1.68 ppm$$

## Q17.25b

Silver chromate is dissolved in 0.1M $$AgNO_3$$. What affect does the common ion have on the solubility? $$K_{sp} = 9 \times 10^{-12}$$ molar solubility =$$1.3\times 10^{-4}\ M$$

## S17.25b

$Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4 (aq)}$

 ICE $$Ag_2CrO_{4(s)}^+$$ $$2Ag^+_{(aq)}$$ $$CrO^{2-}_{4 (aq)}$$ Initial - 0.1 M 0 Change - +2x +x Equilibrium - 0.1M+2x x

$K_{sp}=[Ag_+]^2[CrO_4^{2-}]= (0.1\;M+2x)^2(x)= 9 \times 10^{-12}$

if we assume $$x << (0.1M)^2(x)=9 \times 10^{-12} \approx 9 \times 10^{-10}$$

## Q17.31

Will a precipitate of lead (II) occur if three drops of 0.20MKI are added to 100.0mL of 0.010M $$Pb(NO_3)_2$$? assuming 1 drop= 0.05 mL.

## S17.31

$$PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)}+2I^-_{(aq)}$$

$$K_{sp}=7.1\times 10^{-9}$$

$$amtI-=(3drops)\times \frac{0.05mL}{1drop}\times \frac{L}{1000mL}\times \frac{0.20mol\ KI}{L}\times \frac{mol\ I^-}{mol\ KI}=3.0\times 10^{-5} mol\ I^-$$

$$total\ volume=(0.1L)+3drops\frac{0.05mL}{1drop} \frac{L}{1000mL}=0.10015L$$

$$[I^-]=\frac{3.0\times 10^-5\ mol}{0.10015L}=3.0\times 10^{-4}\ M$$

$$Q_{sp}=[Pb^{2+}][I^-]^2=(0.010M)(3.0\times 10^{-4})^2=9.0\times 10^{-10}$$

Since $$Q_{sp} < K_{sp}$$, therefore, $$PbI_{2(s)}$$ will NOT precipitate

## Q17.33c

Can $$Ce(IO_3)_3$$ precipitate when adding together 100 mL of $$1.0 \times 10^{-2} \; M$$ $$Ce(NO_3)_3$$ and 200 mL of $$2.0 \times 10^{-3} \;M$$ $$KIO_3$$? ($$K_{sp} = 1.9 \times 10^{-10}$$).

## S17.33c

Yes, it will precipitate, since Q is bigger than K.

## Q17.33

4.7 mL of 0.35 M $$KI$$ are added to a 150.0 mL of 0.069 M $$Pb(NO_3)_2$$. Will a precipitate of lead iodide be formed?

The following reactions has a $$K_{sp}$$ of $$7.1 \times 10^{-9}$$:

$PbI_{2 (s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2 I^-_{(aq)}$

## S17.33

Two salt solutions are combined so four different salts may precipitate out. However, we are only interseted in the $$Pb(NO_3)_2$$. So this is a simple compare problem involving calculating $$Q_{sp}$$ and then comparing to $$K_{sp}$$.

$Q_{sp} = [Pb^{2+}][I^-]^2$

Both $$[Pb^{2+}]$$ and $$[I^-]$$ are needed to be known to calculate $$Q_{sp}$$ for the final solution. This is a changing concentration problem too.

$(4.7\; \cancel{mL} ) \left(\dfrac{1\; L}{1000\; \cancel{mL}}\right) \left( \dfrac{ 0.35\; \cancel{mol \;KI}}{1\; L}\right) \left( \dfrac{1\; mol\; I^-}{1 \; \cancel{mol\; KI}}\right) = 1.65 \times 10^{-3} mol I^-$

Convert to concentration

$[I^-] = \dfrac{1.65 \times 10^{-3}\, mol\, I^-}{0.150\; L} = 1.07 \times 10^{-2}\; M$

$Q_{sp}= [Pb^{2+}][I^-]^2 = (0.069)(1.07 \times 10^{-2})^2=8.3 \times 10^{-6} > K_{sp}=7.1 \times 10^{-9}$

Therefore, $$PbI_{2(s)}$$ should precipitate after mixing these two solutions.

## Q17.35a

When 310 ml of 0.460 M $$CaS$$ are added to 310.0 ml of 0.112M $$NaNO_3$$, what percentage of the $$Na^+$$ is left unprecipitated?

## S17.35a

$NaNO_3 + CaS \rightarrow NaS + CaNO_3$

$(310\; mL)(0.460\;M \;CaS) = 142.6\; mmol \;CaS$

$(310\; mL)(0.112\; M\; NaNO_3)= 34.72\; mmol\; NaNO_3$

There are no unprecipitated sodium ions in the solution because there is an excess of CaS.

## Q17.35b

How much $$Mg^{2+}$$ ions will remain in solution after adding $$NaOH$$ to a solution that’s composed of 200.0 mL of 0.010M $$Mg^{2+}$$ and 0.005M $$Ca^{2+}$$?

## S17.35b

The $$Mg^{2+}$$ ions already get used 100%.

## Q17.35c

When 70.0 mL sample of 0.372 M $$Na_2SO_{4(aq)}$$ is added to 100 mL of 0.138 M $$Ca(NO_3)_{2(aq)}$$, what percentage of $$Ca^{2+}$$ remains unprecipitated?

## S17.35c

$$[SO_4^{2-}]= 0.372\ M \times \frac{70\ mL}{(70mL+100mL)}= 0.153\ M$$

$$[Ca^{2+}]= 0.138\ M \times \frac{100mL}{(70mL+100mL}= 0.0776 M$$

We determine the value of the ion product and compare it to the solubility product constant value.

$$Q_{sp}= [Ca^{2+}][NO_3^-]^2 = (0.0776)(0.153)^2= 1.82\times 10^{-3} > 9.1\times 10^{-6} = K_{sp}\ for\ Ca(NO_3)_2$$

Since $$Q_{sp} > K_{sp}$$, $$Ca(NO_3)_2$$ should precipitate.

Equation: $$Ca(NO_3)_{2(s)} \rightleftharpoons \quad Ca^{2+}_{(aq)} + \quad 2NO_{3(aq)}^-$$

Orig. soln: ____ 0.0776 M 0.153M

Form solid: ____ -0.0776 M -0.00219M

Not @ Eq: ____ 0 M 0.151 M

Changes: ____ + X M + 2X M

Equil: ____ X M (0.151 + 2X) M

$$K_{sp}= [Ca^{2+}][NO_3^-]^2 = 9.1\times 10^{-6} = (X)(0.151 + 2X)^2$$

$$X= 2.08 \times 10^{-7}$$

$$[Ca^{2+}]=X=2.08\times 10^{-7}$$

$$\%\ Ca^{2+}\ unprecipitated = \frac{2.08\times 10^{-7}\ M\ final}{0.0776\ M\ initial}\times 100 \%= 2.68\times 10^{-4}\ \% \ unprecipitated$$

## Q17.35d

Silver ion and bromine ion precipitate into $$AgBr_{(s)}$$. The $$Ag^+$$ concentration is about 0.065M If the $$Br^-$$ concentration remains at $$1.0 \times 10^{17}\; M$$, what will be the remaining of $$[Ag^+]$$ in solution when precipitation stops $$(K_{sp}=7.7\times 10^{13})$$?

## S17.35d

$AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)}+Br^-_{(aq)}$

$$Q_{sp}=[Ag^+][Br^-]=(0.065)(1.0\times 10^{15})$$

$$Q_{sp}>K_{sp}$$ therefore, precipitation occurs

$$[Ag^+][Br^-]=K_{sp}$$

$$[Ag^+]\times (1.0\times 10^{15})=7.7\times 10^{13}$$

$$[Ag^+]\ remaining=7.7\times 10^{-4}\ M$$

## Q17.37a

$$[OH^-]$$ is maintained at $$3.0 \times 10^{-2}\; M$$ in a solution and $$[Mg^{2+}] = 0.036\; M$$. How much $$Mg^{2+}$$ remain after $$Mg(OH)_2$$ precipitate? ($$K_{sp} = 1.8 \times 10^{-11}$$)

## S17.37a

$$[Mg^{2+}] = 2.0 \times 10^{-8}\; M$$ remains.

## Q17.37b

If a constant $$[F^-]=0.003\ M$$ is maintained in a solution in which the initial $$[Ca^{2+}] = 0.038\ M$$, what percentage of $$Ca^{2+}$$ will remain in solution after $$CaF_{2(s)}$$ precipitates? What $$[F^-]$$ should be maintained to ensure only 3.4% of $$Ca^{2+}$$ remains unprecipitated?

## S17.37b

We first use the solubility product constant expression to determine $$[Ca^{2+}]$$ in a solution with 0.003 M $$F^-$$

$$K_{sp}= [Ca^{2+}][F^-]^2$$

$$5.3\times 10^{-9} = [Ca^{2+}](0.003)^2$$

$$[Ca^{2+}]= \frac{5.3\times 10^{-9}}{(0.003)^2} = 5.89\times 10^{-4}$$

$$\% \ unprecipitated = \frac{5.89\times 10^{-4}\ M}{0.038\ M}\times 100\% = 1.5\%$$

Now, we want to find $$[F^-]$$ must be maintained to keep $$[Ca^{2+}]_{final}$$ =3.4%

$$[Ca^{2+}]_{initial} = 0.034\times 0.038M = 1.29\times 10^{-3}\ M$$

$$K_{sp}= [Ca^{2+}][F^-]^2=5.3\times 10^{-9} = (1.29\times 10^{-3})[F^-]^2$$

$$[F^-]= \sqrt{\frac{5.3\times 10^{-9}}{1.29\times 10^{-3}}} = 0.0021\ M$$

## Q17.37c

If $$K_{sp}$$ of $$Zn(CN)_{2(s)}$$ is $$3\times 10^{-16}$$, what percentage of $$Zn^{2+}$$ is precipitated when we know that

• $$[Zn^{2+}]_o= 0.36\;M$$
• $$[CN^-]_o= 5 \times 10^{-3}\;M$$
• $$K_{sp}=3\times 10^{-16}$$ for $$Zn(CN)_{2(s)}$$.

## S17.37c

$Zn(CN)_{2(s)} \rightleftharpoons Zn^{2+}_{(aq)} + 2CN^-_{(aq)}$

$$[Zn^{2+}][CN]^2 =3\times 10^{-16}$$

$$[Zn^{2+}]=\frac{3\times 10^{-16}}{(5\times 10^{-3})^2} = 1.2\times 10^{-11}$$

Due to the precipitation reaction, $$[Zn]$$ was reduced to $$3.33\times 10^{-9}$$, signifying that it has essentially fully precipitated out of solution.

## Q17.39a

There are 320 g $$Na^+$$ per kg of water

1. should NaOH precipitate when $$[OH^-] = 1.57\times 10^{-3}\ M$$?
2. is it possible for $$Na$$ and $$Cl$$ to be separated in the water?

1. no
2. no

## Q17.39b

A sample of seawater contains 0.25 M $$Mg^{2+}$$ and 0.0033 M $$Ca^{2+}$$. Which will precipitate first when $$NaOH$$ is added? What’s the concentration of $$X^{2+}$$ ion remaining after $$X(OH)_2$$ precipitate?

## S17.39b

$$Mg(OH)_2$$ will precipitate first, and it gets used almost completely.

## Q17.39c

0.01M of $$Cu^+$$ and 0.01M of $$Ag^+$$ are in solution. A piece of solid $$NaI$$ is put in inside. What is the concentration of $$Ag^+$$ as soon as $$CuI$$ begins to precipitate?

## S17.39c

$$Cu^+ + I^- \rightarrow CuI$$ and $$K_{sp}$$ is $$5.1\times 10^{-12}$$

$$Ag^+ + I^- \rightarrow AgI$$ and $$K_{sp}$$ is $$8.3\times 10^{-17}$$

$$[Ag^+]= \frac{K_{sp}\ of\ AgI}{[I^-]\ in\ CuI}$$

$$[Ag^+]= \frac{8.3 \times 10^{-17}}{5.1\times 10^{-10}} = 1.627\times 10^{-7}$$

## Q17.41

$$KI_{(aq)}$$ is slowly added to a solution with $$[Pb^{2+}] = [Ag^+] = 0.30\; M$$. For $$PbI_2$$, $$K_{sp} = 9.6 \times 10^{-9}$$; for $$AgI$$, $$K_{sp} = 6.2 \times 10^{-18}$$.

1. Which precipitate should form first, $$PbI_2$$ or $$AgI$$?
2. What is fractional precipitation?
3. Can $$Pb^{2+}$$ and $$Ag^+$$ be effectively separated by fractional precipitation of their iodides?

## S17.41

1. AgI will precipitate first because it has the smaller Ksp value.
2. Fractional precipitation is the technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so on.
3. Yes because the ratio of the Ksp values compared to the molar concentration of .3 M for [Pb2+] and [Ag+] shows that both ions will completely dissociate.

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions

## Q17.41

If $$Cl^-$$ is added to a solution that contains $$0.025\;M$$ of $$Ag^+$$ ions and $$0.00448\;M$$ of $$Pb^{2+}$$ ions. What concentration of $$Cl^-$$ is required to precipitate each ion? $$K_{sp} (AgCl) = 1.8 \times 10^{-10}$$ and $$K_{sp} (PbCl_2) = 1.6 \times 10^{-5}$$.

## S17.41

• After adding $$7.2 \times 10^{-9}\; M$$ of $$[Cl^-]$$, silver will precipitate.
• After adding $$3.57 \times 10^{-3}\; M$$ of $$[Cl^-]$$, lead will precipitate.

## Q17.41

0.23 M $$Na^+$$ is added to a solution of 0.45 M $$Cl^-$$ and 0.48 M $$OH^-$$.

1. Should $$NaCl$$ or $$NaOH$$ precipitate first?
2. What concentration of $$[OH^-]$$ is required for the precipitation?
3. Can these ions be effectively precipitated by fractional precipitation?

1. NaCl
2. 0.049 M
3. yes

## Q17.43a

A concentration solution of potassium chromate is added to a solution that is 0.1 M $$Ba^{2+}$$ and 0.1 M $$Sr^{2+}$$ Which ion will precipitate first? What is the percentage of this ion remaining when the other ion begins to precipitate?

## S17.43a

$$Q_c=K_{sp}$$

a)

$$BaCrO_{4(s)} \rightleftharpoons Ba^{2+}+CrO_4^{2-}$$

$$K_{sp}=[Ba^{2+}][CrO_4^{2-}]$$$$[CrO_4^{2-}]=1.2\times 10^{-9}\ M$$

$$1.2\times 10^{-10}=(0.1\ M)[CrO_4^{2-}]$$

$$SrCrO_{4(s)} \rightleftharpoons Sr^{2+}+CrO_4^{-2}$$

$$K_{sp}=[Sr^{2+}][CrO_4^{2-}]$$$$[CrO_4^{2-}]=3.5\times 10^{-4}\ M$$

$$3.5\times 10^{-5}=(0.1\ M)[CrO_4^{2-}]$$
Therefore, $$BaCrO_4$$ PRECIPITATES FIRST

b)

$$BaCrO_{4(s)} \rightleftharpoons Ba^{2+}+CrO_4^{2-}$$

$$K_{sp}=[Ba^{2+}][CrO_4^{2-}]$$

$$1.2\times 10^{-10}=[Ba^{2+}][3.5\times 10^{-4}]$$

$$[Ba^{2+}]=3.4\times 10^{-7}\ M$$

$$\% \ Ba^{2+}\ remaining= \frac{3.4\times 10^{-7}\ M}{0.1}\times 100\% = 0.000343\%$$

## Q17.43b

0.4 M $$Cu^{2+}$$ is added to a solution of 0.039 M $$CO_2^3$$ and 0.039 M $$Cl^-$$.

1. Which precipitates first?
2. When the second ion starts to precipitate, what is the concentration of the first ion?
3. Can these be effectively separated by fractional precipitation?

## S17.43b

1. $$CuCO_3$$
2. 0.028 M
3. yes

## Q17.43c

An aqueous solution that 2.00 M in $$Cu(CO_3)$$ is slowly added from a buret to a aqueous solution that is 0.020 M in $$Cl^-$$ and 0.30 M in $$Br^-$$:

1. Which ion will precipitate first?
2. What is the remaining concentration of the first ion when the second ion begins to precipitate?
3. Is the separation of $$F^-$$ and $$Br^-$$ feasible by fractional precipitation in this solution?

## S17.43c

From the table of $$K_{sp}$$ values (Table E3):

$K_{sp} \; (CuCl)= 1.7 \times 10^{-7}$

$K_{sp}\, (CuBr) = 6.3 \times 10^{-9}$

$$CuBr$$ will be the first to precipitate due to its smaller $$K_{sp}$$ value.

$$CuCl$$ will precipitate when $$Q_{sp} > K_{sp}$$. First find the $$Cu$$ concentration when $$CuCl$$ first precipitates.

$1.7 \times 10^{-7}= 0.02 [Cu^+]$

$[Cu^+]=8.5 \times 10^{-7}$

$6.3 \times 10^{-9}=8.5 \times 10^{-7} [Br^-]$

$0.007412 = [Br^-]$

$$[Br^-]$$ drop from 0.3 M to 0.007412 M before $$CuCl$$ precipitate begins to form. This means that \frac{0.007412}{0.3}= 0.0247= 2.47\% \). This small percentage of $$Br^-$$ ions that remain unreacted in the solution implies that fractional precipitation is feasible.

## Q17.43d

If a solution contains $$5.2 \times 10^{-3} \; M$$ of $$Pb^{2+}$$ and $$4.8 \times 10^{-2} \; M$$ of $$Cu^+$$. What concentration of $$I^-$$ is needed to be added to start a precipitation reaction?

## S17.43d

$$1.10\times 10^{-10}\ M$$ of $$I^-$$ is needed to begin precipitation.

## Q17.59

a) Can $$NaOH$$ be separated by precipitating $$NaCl$$ but not $$HCl$$ if all concentrations are 0.2 M?

b) Will the separation be complete?

a) Yes

b) No

## Q17.59

Which of the following solids are more soluble in an acidic solution than in pure water: $$NaCl$$, $$Fe(OH)_2$$, $$CaF_2$$, $$ZnCO_3$$, or $$C_5H_4CO_2H$$?

## S17.59

• $$NaCl$$: They will have similar solubilities in both solvents because $$NaCl$$ does not produce any ions that will directly react with an acid.
• $$Fe(OH)_2$$: This will be more soluble in an acidic solution because the $$OH^-$$ ions created will form water with $$H^+$$ ions present in the acidic solution.
• $$CaF_2$$: This will be more soluble in an acidic solution because the $$F^-$$ ions created will react with the $$H^+$$ ions present in the acidic solution to increase the pH.
• $$ZnCO_3$$: This will be more soluble in an acidic solution because the $$CO_3$$ will lose an oxygen to form water with 2 protons already present in the acidic solution.
• $$C_5H_4CO_2H$$: This will be more soluble in pure water because in this molecule there are $$H_3O^+$$ ions that will not react with the acidic solution at all, and will therefore be more soluble in water.

## Q17.59

Is $$Ce(IO_3)­_3$$ able to precipitate from a solution that contains 500 mL of 0.10 M $$KIO_3$$ and 200 mL of 0.05 M $$Ce(NO_3)_3$$? ($$K_{sp} = 1.9 \times 10^{-10}$$)

## S17.59

It’s going to precipitate.

## Q17.61

Can $$Pb(OH)_2$$ precipitate from a solution that contains $$3.2 \times 10^{-3}\; M \;PbCl_2$$ and $$0.05\; M \;NH_3$$?

## S17.61

No precipitation would occur.

## Q17.61

A saturated solution of $$Ca_3(PO_4)_2$$ has $$[Ca^{2+}] = [PO_4^{3-}] = 2.9\times 10^{-7}\ M$$. Will $$Ca_3(PO_4)_2$$ precipitate?

no

## Q17.61

The solubility of $$Ba(OH)_2$$ in a particular buffer solution is $$0.5\frac{g}{L}$$. What is the pH of the buffer solution? $$K_{sp}= 5.0 \times 10^{-3}$$

## S17.61

First find $$[Ba^{2+}]$$ using $$0.5\frac{g}{L}$$

$\dfrac{0.5\; grams}{171.327\; \frac{g}{mol}} = 0.00584\; \frac{mol}{L}= [Ba^{2+}]$

$5 \times 10^{-3}=[Ba^{2+}][OH^-]^2$

$[OH-]=0.925$

$pOH= -log(0.925)= 0.034$

$pH= 13.9663$

## Q17.81

The solubility of $$Na^{2+}$$ in 2 M $$CaCl$$ is $$1.3\times 10^{-2} \frac{mol}{L}$$. If $$K_{sp}$$ of $$Na^{2+}$$ is $$3.1\times 10^{-10}$$, what is the value of $$K_f$$ for $$NaCl$$?

## S17.81

$$K_f=2.88\times 10^{-2}$$

## Q17.81

If a 0.20 mol sample of $$Zn(NH_3)_4SO_4$$ is dissolved in 1.00L of an aqueous solution of 1.0 M $$NH_3$$, what’s the concentration of $$[Zn^{2+}]$$? Hint: Use equilibrium constants in Table E4.

## S17.81

Many complex ions such as tetrammine zinc (II), $$Zn(NH_3)_4^{+2}$$, are loosely aggregated and tend to dissociate in a water solution until an equilibrium is established between the complex ion and its components.

$Zn(NH_3)_4^{+2} \rightleftharpoons Zn^{2+} + 4 NH_3$

$K= \dfrac{[Zn^{2+}][NH_3]^4}{[Zn(NH_3)^{+2}]^4} = 7.8 \times 10^8$

Video Solution

## Q17.81

The solubility of $$AlF_{3(s)}$$ in 1.5 M $$OH^{-}_{(aq)}$$ is $$2.0\times 10^{-4}\frac{mol}{L}$$. Given that $$K_{sp}$$ of $$AlF_3$$ is $$3.7 \times 10^{-9}$$, what is the value of $$K_f$$ for $$Al(OH)_3$$?

## S17.81

To get the net reaction we combine the solubility product expression for $$AlH_{3(s)}$$ with the formation expression of $$Al(OH)_{3(s)}$$.

Solubility: $$AlF_{3(s)} \rightleftharpoons Al^{3+}_{(aq)}+3F^-_{(aq)}$$ , $$K_{sp}=3.7\times 10^{-9}$$

Formation: $$Al^{3+}_{(aq)} + 3OH^-_{(aq)} \rightleftharpoons Al(OH)_{3(aq)}$$ , $$K_f = ?$$

Net Reaction: $$AlF_{3(s)}+3OH^-_{(aq)} \rightleftharpoons Al(OH)_{3(aq)}+3F^-_{(aq)}$$ , $$K_{overall}=K_{sp}K_f$$

$$K_{overall}=\frac{[Al(OH)_3 ][F^-]^3}{[OH^-]^3} =\frac{(2.0\times 10^{-4})^4}{(1.5)^3} =4.74\times 10^{-16}=K_f\times (3.7\times 10^{-9})$$

$$K_f= \frac{ 4.74\times 10^{-16} }{ 3.7\times 10^{-9} } =1.27\times 10^{-7}$$

## Q17.83

A mixture of $$PbSO_4(s)$$ and $$PbS_2O_3(s)$$ is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is $$[Pb^{2+}]$$ in the saturated solution? For $$PbSO_4$$, $$K_{sp} = 3.4 \times 10^{-7}$$; for $$PbS_2O_3$$, $$K_{sp} = 7.6 \times 10^{-4}$$.

## S17.83

$$K_{sp}(PbSO_4) = [Pb^{2+}][SO_4^{2-}]$$ with $$[Pb^{2+}] = [SO_4^{2-}]$$

So $$K_{sp}(PbSO_4)= [Pb^{2+}]^2$$

$$K_{sp}(PbS_2O_3) = [Pb^{2+}][S_2O_3^{2-}]$$ with $$[Pb^{2+}] = [S_2O_3^{2-}]$$

So $$K_{sp}(PbS_2O_3) = [Pb^{2+}]^2$$

$$K_{sp}(PbSO4) \times K_{sp}(PbS_2O_3) = [Pb^{2+}]^4$$

$$(3.4\times 10^{-7})(7.6\times 10^{-4}) = [Pb^{2+}]^4$$

$$\mathbf{[Pb^{2+}] = .0040 M}$$

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Solubilty/Relating_Solubility_and_Kspv

## Q17.83

A mixture of $$AlPO_4(s)$$ and $$AlPO_3(s)$$ is shaken with water until a saturated solution is formed. Both of the solids remain in excess. What is $$[Al^{3+}]$$ in the saturated solution? For $$AlPO_4$$, $$K_{sp} = 1.3 \times 10^{-6}$$ and for $$AlPO_3$$, $$K_{sp} = 3.5 \times 10^{-4}$$.

## S17.83

There are three relationships you need for this question:

$$1.3 \times 10^{-6}=[Al^{3+} ][PO_4^{3-}]$$

$$3.5 \times 10^{-4}=[Al^{3+} ][PO_3^{3-} ]$$

The third is the electro neutrality equation, which states that the total positive charge concentration must equal the total negative charge concentration.

$$[Al^{3+} ]=[PO_4^{3-} ]+[PO_3^{3-}]$$

Use the first two expressions to solve for the concentration of each anion, then substitute them into the electroneutrality expression and solve for $$[Al^{3+}]$$.

$$[PO_4^{3-}]=\frac{1.3 \times 10^{-6}}{[Al^{3+}]}$$

$$[PO_3^{3-}]=\frac{3.5 \times 10^{-4}}{[Al^{3+}]}$$

$$[Al^{3+}]=\frac{1.3\times 10^{-6}}{[Al^{3+}]}+\frac{3.5 \times 10^{-4}}{[Al^{3+}]}$$

$$[Al^{3+} ]^2= 1.3\times 10^{-6}+3.5\times 10^{-4}=3.5\times 10^{-4}$$

$$[Al^{3+}]= \sqrt{(3.5\times 10^{-4}}=0.019\ M$$

## Q17.83

A mixture of $$CaSO_4$$ and $$AgSO_4$$ is shaken with water until the solution is saturated. What is $$[Ag^{2+}]$$?

0.0314 M

## Q17.83 - (Answer is not given)

At what pH does $$Cr(OH)_3$$ just starts to precipitate from an 0.10 M $$Cr^{3+}$$ solution? From this calculation justify why it is impossible to prepare an 0.10 M $$Cr^{3+}$$ solution at pH=7. The $$K_{sp}$$ of $$Cr(OH)_3$$ is $$6.3 \times 10^{-31}$$.

## S17.83

17: Solubility and Complex ion Equilibria is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.