Chapter 12, Problem 12.61PAE

(a)

Interpretation Introduction

To determine: Solubility of ${\text{ZnCO}}_3$ in water

Explanation of Solution

Let solubility of ${\text{ZnCO}}_3$ is $3$ then

	${\text{ZnCO}}_3\left(S\right)$	$Z{n}^{2+}\left(aq\right)M$	$C{O}_3^{2-}\left(aq\right)M$
Initial concentration Change in concentration ${\text{due to ZnCO}}_3$ dissolved per L	Solid Solid	$0$ $S$	$0$ $S$
Final concentration	solid	$S$	$S$

We know that

$K_{SP}=\left[Z{n}^{2+}\right]\left[C{O}_3^{2-}\right]$

$\begin{array}{c}{K}_{SP}=S\times S\\ K_{SP}=1.5\times {10}^{-11}\end{array}$

Putting these value you get

$\begin{array}{c}1.5\times {10}^{-11}=S\times S\\ S=\sqrt{1.5\times {10}^{-11}}\\ =3.9\times {10}^{-6}\end{array}$

Hence molar solubility of Zinc carbonate in saturated solution $=3.9\times {10}^{-6}$ M

(b)

Interpretation Introduction

To determine: Calculate the solubility of ${\text{ZnCO}}_3$ in $0.050MZn{\left(N{O}_3\right)}_2$.

Explanation of Solution

$\begin{array}{l}Zn{\left(N{O}_3\right)}_2\stackrel{}{\rightleftharpoons }Z{n}^{2+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)\\ \text{0}\text{.050M 2}\times \left(0.050\right)M\end{array}$

Molarity (M)

$Zn{\left(N{O}_3\right)}_2=0.050M$

$ZnC{O}_3\left(S\right)\stackrel{}{\rightleftharpoons }Z{n}^{2+}\left(aq\right)+C{O}^{2-}\left(aq\right)$

Let solubility of

$\begin{array}{l}ZnC{O}_3\left(S\right)\text{ is S M}\\ \text{ S S}\end{array}$

	$ZnC{O}_3\left(S\right)$	$Z{n}^{2+}\left(aq\right)M$	$C{O}_3^{2-}\left(aq\right)M$
Initial concentration	Solid	$0.050$	$0$
Change in concentration due to $ZnC{O}_3\left(S\right)$	Solid	$S$	$S$
Final concentration	solid	$$ (0.050+S)	$S$

$K_{SP}=\left[Z{n}^{2+}\left(aq\right)\right]\left[C{O}_3^{2-}\left(aq\right)\right]$

Putting the value of final concentration

$K_{SP}=\left(0.050+S\right)\left(S\right)$

Putting the value of $K_{SP}=1.5\times {10}^{-11}$

$1.5\times {10}^{-11}=\left(0.050+S\right)\left(S\right)$

We assume $0.050+S\approx 0.05$

$\begin{array}{c}1.5\times {10}^{-11}=0.050+S\\ S=\frac{1.5\times {10}^{-11}}{5\times {10}^{-2}}\\ =0.3\times {10}^{-9}\\ =3\times {10}^{-10}M\end{array}$

Hence my assumption $0.050+S\approx 0.05$

Therefore, the molar solubility of ${\text{ZnCO}}_3$ in $0.050MZn{\left(N{O}_3\right)}_2$. Solution is $3\times {10}^{-10}M$.

(c)

Interpretation Introduction

To determine: Calculate the solubility of $ZnC{O}_3$ in $0.050M,K_{2}C{O}_3$.

Explanation of Solution

$K_{2}C{O}_3\left(S\right)\stackrel{}{\rightleftharpoons }2K^{+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$

So, $0.050M{K}_{2}C{O}_3$, gives $2\times 0.050K^{+}\left(aq\right)$ and $1\times 0.050C{O}_3^{2-}\left(aq\right)$.

Let solubility of $ZnC{O}_3\left(S\right)=S$

The $S$ mole of $ZnC{O}_3\left(S\right)$ is dissolved in per liter of solution, so it gives $S$ mole of $C{O}_3^{2-}\left(aq\right)$ per liter of solution.

The potassium ions would not have any effect on solubility of $ZnC{O}_3\left(S\right)$, but $C{O}_3^{-}{\text{ ion of K}}_{2}C{O}_3\left(S\right)$ acts as initial source of $C{O}_3^{-}$ ion. And effect the solubility of $ZnC{O}_3\left(S\right)$.

	$ZnC{O}_3\left(S\right)$	$Z{n}^{2+}\left(aq\right)M$	$C{O}_3^{2-}\left(aq\right)M$
Initial concentration Due to $$ K2 CO 3. (S )	Solid	$0$	$0.05$
Change in concentration due to $ZnC{O}_3$ (S )	Solid	$$ $S$	$S$
Final concentration	solid	$S$	$\left(0.050+S\right)$

$K_{SP}=\left[Z{n}^{2+}\right]\left[C{O}_3^{2-}\right]$

Putting the value

$\begin{array}{c}1.5\times {10}^{-11}=S\left(0.050+S\right)\\ S<<0.05\text{ so 0}\text{.05}+\text{S}=\text{0}\text{.05}\\ S=\frac{1.5\times {10}^{-11}}{0.05}\\ =0.3\times {10}^{-9}\\ =3\times {10}^{-10}M\end{array}$

So solubility of $ZnC{O}_3\left(S\right)\text{ in 0}\text{.050M}{\text{K}}_{2}C{O}_3\text{ Solution is 3}\times {\text{10}}^{-10}M$.

Conclusion

Initial presence of common anion or cation reduce the solubility of salt.