Chapter 6: Equilibrium

Ans:  Let the maximum concentration of each solution be ${\text{xmol}}/{\text{L}}$. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., ${\text{x}}/2$.

$\therefore \left[ {FeS{O^4}} \right] = \left[ {N{a_2}S} \right] = \dfrac{x}{2}M$

Then, $\left[ {F{e^{2 + }}} \right] = \left[ {FeS{O^4}} \right] = \dfrac{x}{2}M$

Also, $\left[ {{S^{2 - }}} \right] = \left[ {N{a_2}S} \right] = \dfrac{x}{2}M$

$Fe{S_{(x)}} \leftrightarrow Fe_{(aq)}^{2 + } + S_{(aq)}^{2 - }$

${{K_{sp}} = \left[ {F{e^{2 + }}} \right]\left[ {{S^{2 - }}} \right]} $ 

${6.3 \times {{10}^{ - 18}} = \left( {\dfrac{x}{2}} \right)\left( {\dfrac{x}{2}} \right)} $ 

${\dfrac{{{x^2}}}{4} = 6.3 \times {{10}^{ - 18}}} $ 

${ \Rightarrow x = 5.02 \times {{10}^{ - 9}}} $

If the concentrations of both solutions are equal to or less than $5.02 \times {10^{ - 9}}{\text{M}}$, then there will be no precipitation of iron sulphide.

Ans:  ${{\text{CaS}}{{\text{O}}_{4(s)}} \leftrightarrow {\text{C}}{{\text{a}}^{2 + }}_{(aq)} + SO_{4(aq)}^{2 - }} $ 

${{K_{sp}} = \left[ {{\text{C}}{{\text{a}}^{2 + }}} \right]\left[ {{\text{SO}}_4^{2 - }} \right]} $

Let the solubility of ${\text{CaS}}{{\text{O}}_4}$ be ${\text{s}}$.

Then, ${K_{sp}} = {s^2}$

${9.1 \times {{10}^{ - 6}} = {s^2}} $ 

${s = 3.02 \times {{10}^{ - 3}}\;{\text{mol}}/{\text{L}}} $

Molecular mass of ${\text{CaS}}{{\text{O}}_4} = 136\;{\text{g}}/{\text{mol}}$

Solubility of ${\text{CaS}}{{\text{O}}_4}$ in gram/L

${ = 3.02 \times {{10}^{ - 3}} \times 136} $ 

${ = 0.41\;{\text{g}}/{\text{L}}} $

This means that we need $1\;{\text{L}}$ of water to dissolve $0.41\;{\text{g}}$ of ${\text{CaS}}{{\text{O}}_4}$

Therefore, to dissolve $1\;{\text{g}}$ of ${\text{CaS}}{{\text{O}}_4}$ we require $ = \dfrac{1}{{0.41}}\;{\text{L}} = 2.44\;{\text{L}}$ of water.

Ans: For precipitation to take place, it is required that the calculated ionic product exceeds the ${{\text{K}}_{{\text{sp}}}}$ value.

Before mixing:

${\left[ {{S^{2 - }}} \right] = 1.0 \times {{10}^{ - 19}}M\left[ {{M^{2 + }}} \right] = 0.04M} $

${{\text{ volume }} = 10\;{\text{mL}}\quad {\text{ volume }} = 5\;{\text{mL}}} $

After mixing:

$\left[ {{S^{2 - }}} \right] = ?\quad \left[ {{M^{2 + }}} \right] = ?$

${{\text{ volume }} = (10 + 5) = 15\;{\text{mL}}\quad {\text{ volume }} = 15\;{\text{mL}}} $

${\left[ {{S^{2 - }}} \right] = \dfrac{{1.0 \times {{10}^{ - 19}} \times 10}}{{15}} = 6.67 \times {{10}^{ - 20}}{\text{M}}} $

${\left[ {{M^{2 + }}} \right] = \dfrac{{0.04 \times 5}}{{15}} = 1.33 \times {{10}^{ - 2}}{\text{M}}} $

${{\text{ Ionic product }} = \left[ {{M^{2 + }}} \right]\left[ {{S^{2 - }}} \right]} $

${ = \left( {1.33 \times {{10}^{ - 2}}} \right)\left( {6.67 \times {{10}^{ - 20}}} \right)} $

${ = 8.87 \times {{10}^{ - 22}}} $

This ionic product exceeds the ${{\text{K}}_{{\text{spof }}}}{\text{Zns}}$ and ${\text{CdS}}$. Therefore, precipitation will occur in ${\text{CdC}}{{\text{l}}_2}$ and ${\text{ZnC}}{{\text{l}}_2}$ solutions.