Chapter 15, Problem 70QRT

(a)

Interpretation Introduction

Interpretation:

The solubility of ${\text{ZnCO}}_{\text{3}}$ has to be calculated in water.

Concept Introduction:

Solubility product relates the solubility of a salt with the concentration of ions present in the salt.  Solubility product holds a direct relation with the solubility of the salt.  The solubility product is the ability of the solid to dissolve in aqueous solution.  The more the solubility product the more the solid dissolves in solution. It is denoted as $K_{\text{sp}}$.  It is known as solubility product constant.

Answer to Problem 70QRT

The solubility of ${\text{ZnCO}}_{\text{3}}$ in water is $\underset{\_}{3.74\times 10^{-6}M}$.

Explanation of Solution

The dissociation reaction of ${\text{ZnCO}}_{\text{3}}$ is shown below.

  ${\text{ZnCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Zn}}^{2+}\left(\text{aq}\right)+{\text{CO}}_3^{2-}\left(\text{aq}\right)$

The $K_{\text{sp}}$ for ${\text{ZnCO}}_{\text{3}}$ is given as $1.4\times {10}^{-11}$.

The expression of molar solubility of ${\text{ZnCO}}_{\text{3}}$ in pure water is as follows.

    $\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Zn}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\left(S\right)\left(S\right)\\ =S^2\end{array}$                                                                                    ……(1)

Where,

• $K_{\text{sp}}$ is the solubility product.
• $S$ is the solubility.

Substitute $K_{\text{sp}}$ of ${\text{ZnCO}}_{\text{3}}$ in the equation (1).

    $\begin{array}{c}{K}_{\text{sp}}=S^2\\ 1.4\times {10}^{-11}=S^2\\ S=\sqrt{1.4\times {10}^{-11}}\\ =\underset{\_}{3.74\times 10^{-6}M}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The solubility of ${\text{ZnCO}}_{\text{3}}$ has to be calculated in $\text{0}\text{.050}\text{M}\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

Concept Introduction:

Refer to concept of part (a).

Answer to Problem 70QRT

The solubility of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\underset{\_}{2.8\times 10^{-10}M}$.

Explanation of Solution

The concentration of $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ solution in which ${\text{ZnCO}}_{\text{3}}$ is present is $\text{0}\text{.050}\text{M}$.

The dissociation reaction of $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows.

    $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\text{aq}\right)\rightleftharpoons {\text{Zn}}^{2+}\left(\text{aq}\right)+2{\text{NO}}_3^{-}\left(\text{aq}\right)$

This means one mole of $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ gives one moles of ${\text{Zn}}^{2+}$.  Therefore, the concentration of ${\text{Zn}}^{2+}$ ions already present in $\text{0}\text{.050}\text{M}$ $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\text{0}\text{.050}\text{M}$.

The ICE table for the dissociation of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}$ $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows.

    $\begin{array}{l}{\text{ ZnCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Zn}}^{2+}\left(\text{aq}\right)+{\text{CO}}_3^{2-}\left(\text{aq}\right)\\ \text{I }\text{0}\text{.050}\text{M }0\\ \text{C }SS\\ \text{E }\text{0}\text{.050}\text{M}+SS\end{array}$

Where,

• $S$ is the solubility.
• I is the Initial concentration.
• C is the change in concentration.
• E is the equilibrium concentration.

The expression of molar solubility of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}$ $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is as follows.

    $\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Zn}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\left(\text{0}\text{.050}\text{M}+S\right)\left(S\right)\end{array}$

Where,

• $K_{\text{sp}}$ is the solubility product.
• $S$ is the solubility.

The value of $\left(\text{0}\text{.050}\text{M}+S\right)$ is approximately equal to the $\text{0}\text{.050}\text{M}$ as the value of $S$ is very small as compared to $\text{0}\text{.050}\text{M}$.

So, the expression of molar solubility of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}$ $\text{Zn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is written as shown below.

    $K_{\text{sp}}=\left(\text{0}\text{.050}\text{M}\right)\left(S\right)$

Substitute $K_{\text{sp}}$ of ${\text{ZnCO}}_{\text{3}}$ in the above equation.

  $\begin{array}{c}{K}_{\text{sp}}=\left(\text{0}\text{.050}\text{M}\right)\left(S\right)\\ 1.4\times {10}^{-11}=\left(\text{0}\text{.050}\text{M}\right)\left(S\right)\\ S=\frac{1.4\times {10}^{-11}}{\text{0}\text{.050}\text{M}}\\ =\underset{\_}{2.8\times 10^{-10}M}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The solubility of ${\text{ZnCO}}_{\text{3}}$ has to be calculated in $\text{0}\text{.050}\text{M}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$.

Concept Introduction:

Refer to concept of part (a).

Answer to Problem 70QRT

The solubility of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}{\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\underset{\_}{2.8\times 10^{-10}M}$.

Explanation of Solution

The concentration of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ solution in which ${\text{ZnCO}}_{\text{3}}$ is present is $\text{0}\text{.050}\text{M}$.

The dissociation reaction of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows.

    ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)\rightleftharpoons 2{\text{K}}^{+}\left(\text{aq}\right)+{\text{CO}}_3^{2-}\left(\text{aq}\right)$

This means one mole of ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ gives one moles of ${\text{CO}}_3^{2-}$.  Therefore, the concentration of ${\text{CO}}_3^{2-}$ ions already present in $\text{0}\text{.050}\text{M}$ ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{0}\text{.050}\text{M}$.

The ICE table for the dissociation of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}$ ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows.

    $\begin{array}{l}{\text{ ZnCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Zn}}^{2+}\left(\text{aq}\right)+{\text{CO}}_3^{2-}\left(\text{aq}\right)\\ \text{I }\text{ 0}\text{0}\text{.050}\text{M}\\ \text{C }SS\\ \text{E }S\text{0}\text{.050}\text{M}+S\end{array}$

Where,

• $S$ is the solubility.
• I is the Initial concentration.
• C is the change in concentration.
• E is the equilibrium concentration.

The expression of molar solubility of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}$ ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows.

    $\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Zn}}^{2+}\right]\left[{\text{CO}}_3^{2-}\right]\\ =\left(S\right)\left(\text{0}\text{.050}\text{M}+S\right)\end{array}$

Where,

• $K_{\text{sp}}$ is the solubility product.
• $S$ is the solubility.

The value of $\left(\text{0}\text{.050}\text{M}+S\right)$ is approximately equal to the $\text{0}\text{.050}\text{M}$ as the value of $S$ is very small as compared to $\text{0}\text{.050}\text{M}$.

So, the expression of molar solubility of ${\text{ZnCO}}_{\text{3}}$ in $\text{0}\text{.050}\text{M}$ ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ is written as shown below.

    $K_{\text{sp}}=\left(\text{0}\text{.050}\text{M}\right)\left(S\right)$

Substitute $K_{\text{sp}}$ of ${\text{ZnCO}}_{\text{3}}$ in the above equation.

  $\begin{array}{c}{K}_{\text{sp}}=\left(\text{0}\text{.050}\text{M}\right)\left(S\right)\\ 1.4\times {10}^{-11}=\left(\text{0}\text{.050}\text{M}\right)\left(S\right)\\ S=\frac{1.4\times {10}^{-11}}{\text{0}\text{.050}\text{M}}\\ =\underset{\_}{2.8\times 10^{-10}M}\end{array}$