Chapter 22, Problem 22.38P

(a)

Interpretation Introduction

Interpretation:

On mixing $\text{10}\text{.0 mL}\text{of}\text{0}\text{.30 M Zn}{\left({\text{NO}}_3\right)}_2$ with $\text{10}\text{.0 mL}\text{of}\text{2}\text{.0}\times {\text{10}}^{-4}{\text{M Na}}_2\text{S}$, the number of milligrams of $\text{ZnS}\left(\text{s}\right)$ precipitate has to be calculated.

Concept Introduction:

• Reaction quotient (Q) is the ratio of product of concentration of products to that of product of concentration of reactants, raised to the power of coefficients.
• ${\text{Q}}_{\text{sp}}{\text{ > K}}_{\text{sp}}$ , precipitation will happen.
• ${\text{Q is less than K}}_{\text{sp}}$, more solute can dissolve in solution.
• ${\text{Q is equal to K}}_{\text{sp}}$, no more solute can dissolve or precipitate in solution.

Answer to Problem 22.38P

Mass of zinc sulphide formed is calculated and is given below:

  $\text{Mass}=0.19\text{mg}$

Explanation of Solution

Given data is as follows:

  $\begin{array}{l}\text{10}\text{.0 mL}\text{of}\text{0}\text{.30 M Zn}{\left({\text{NO}}_3\right)}_2\\ \\ \text{10}\text{.0 mL}\text{of}\text{2}\text{.0}\times {\text{10}}^{-4}{\text{M Na}}_2\text{S}\end{array}$

The chemical equation for the precipitation reaction is,

  ${\text{Zn}}^{\text{+2}}\left(\text{aq}\right)+{\text{ S}}^{\text{2-}}\left(\text{aq}\right)\rightleftharpoons \text{ ZnS}\left(\text{s}\right)$

The concentration of ${\text{Zn}}^{\text{+2}}$ and ${\left[{\text{S}}^{\text{2-}}\right]}_{\text{0}}$ immediately after mixing is calculated as follows,

  $\begin{array}{l}{\left[{\text{Zn}}^{\text{+2}}\right]}_{\text{0}}\text{ =}\frac{\left(\text{10}\text{.0 mL}\right)\left(\text{0}\text{.30 M}\right)}{20.0\text{ mL}}=0.15\text{ M}\\ \\ {\left[{\text{S}}^{\text{-2}}\right]}_{\text{0}}\text{=}\frac{\left(\text{10}\text{.0 mL}\right)\left(\text{2}\text{.0}\times {\text{10}}^{-4}\text{ M}\right)}{20.0\text{ mL}}=1\times {10}^{-4}\text{ M}\end{array}$

  $\begin{array}{l}{\text{ Zn}}^{\text{+2}}\left(\text{aq}\right)+{\text{ S}}^{\text{2-}}\left(\text{aq}\right)\rightleftharpoons \text{ ZnS}\left(\text{s}\right)\\ \\ 0.15{\text{ M Zn}}^{\text{+2}}\text{reacts with }1\times {10}^{-4}{\text{ M S}}^{\text{2-}}\text{to form }1\times {10}^{-4}\text{ M ZnS}\end{array}$

Number of moles of zinc sulphide formed is,

  $\begin{array}{c}\text{Number of moles = Molarity × Volume}\\ \\ \text{Number of moles=}1\times {10}^{-4}\text{ M ZnS}\times 0.02\text{L}\\ \\ \text{Number of moles}=\text{2×1}{0}^{-6}\text{ mol}\end{array}$

Mass of zinc sulphide formed is,

  $\begin{array}{c}\text{Mass =Number of moles}\times \text{Molar mass}\\ \text{=2×1}{0}^{-6}\text{ mol}\times \text{97}\text{.45}\text{g}/\text{mol}\\ =1.{\text{9×10}}^{-4}\text{g}\\ =1.{\text{9×10}}^{-4}\text{g}\times \frac{\text{1000 mg}}{\text{1g}}\\ \text{Mass}=0.19\text{mg}\end{array}$

Mass of zinc sulphide formed is calculated as above.

(b)

Interpretation Introduction

Interpretation:

The concentration of $\left[{\text{Zn}}^{\text{+2}}\right]\text{and }\left[{\text{S}}_{}{}^{2-}\right]$ at equilibrium should be determined.

Answer to Problem 22.38P

The concentration of $\left[{\text{Zn}}^{\text{+2}}\right]\text{and }\left[{\text{S}}_{}{}^{2-}\right]$ at equilibrium is calculated to be $\text{0}\text{.15M}$ and $1.1\times {10}^{-23}\text{ M}$ respectively.

Explanation of Solution

Given data is as follows:

  $\begin{array}{l}\text{10}\text{.0 mL}\text{of}\text{0}\text{.30 M Zn}{\left({\text{NO}}_3\right)}_2\\ \\ \text{10}\text{.0 mL}\text{of}\text{2}\text{.0}\times {\text{10}}^{-4}{\text{M Na}}_2\text{S}\end{array}$

The chemical equation for the precipitation reaction is,

  ${\text{Zn}}^{\text{+2}}\left(\text{aq}\right)+{\text{ S}}^{\text{2-}}\left(\text{aq}\right)\rightleftharpoons \text{ ZnS}\left(\text{s}\right)$

The precipitation reaction has to run to completion.

  $\begin{array}{l}{\text{ Zn}}^{\text{+2}}+{\text{ S}}^{2-}\to \text{ ZnS}\\ \\ \text{Before 10mL}\times \text{0}\text{.30M 10mL}\times \text{2}\times {\text{10}}^{-4}\text{M}\\ \text{reaction =3 mmol = 2}\times {\text{10}}^{-3}\text{ mmol}\\ \\ \text{After 3}-\left(\text{2}\times {\text{10}}^{-3}\right)=\text{2}\times {\text{10}}^{-3}-\text{2}\times {\text{10}}^{-3}\\ \text{reaction = 3 mmol = 0}\end{array}$

There will be excess ${\text{Zn}}^{\text{+2}}$ after the precipitation reaction goes to completion and its concentration in given volume can be determined as given below:

  $\begin{array}{c}\text{Molarity =}\frac{\text{Number of moles}}{\text{Volume}}\\ \\ \text{=}\frac{\text{3 mmol}}{\text{20 mL}}\\ \\ {\left[{\text{Zn}}^{\text{+2}}\right]}_{\text{excess}}\text{= 0}\text{.15M}\end{array}$

The concentration of ${\text{S}}^{\text{2-}}$ will not be zero at equilibrium even though it is assumed that ${\text{S}}^{\text{2-}}$ is completely consumed. The concentration of sulphide ion at equilibrium can be calculated by allowing $\text{ZnS}$ redissolve to satisfy the expression for solubility product.

$\begin{array}{l}\text{Initial concentration }\text{Equilibrium concentration}\\ \\ {\left[{\text{Zn}}^{\text{+2}}\right]}_0=0.15M\stackrel{\text{x mol}}{\to }\left[{\text{Zn}}^{\text{+2}}\right]=0.15+x\\ \\ {\left[{\text{S}}^{\text{2-}}\right]}_0\text{=0 }\left[{\text{S}}^{\text{2-}}\right]\text{=x}\\ \end{array}$

Equilibrium concentration can be determined as follows:

  $\begin{array}{l}{K}_{sp}=\left[{\text{Zn}}^{\text{+2}}\right]\left[{\text{S}}^{\text{2-}}\right]=1.6\times {10}^{-24}{M}^2\\ \\ 1.6\times {10}^{-24}{M}^2\text{=}\left[\text{0}\text{.15+x}\right]\text{x }\left[\text{x can be neglected from 0}\text{.15+x}\right]\\ \\ 1.6\times {10}^{-24}{M}^2\text{=}\left[\text{0}\text{.15}\right]\text{x}\\ \\ \text{x = 1}\text{.1}\times {\text{10}}^{-23}\text{ M }\\ \\ \left[{\text{Zn}}^{\text{+2}}\right]=0.15+1.1\times {10}^{-23}\text{ M = 0}\text{.15M}\\ \left[{\text{S}}^{\text{2-}}\right]=\text{1}\text{.1}\times {\text{10}}^{-23}\text{ M }\end{array}$

The concentration of $\left[{\text{Zn}}^{\text{+2}}\right]\text{and }\left[{\text{S}}_{}{}^{2-}\right]$ at equilibrium is calculated to be $\text{0}\text{.15M}$ and $1.1\times {10}^{-23}\text{ M}$ respectively.