Chapter 15, Problem 32QAP

A solution is prepared by mixing 45.00 mL of 0.022 M AgNO₃ with 13.00 mL of 0.0014 M Na₂CO₃. Assume that volumes are additive.

(a) Will precipitation occur?

(b) Calculate [Ag⁺], [CO₃^2-], [Na⁺], and [NO₃^-] after equilibrium is established.

Interpretation Introduction

(a)

Interpretation:

Whether the reaction undergoes precipitation or not needs to be determined.

Concept introduction:

For a reaction, if ionic ( Q) product of the compound is greater than solubility product of compound, then precipitation occurs.

Answer to Problem 32QAP

The reaction undergoes precipitation.

Explanation of Solution

It is given that volume of $\text{0}\text{.022}\text{M}{\text{AgNO}}_{\text{3}}$ is $\text{45}\text{.00}\text{mL}$.

Volume of $\text{0}\text{.0014}\text{M}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{13}\text{.00}\text{mL}$.

Reaction of ${\text{AgNO}}_{\text{3}}$ with ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is as follows:

${\text{AgNO}}_{\text{3}}{\text{+Na}}_{\text{2}}{\text{CO}}_{\text{3}}\to {\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}+{\text{NaNO}}_{\text{3}}$

$\begin{array}{c}\text{Total}\text{volume}\text{of}\text{solution=13}\text{mL+45 mL}\\ \text{=58}\text{mL}\end{array}$

Molar concentration of $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{CO}}_{\text{3}}{}^{2-}\right]$ in $\text{58}\text{mL}$ solution are as follows:

$\begin{array}{c}\left[{\text{Ag}}^{+}\right]=\frac{0.022\text{M}\times \text{45}\text{mL}}{58\text{mL}}\\ =1.7\times {10}^{-2}\text{M}\end{array}$

So, $\left[{\text{Ag}}^{+}\right]$ ion concentration is $1.7\times {10}^{-2}\text{M}$.

$\begin{array}{c}\left[{\text{CO}}_{\text{3}}{}^{2-}\right]=\frac{0.0014\text{M}\times 13\text{mL}}{58\text{mL}}\\ =3.13\times {10}^{-4}\text{M}\end{array}$

So, $\left[{\text{CO}}_{\text{3}}{}^{2-}\right]$ ion concentration is $3.13\times {10}^{-4}\text{M}$.

Calculate ionic product (Q) for ${\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}$

$\begin{array}{c}Q={\left[{\text{Ag}}^{+}\right]}^2\left[{\text{CO}}_3{}^{2-}\right]\\ ={\left(1.7\times {10}^{-2}\right)}^2\times 3.13\times {10}^{-4}\\ \text{=9}\text{.05}\times {\text{10}}^{-8}\end{array}$

So, ionic product ( Q)of silver carbonate $\left({\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ is $\text{9}\text{.05}\times {\text{10}}^{-8}$.

Solubility of silver carbonate $\left({\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ is $8.1\times {10}^{-12}$

$Q>\text{Solubility}\text{product}$

So, precipitation of silver carbonate $\left({\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ occurs.

Interpretation Introduction

(b)

Interpretation:

The equilibrium concentration of $\left[{\text{Ag}}^{+}\right],\left[{\text{CO}}_{\text{3}}{}^{2-}\right],\left[{\text{Na}}^{+}\right],\left[{\text{NO}}_3{}^{-}\right]$ needs to be determined.

Concept introduction:

The equilibrium reaction after mixing solution of ${\text{AgNO}}_3$ and ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is,

${\text{AgNO}}_3\left(\text{aq}\right){\text{+Na}}_2{\text{CO}}_{\text{3}}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{NaNO}}_{\text{3}}\left(\text{aq}\right)$

The molarity of solution is calculated as follows:

$M=\frac{n}{V}$

Here, n is number of moles of solute and V is volume of solution in L.

Answer to Problem 32QAP

$\begin{array}{l}\left[{\text{Ag}}^{+}\right]=0.0164\text{M}\\ \left[{\text{CO}}_{\text{3}}{}^{2-}\right]=3.125\times {10}^{-8}\text{M}\\ \left[{\text{Na}}^{+}\right]=6.22\times {10}^{-4}\text{M}\\ \left[{\text{NO}}_3{}^{-}\right]=1.72\times {10}^{-2}\text{M}\end{array}$

Explanation of Solution

The equilibrium reaction after mixing solution of ${\text{AgNO}}_3$ and ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is,

${\text{AgNO}}_3\left(\text{aq}\right){\text{+Na}}_2{\text{CO}}_{\text{3}}\left(\text{aq}\right)\rightleftharpoons {\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)+2{\text{NaNO}}_{\text{3}}\left(\text{aq}\right)$   ....... (1)

$\begin{array}{c}\text{Moles}\text{of}{\text{AgNO}}_3=0.022\text{M}\times \text{45}\text{ml}\\ \text{=0}\text{.99}\text{mol}\end{array}$

$\begin{array}{c}\text{Moles}\text{of}{\text{Na}}_{\text{2}}{\text{CO}}_3=0.0014\text{M}\times 13\text{ml}\\ \text{=0}\text{.0182}\text{mol}\end{array}$

As 0.0182 mol of ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is completely utilized to react with 0.99 mol of ${\text{AgNO}}_3$. ${\text{Na}}_2{\text{CO}}_{\text{3}}$ is the limiting reagent.

ICE table for equation (1) is,

$\begin{array}{l}2{\text{AgNO}}_3\left(\text{aq}\right){\text{+Na}}_2{\text{CO}}_{\text{3}}\left(\text{aq}\right)\rightleftharpoons 2{\text{NaNO}}_{\text{3}}\left(\text{aq}\right)\\ \text{Initial}\text{0}\text{.99}\text{0}\text{.0182}\text{0}\\ \text{Change}\text{0}\text{.0364}\text{0}\text{.0182}\text{0}\text{.0182}\\ \text{Equilibrium}0.9540\text{0}\text{.0182}\end{array}$

$\begin{array}{c}\text{Total}\text{volume}\text{of}\text{solution=13}\text{mL+45}\text{mL}\\ \text{=58 mL}\end{array}$

Equilibrium concentration of Ag ion is,

$\begin{array}{c}\left[{\text{Ag}}^{+}\right]=\frac{0.954\text{mol}}{58\text{mL}}\\ =0.0164\end{array}$

Equilibrium concentration of NO₃ ion is,

$\begin{array}{c}\left[{\text{NO}}_{\text{3}}{}^{-}\right]=\frac{0.022\text{M}\times \text{45 mL}}{58\text{mL}}\\ =1.72\times {10}^{-2}\text{M}\end{array}$

Equilibrium concentration of NO₃ ion is,

$\begin{array}{c}\left[{\text{Na}}^{+}\right]=\frac{0.0182\text{mol}\times 2}{58\text{mL}}\\ =6.22\times {10}^{-4}\text{M}\end{array}$

Equilibrium concentration of ${\text{CO}}_{\text{3}}$ ion is,

$\begin{array}{c}{K}_{sp}\left({\text{Ag}}_2{\text{CO}}_{\text{3}}\right)=8\times {10}^{-12}\\ K_{sp}\left({\text{Ag}}_2{\text{CO}}_{\text{3}}\right)={\left[{\text{Ag}}^{+}\right]}^2\left[{\text{CO}}_{\text{3}}{}^{2-}\right]\end{array}$

$\begin{array}{c}\left[{\text{CO}}_{\text{3}}{}^{2-}\right]=\frac{K_{sp}\left({\text{Ag}}_{\text{2}}{\text{CO}}_{\text{3}}\right)}{{\left[{\text{Ag}}^{+}\right]}^2}\\ =\frac{8\times {10}^{-12}}{{\left(1.6\times {10}^{-2}\right)}^2}\\ =3.125\times {10}^{-8}\text{M}\end{array}$