Chapter 11, Problem 114QP

(a)

Interpretation Introduction

Interpretation:

The mass of silver carbonate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Given Information:

The molarity of silver nitrate solution is $0.200\text{M}$ and of sodium carbonate solution is $0.200\text{ M}$ . The volume of sodium carbonate solution is $100\text{ ml}$ and of silver nitrate solution is $100\text{ ml}$ .

Titration is a method to determine the concentration of a substance in the solution by making it react with a solution of known concentration of other substance, just beyond the point where the reaction between both the substances completes. In precipitation reactions, on the reaction of reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of silver carbonate on the reaction of silver nitrate and sodium carbonate is,

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

Thus, two moles of silver nitrate react with one mole of sodium carbonate to form a mole of silver carbonate.

Convert volume units from milliliters to liters as follows:

$\begin{array}{c}1\text{ L}=1000\text{ mL}\\ \text{1 mL}=\frac{1\text{ L}}{1000\text{ mL}}\end{array}$

Convert $100\text{ ml}$ volume of silver nitrate solution from milliliters to liters as follows:

$\begin{array}{c}\text{100 mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 100\text{ mL}\\ =0.1\text{L}\end{array}$

Convert $100\text{ ml}$ volume of sodium carbonate solution from milliliters to liters as follows:

$\begin{array}{c}100\text{ mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 100\text{ mL}\\ =0.1\text{L}\end{array}$

Molarity of the solution $\left(\text{M}\right)$ is equal to the ratio of the number of moles $\left(n\right)$ of a solute to the volume $\left(V\right)$ of the solution in litres.

$\text{M}=\frac{n}{V}$ …… (1)

Substitute $\text{M}$ as $\text{0}{\text{.2 mol L}}^{-1}$ and $V$ as $0.1\text{ L}$ in equation (1) to determine the number of moles of silver nitrate as follows:

$\begin{array}{c}\text{0}{\text{.2 mol L}}^{-1}=\frac{n}{0.1\text{L}}\\ n=0.2{\text{ mol L}}^{-1}\times 0.1\text{L}\\ =0.02\text{ moles}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.2 mol L}}^{-1}$ and $V$ as $0.01\text{ L}$ in equation (1) to determine the number of moles of sodium carbonate as follows:

$\begin{array}{c}\text{0}{\text{.2 mol L}}^{-1}=\frac{n}{0.1\text{ L}}\\ n=0.2{\text{ mol L}}^{-1}\times 0.1\text{L}\\ =0.02\text{ moles}\end{array}$

From the equation, it can be summarized that two moles of silver nitrate react with one mole of sodium carbonate to produce one mole of silver carbonate. Therefore,

$\begin{array}{c}{\text{2 mol AgNO}}_3\text{ reacts with}=1{\text{ mol Na}}_2{\text{CO}}_3\\ \text{0}{\text{.02 mol AgNO}}_3\text{ reacts with}=\frac{1{\text{ mol Na}}_2{\text{CO}}_3}{{\text{2 mol AgNO}}_3}\times \text{0}{\text{.02 mol AgNO}}_3\\ =0.01{\text{ mol Na}}_2{\text{CO}}_3\end{array}$

Silver carbonate formed is as follows:

$\begin{array}{c}1{\text{ mol Na}}_2{\text{CO}}_3=1{\text{ mol Ag}}_2{\text{CO}}_3\\ 0.01{\text{ mol Na}}_2{\text{CO}}_3=\frac{1{\text{ mol Ag}}_2{\text{CO}}_3}{1{\text{ mol Na}}_2{\text{CO}}_3}\times 0.01{\text{ mol Na}}_2{\text{CO}}_3\\ =0.01{\text{ mol Ag}}_2{\text{CO}}_3\end{array}$

The molar mass of ${\text{Ag}}_2{\text{CO}}_3$ is as follows:

$\begin{array}{c}{\text{Ag}}_2{\text{CO}}_3=2\times \left(107.87{\text{ g mol}}^{-1}\right)+12{\text{ g mol}}^{-1}+3\times \left(16{\text{ g mol}}^{-1}\right)\\ =275.74{\text{ g mol}}^{-1}\end{array}$

The number of moles $\left(n\right)$ of a substance is equal to the ratio of the mass $\left(m\right)$ of that substance to the molar mass $\left(MM\right)$ of the substance.

$n=\frac{m}{MM}$ …… (2)

Substitute $n$ as $\text{0}\text{.01 moles}$ and $MM$ as $275.74{\text{ g mol}}^{-1}$ in equation (2) to determine the mass of ${\text{Ag}}_2{\text{CO}}_3$ as follows:

$\begin{array}{c}\text{0}\text{.01 mol}=\frac{m}{\text{275}{\text{.74 g mol}}^{-1}}\\ m=0.01\text{ mol}\times 275.74{\text{g mol}}^{-1}\\ =2.76\text{ g}\end{array}$

Thus, the mass of silver carbonate formed is $2.76\text{ g}$ .

(b)

Interpretation Introduction

Interpretation:

The mass of silver carbonate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Given Information:

The molarity of silver nitrate solution is $0.200\text{ M}$ and of sodium carbonate solution is $0.200\text{ M}$ . The volume of sodium carbonate solution is $100\text{ ml}$ and of silver nitrate solution is $200\text{ ml}$ .

The chemical reaction for the formation of silver carbonate on the reaction of silver nitrate and sodium carbonate is,

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

Thus, two moles of silver nitrate react with one mole of sodium carbonate to form a mole of silver carbonate.

Convert volume units from milliliters to liters as follows:

$\begin{array}{c}1\text{ L}=1000\text{ mL}\\ \text{1 mL}=\frac{1\text{ L}}{1000\text{ mL}}\end{array}$

Convert $200\text{ ml}$ volume of silver nitrate solution from milliliters to liters as follows:

$\begin{array}{c}\text{200 mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 200\text{ mL}\\ =0.2\text{L}\end{array}$

Convert $100\text{ ml}$ volume of sodium carbonate solution from milliliters to liters as follows:

$\begin{array}{c}100\text{ mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 100\text{ mL}\\ =0.1\text{L}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.2 mol L}}^{-1}$ and $V$ as $0.2\text{ L}$ in equation (1) to determine the number of moles of silver nitrate as follows:

$\begin{array}{c}\text{0}{\text{.2 mol L}}^{-1}=\frac{n}{0.2\text{L}}\\ n=0.2{\text{ mol L}}^{-1}\times 0.2\text{L}\\ =0.04\text{ moles}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.2 mol L}}^{-1}$ and $V$ as $0.01\text{ L}$ in equation (1) to determine the number of moles of sodium carbonate as follows:

$\begin{array}{c}\text{0}{\text{.2 mol L}}^{-1}=\frac{n}{0.1\text{ L}}\\ n=0.2{\text{ mol L}}^{-1}\times 0.1\text{L}\\ =0.02\text{ moles}\end{array}$

From the equation, it can be summarized that two moles of silver nitrate react with one mole of sodium carbonate to produce one mole of silver carbonate.

$\begin{array}{c}{\text{2 mol AgNO}}_3\text{ reacts with}=1{\text{ mol Na}}_2{\text{CO}}_3\\ \text{0}{\text{.04 mol AgNO}}_3\text{ reacts with}=\frac{1{\text{ mol Na}}_2{\text{CO}}_3}{{\text{2 mol AgNO}}_3}\times \text{0}{\text{.04 mol AgNO}}_3\\ =0.02{\text{ mol Na}}_2{\text{CO}}_3\end{array}$

Silver carbonate formed is as follows:

$\begin{array}{c}1{\text{ mol Na}}_2{\text{CO}}_3=1{\text{ mol Ag}}_2{\text{CO}}_3\\ 0.02{\text{ mol Na}}_2{\text{CO}}_3=\frac{1{\text{ mol Ag}}_2{\text{CO}}_3}{1{\text{ mol Na}}_2{\text{CO}}_3}\times 0.02{\text{ mol Na}}_2{\text{CO}}_3\\ =0.02{\text{ mol Ag}}_2{\text{CO}}_3\end{array}$

The molar mass of ${\text{Ag}}_2{\text{CO}}_3$ is as follows:

$\begin{array}{c}{\text{Ag}}_2{\text{CO}}_3=2\times \left(107.87{\text{ g mol}}^{-1}\right)+12{\text{ g mol}}^{-1}+3\times \left(16{\text{ g mol}}^{-1}\right)\\ =275.74{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $\text{0}\text{.02 moles}$ and $MM$ as $275.74{\text{ g mol}}^{-1}$ in equation (2) to determine the mass of ${\text{Ag}}_2{\text{CO}}_3$ as follows:

$\begin{array}{c}\text{0}\text{.02 mol}=\frac{m}{\text{275}{\text{.74 g mol}}^{-1}}\\ m=0.02\text{ mol}\times 275.74{\text{g mol}}^{-1}\\ =5.52\text{ g}\end{array}$

Thus, the mass of silver carbonate formed is $5.52\text{ g}$ .

(c)

Interpretation Introduction

Interpretation:

The mass of silver carbonate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Given Information:

The molarity of silver nitrate solution is $0.200\text{ M}$ and of sodium carbonate solution is $0.200\text{ g}$ . The volume of sodium carbonate solution is $200\text{ ml}$ and of silver nitrate solution is $100\text{ ml}$ .

The chemical reaction for the formation of silver carbonate on the reaction of silver nitrate and sodium carbonate is,

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

Thus, two moles of silver nitrate react with one mole of sodium carbonate to form a mole of silver carbonate.

Convert volume units from milliliters to liters as follows:

$\begin{array}{c}1\text{ L}=1000\text{ mL}\\ \text{1 mL}=\frac{1\text{ L}}{1000\text{ mL}}\end{array}$

Convert $100\text{ ml}$ volume of silver nitrate solution from milliliters to liters as follows:

$\begin{array}{c}\text{100 mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 100\text{ mL}\\ =0.1\text{L}\end{array}$

Convert $200\text{ ml}$ volume of sodium carbonate solution from milliliters to liters as follows:

$\begin{array}{c}200\text{ mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 200\text{ mL}\\ =0.2\text{L}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.2 mol L}}^{-1}$ and $V$ as $0.1\text{ L}$ in equation (1) to determine the number of moles of silver nitrate as follows:

$\begin{array}{c}\text{0}{\text{.2 mol L}}^{-1}=\frac{n}{0.1\text{L}}\\ n=0.2{\text{ mol L}}^{-1}\times 0.1\text{L}\\ =0.02\text{ moles}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.2 mol L}}^{-1}$ and $V$ as $0.02\text{ L}$ in equation (1) to determine the number of moles of sodium carbonate as follows:

$\begin{array}{c}\text{0}{\text{.2 mol L}}^{-1}=\frac{n}{0.2\text{ L}}\\ n=0.2{\text{ mol L}}^{-1}\times 0.2\text{L}\\ =0.04\text{ moles}\end{array}$

From the equation, it can be summarized that two moles of silver nitrate react with one mole of sodium carbonate to produce one mole of silver carbonate.

$\begin{array}{c}{\text{2 mol AgNO}}_3\text{ reacts with}=1{\text{ mol Na}}_2{\text{CO}}_3\\ \text{0}{\text{.02 mol AgNO}}_3\text{ reacts with}=\frac{1{\text{ mol Na}}_2{\text{CO}}_3}{{\text{2 mol AgNO}}_3}\times \text{0}{\text{.02 mol AgNO}}_3\\ =0.01{\text{ mol Na}}_2{\text{CO}}_3\end{array}$

Silver carbonate formed is as follows:

$\begin{array}{c}1{\text{ mol Na}}_2{\text{CO}}_3=1{\text{ mol Ag}}_2{\text{CO}}_3\\ 0.01{\text{ mol Na}}_2{\text{CO}}_3=\frac{1{\text{ mol Ag}}_2{\text{CO}}_3}{1{\text{ mol Na}}_2{\text{CO}}_3}\times 0.01{\text{ mol Na}}_2{\text{CO}}_3\\ =0.01{\text{ mol Ag}}_2{\text{CO}}_3\end{array}$

The molar mass of ${\text{Ag}}_2{\text{CO}}_3$ is as follows:

$\begin{array}{c}{\text{Ag}}_2{\text{CO}}_3=2\times \left(107.87{\text{ g mol}}^{-1}\right)+12{\text{ g mol}}^{-1}+3\times \left(16{\text{ g mol}}^{-1}\right)\\ =275.74{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $\text{0}\text{.01 moles}$ and $MM$ as $275.74{\text{ g mol}}^{-1}$ in equation (2) to determine the mass of ${\text{Ag}}_2{\text{CO}}_3$ as follows:

$\begin{array}{c}\text{0}\text{.01 mol}=\frac{m}{\text{275}{\text{.74 g mol}}^{-1}}\\ m=0.01\text{ mol}\times 275.74{\text{g mol}}^{-1}\\ =2.76\text{ g}\end{array}$

Thus, the mass of silver carbonate formed is $2.76\text{ g}$ .