Chapter 11, Problem 115QP

(a)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the chemical reaction is to be determined.

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100\text{ M}$ and of potassium sulfate solution is $0.500\text{ M}$ . The volume of potassium sulfate solution is $90\text{ ml}$ and of barium chloride solution is $500\text{ ml}$ .

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 the reactants, an insoluble end product is formed which precipitates out from the solution.

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

${\text{BaCl}}_2\left(aq\right)+{\text{K}}_2{\text{SO}}_4\left(aq\right)\to {\text{BaSO}}_4\left(s\right)+\text{2KCl}\left(aq\right)$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

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 $90\text{ ml}$ volume of potassium sulfate solution from milliliters to liters as follows:

$\begin{array}{c}\text{90 mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 90\text{ mL}\\ =0.09\text{L}\end{array}$

Convert $500\text{ ml}$ volume of barium chloride solution from milliliters to liters as follows:

$\begin{array}{c}500\text{ mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 500\text{ mL}\\ =0.5\text{L}\end{array}$

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 $\text{0}{\text{.1 mol L}}^{-1}$ and $V$ as $0.5\text{ L}$ in equation (1) to determine the number of moles of barium chloride as follows:

$\begin{array}{c}\text{0}{\text{.1 mol L}}^{-1}=\frac{n}{0.5\text{L}}\\ n=0.1{\text{ mol L}}^{-1}\times 0.5\text{L}\\ =0.05\text{ moles}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.5 mol L}}^{-1}$ and $V$ as $0.09\text{ L}$ in equation (1) to determine the number of moles of potassium sulphate as follows:

$\begin{array}{c}\text{0}{\text{.5 mol L}}^{-1}=\frac{n}{0.09\text{ L}}\\ n=0.5{\text{ mol L}}^{-1}\times 0.09\text{L}\\ =0.045\text{ moles}\end{array}$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate. Therefore,

$\begin{array}{c}{\text{1 mol BaCl}}_2\text{ reacts with}=1{\text{ mol K}}_2{\text{SO}}_4\\ \text{0}{\text{.05 mol BaCl}}_2\text{ reacts with}=\frac{1{\text{ mol K}}_2{\text{SO}}_4}{{\text{1 mol BaCl}}_2}\times \text{0}{\text{.05 mol BaCl}}_2\\ =\text{0}{\text{.05 mol K}}_2{\text{SO}}_4\end{array}$

Barium sulfate formed is as follows:

$\begin{array}{c}{\text{1 mol K}}_2{\text{SO}}_4\text{ forms}=1{\text{ mol BaSO}}_4\\ 0.05{\text{ mol K}}_2{\text{SO}}_4\text{ forms}=\frac{1{\text{ mol BaSO}}_4}{{\text{1 mol K}}_2{\text{SO}}_4}\times 0.05{\text{ mol K}}_2{\text{SO}}_4\\ =0.05{\text{ mol BaSO}}_4\end{array}$

However, only $0.045\text{ mol}$ of ${\text{K}}_2{\text{SO}}_4$ is present. Therefore, $0.045\text{ mol}$ of ${\text{K}}_2{\text{SO}}_4$ reacts with $0.045\text{ mol}$ of ${\text{BaCl}}_2$ to produce $0.045\text{ mol}$ of ${\text{BaSO}}_4$ .

The molar mass of ${\text{BaSO}}_4$ is as follows:

$\begin{array}{c}{\text{BaSO}}_4=137.38{\text{ g mol}}^{-1}+32{\text{ g mol}}^{-1}+4\times \left(16{\text{ g mol}}^{-1}\right)\\ =233.38{\text{ g mol}}^{-1}\end{array}$

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 $\text{0}\text{.045 moles}$ and $MM$ as $233.38{\text{ g mol}}^{-1}$ in equation (2) to determine the mass of ${\text{BaSO}}_4$ as follows:

$\begin{array}{c}\text{0}\text{.045 mol}=\frac{m}{\text{233}{\text{.38 g mol}}^{-1}}\\ m=0.045\text{ mol}\times 233.38{\text{g mol}}^{-1}\\ =10.5021\text{ g}\end{array}$

Thus, the mass of barium sulfate formed is $\text{l0}\text{.502l g}$ .

(b)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100\text{ M}$ and of potassium sulfate solution is $0.500\text{ M}$ . The volume of potassium sulfate solution is $100\text{ ml}$ and of barium chloride solution is $100\text{ ml}$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

${\text{BaCl}}_2\left(aq\right)+{\text{K}}_2{\text{SO}}_4\left(aq\right)\to {\text{BaSO}}_4\left(s\right)+\text{2KCl}\left(aq\right)$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

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 potassium sulfate solution from milliliters to liters as follows:

$\begin{array}{c}\text{100 mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 100\text{ mL}\\ =0.01\text{L}\end{array}$

Convert $100\text{ ml}$ volume of barium chloride 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 $\text{0}{\text{.1 mol L}}^{-1}$ and $V$ as $0.1\text{ L}$ in equation (1) to determine the number of moles of barium chloride as follows:

$\begin{array}{c}\text{0}{\text{.1 mol L}}^{-1}=\frac{n}{0.1\text{L}}\\ n=0.1{\text{ mol L}}^{-1}\times 0.1\text{ L}\\ =0.01\text{ moles}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.5 mol L}}^{-1}$ and $V$ as $0.1\text{ L}$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$\begin{array}{c}\text{0}{\text{.5 mol L}}^{-1}=\frac{n}{0.1\text{L}}\\ n=0.5{\text{ mol L}}^{-1}\times 0.1\text{ L}\\ =0.05\text{ moles}\end{array}$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$\begin{array}{c}{\text{1 mol BaCl}}_2\text{ reacts}=1{\text{ mol K}}_2{\text{SO}}_4\\ 0.01{\text{ mol BaCl}}_2\text{ reacts}=\frac{1{\text{ mol K}}_2{\text{SO}}_4}{{\text{1 mol BaCl}}_2}\times 0.01{\text{ mol BaCl}}_2\\ =0.01{\text{ mol K}}_2{\text{SO}}_4\end{array}$

The number of moles of ${\text{BaSO}}_4$ produced as follows:

$\begin{array}{c}1{\text{ mol K}}_2{\text{SO}}_4\text{ produces}=1{\text{ mol BaSO}}_4\\ 0.01{\text{ mol K}}_2{\text{SO}}_4\text{ produces}=\frac{1{\text{ mol BaSO}}_4}{1{\text{ mol K}}_2{\text{SO}}_4}\times 0.01{\text{ mol K}}_2{\text{SO}}_4\\ =0.01{\text{ mol BaSO}}_4\end{array}$

The molar mass of ${\text{BaSO}}_4$ is as follows:

$\begin{array}{c}{\text{BaSO}}_4=137.38{\text{ g mol}}^{-1}+32{\text{ g mol}}^{-1}+4\times \left(16{\text{ g mol}}^{-1}\right)\\ =233.38{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $\text{0}\text{.01 moles}$ and $MM$ as $233.38{\text{ g mol}}^{-1}$ in equation (2) to determine the mass of ${\text{BaSO}}_4$ as follows:

$\begin{array}{c}\text{0}\text{.01 mol}=\frac{m}{\text{233}{\text{.38 g mol}}^{-1}}\\ m=0.01\text{ mol}\times 233.38{\text{g mol}}^{-1}\\ =2.3338\text{ g}\end{array}$

Thus, the mass of barium sulfate formed is $2.3338\text{ g}$ .

(c)

Interpretation Introduction

Interpretation:

The mass of barium sulfate formed after the completion of the given chemical reaction is to be determined.

Explanation of Solution

Given Information:

The molarity of barium chloride solution is $0.100\text{ M}$ and of potassium sulfate solution is $0.500\text{ M}$ . The volume of potassium sulfate solution is $500\text{ ml}$ and of barium chloride solution is $100\text{ ml}$ .

The chemical reaction for the formation of barium sulfate on the reaction of barium chloride and potassium sulfate is,

${\text{BaCl}}_2\left(aq\right)+{\text{K}}_2{\text{SO}}_4\left(aq\right)\to {\text{BaSO}}_4\left(s\right)+\text{2KCl}\left(aq\right)$

Thus, one mole of barium chloride reacts with one mole of potassium sulfate to form a mole of barium sulfate.

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 $500\text{ ml}$ volume of potassium sulfate solution from milliliters to liters as follows:

$\begin{array}{c}\text{500 mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 500\text{ mL}\\ =0.5\text{L}\end{array}$

Convert $100\text{ ml}$ volume of barium chloride 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 $\text{0}{\text{.1 mol L}}^{-1}$ and $V$ as $0.1\text{ L}$ in equation (1) to determine the number of moles of barium chloride as follows:

$\begin{array}{c}\text{0}{\text{.1 mol L}}^{-1}=\frac{n}{0.1\text{L}}\\ n=0.1{\text{ mol L}}^{-1}\times 0.1\text{ L}\\ =0.01\text{ moles}\end{array}$

Substitute $\text{M}$ as $\text{0}{\text{.5 mol L}}^{-1}$ and $V$ as $0.5\text{ L}$ in equation (1) to determine the number of moles of potassium sulfate as follows:

$\begin{array}{c}\text{0}{\text{.5 mol L}}^{-1}=\frac{n}{0.5\text{L}}\\ n=0.5{\text{ mol L}}^{-1}\times 0.5\text{ L}\\ =0.25\text{ moles}\end{array}$

From the equation, it can be summarized that one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

$\begin{array}{l}\therefore \text{ 0}\text{.25 moles of K2SO4 must react with 0}\text{.25 mole of BaCl2 to form 0}\text{.25 mole of BaSO4}\\ \text{But there are only 0}\text{.01 mole of BaCl2}\text{. }\\ \therefore \text{ 0}\text{.01 moles of BaCl2 reacts with 0}\text{.01 moles of K2 SO4 to form 0}\text{.01 moles of BaSO4}\end{array}$

$\begin{array}{c}{\text{1 mol BaCl}}_2\text{ reacts}=1{\text{ mol K}}_2{\text{SO}}_4\\ 0.25{\text{ mol K}}_2{\text{SO}}_4\text{ reacts}=\frac{1{\text{ mol BaCl}}_2}{{\text{1 mol K}}_2{\text{SO}}_4}\times 0.25{\text{ mol K}}_2{\text{SO}}_4\\ =0.25{\text{ mol BaCl}}_2\end{array}$

However, the number of moles of ${\text{BaCl}}_2$ is only $0.01\text{ mol}$ .

$\begin{array}{c}\text{0}{\text{.01 mol BaCl}}_2\text{ reacts}=\frac{1{\text{ mol K}}_2{\text{SO}}_4}{{\text{1 mol BaCl}}_2}\times \text{0}{\text{.01 mol BaCl}}_2\\ =\text{0}{\text{.01 mol K}}_2{\text{SO}}_4\end{array}$

The amount of ${\text{BaSO}}_4$ formed is as follows:

$\begin{array}{c}1{\text{ mol K}}_2{\text{SO}}_4\text{ produces}=1{\text{ mol BaSO}}_4\\ \text{0}{\text{.01 mol K}}_2{\text{SO}}_4\text{ produces}=\frac{1{\text{ mol BaSO}}_4}{1{\text{ mol K}}_2{\text{SO}}_4}\times \text{0}{\text{.01 mol K}}_2{\text{SO}}_4\\ =\text{0}{\text{.01 mol BaSO}}_4\end{array}$

The molar mass of ${\text{BaSO}}_4$ is as follows:

$\begin{array}{c}{\text{BaSO}}_4=137.38{\text{ g mol}}^{-1}+32{\text{ g mol}}^{-1}+4\times \left(16{\text{ g mol}}^{-1}\right)\\ =233.38{\text{ g mol}}^{-1}\end{array}$

Substitute $n$ as $\text{0}\text{.01 moles}$ and $MM$ as $233.38{\text{ g mol}}^{-1}$ in equation (2) to determine the mass of ${\text{BaSO}}_4$ as follows:

$\begin{array}{c}\text{0}\text{.01 mol}=\frac{m}{\text{233}{\text{.38 g mol}}^{-1}}\\ m=0.01\text{ mol}\times 233.38{\text{g mol}}^{-1}\\ =2.3338\text{ g}\end{array}$

Thus, the mass of barium sulfate formed is $2.3338\text{ g}$ .