Chapter 4, Problem 102QP

(a)

Interpretation Introduction

Interpretation:

The molarity of the nitric acid solution.

Explanation of Solution

The number of moles is calculated from the following expression:

$n=\frac{m}{M_m}$ …… (1)

Here, $m$ is the mass of solute, $M_m$ is the molar mass of solute, and $n$ is the number of moles of solute.

The conversion of volume from milliliters to liters is given as follows:

$\text{Volume}\left(\text{L}\right)=x\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}$ …… (2)

The molarity of the solution is determined by the following expression:

$M=\frac{n}{V}$ …… (3)

Here, $n$ is the number of moles of solute, $V$ is the volume of the solution in liters, and $M$ is the molarity of the solution.

The molar mass of nitric acid is $63.01{\text{ gmol}}^{-1}$ .

Substitute $\text{6}\text{.30 g}$ for $m$ and $63.01{\text{ gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of ${\text{HNO}}_{\text{3}}$ as follows:

$\begin{array}{c}{n}_{{\text{HNO}}_3}=6.30\cancel{{\text{g HNO}}_3}\times \frac{{\text{1 mol HNO}}_3}{63.01\cancel{{\text{g HNO}}_3}}\\ =0.100{\text{ mol HNO}}_3\end{array}$

Substitute $255\text{ mL}$ for $x$ in equation (2) to determine the volume of ${\text{HNO}}_{\text{3}}$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=255\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}\\ =0.255\text{ L}\end{array}$

Substitute $0.100\text{ mol}$ for $n$ and $0.255\text{ L}$ for $V$ in equation (3) to determine the molarity of ${\text{HNO}}_{\text{3}}$ solution as follows:

$\begin{array}{c}{M}_{{\text{HNO}}_{\text{3}}}=\frac{0.100{\text{ mol HNO}}_{\text{3}}}{\text{0}\text{.255 L solution}}\\ =0.392{\text{ M HNO}}_{\text{3}}\end{array}$

Therefore, the molarity of ${\text{HNO}}_{\text{3}}$ solution is $0.392\text{ M}$ .

(b)

Interpretation Introduction

Interpretation:

The molarity of the sulfuric acid solution.

Explanation of Solution

The molar mass of sulfuric acid is $\text{98}{\text{.08 gmol}}^{-1}$ .

Substitute $\text{49}\text{.0 g}$ for $m$ and $\text{98}{\text{.08 gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ as follows:

$\begin{array}{c}{n}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}=49.0\cancel{{\text{g H}}_{\text{2}}{\text{SO}}_{\text{4}}}\times \frac{{\text{1 mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.08 }\cancel{{\text{g H}}_{\text{2}}{\text{SO}}_{\text{4}}}}\\ =0.500{\text{ mol H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Substitute $125\text{ mL}$ for $x$ in equation (2) to determine the volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=125\cancel{\text{mL}}\times \frac{\text{1 L}}{1000\cancel{\text{mL}}}\\ =0.125\text{ L}\end{array}$

Substitute $0.500\text{ mol}$ for $n$ and $0.125\text{ L}$ for $V$ in equation (3) to determine the molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution as follows:

$\begin{array}{c}{M}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}=\frac{0.500{\text{ mol H}}_{\text{2}}{\text{SO}}_{\text{4}}}{0.125\text{ L solution}}\\ =4.00{\text{ M H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Therefore, the molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $4.00\text{ M}$ .

(c)

Interpretation Introduction

Interpretation:

The molarity of potassium hydroxide solution.

Explanation of Solution

The molar mass of potassium hydroxide is $\text{56}{\text{.11 gmol}}^{-1}$ .

Substitute $\text{2}\text{.80 g}$ for $m$ and $\text{56}{\text{.11 gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of $\text{KOH}$ as follows:

$\begin{array}{c}{n}_{\text{KOH}}=2.80\cancel{\text{ g KOH}}\times \frac{\text{1 mol KOH}}{\text{56}\text{.11 }\cancel{\text{g KOH}}}\\ =0.05\text{ mol KOH}\end{array}$

Substitute $525\text{ mL}$ for $x$ in equation (2) to determine the volume of $\text{KOH}$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=525\cancel{\text{ml}}\times \frac{\text{1 L}}{1000\cancel{\text{ml}}}\\ =0.525\text{ L}\end{array}$

Substitute $0.05\text{ mol}$ for $n$ and $0.525\text{ L}$ for $V$ in equation (3) to determine the molarity of $\text{KOH}$ solution as follows:

$\begin{array}{c}{M}_{\text{KOH}}=\frac{0.05\text{ mol KOH}}{0.525\text{ L solution}}\\ =0.095\text{ M KOH}\end{array}$

Therefore, the molarity of $\text{KOH}$ solution is $0.095\text{ M}$ .

(d)

Interpretation Introduction

Interpretation:

The molarity of calcium hydroxide solution.

Explanation of Solution

The molar mass of calcium hydroxide is $\text{74}{\text{.10 gmol}}^{-1}$ .

Substitute $\text{7}\text{.40 g}$ for $m$ and $\text{74}{\text{.10 gmol}}^{-1}$ for $M_m$ in equation (1) to determine the number of moles of $\text{Ca}{\left(\text{OH}\right)}_2$ as follows:

$\begin{array}{c}{n}_{\text{Ca}{\left(\text{OH}\right)}_2}=7.40\cancel{\text{g Ca}{\left(\text{OH}\right)}_2}\times \frac{\text{1 mol Ca}{\left(\text{OH}\right)}_2}{\text{74}\text{.10 }\cancel{\text{g Ca}{\left(\text{OH}\right)}_2}}\\ =0.10\text{ mol Ca}{\left(\text{OH}\right)}_2\end{array}$

Substitute $\text{200}\text{.0 mL}$ for $x$ in equation (2) to determine the volume of $\text{Ca}{\left(\text{OH}\right)}_2$ solution in liters as follows:

$\begin{array}{c}\text{Volume}\left(\text{L}\right)=200.0\cancel{\text{ml}}\times \frac{\text{1 L}}{1000\cancel{\text{ml}}}\\ =0.200\text{ L}\end{array}$

Substitute $0.10\text{ mol}$ for $n$ and $0.200\text{ L}$ for $V$ in equation (3) to determine the molarity of $\text{Ca}{\left(\text{OH}\right)}_2$ solution as follows:

$\begin{array}{c}{M}_{\text{Ca}{\left(\text{OH}\right)}_2}=\frac{0.10\text{ mol Ca}{\left(\text{OH}\right)}_2}{0.200\text{ L solution}}\\ =0.50\text{ M Ca}{\left(\text{OH}\right)}_2\end{array}$

Therefore, the molarity of $\text{Ca}{\left(\text{OH}\right)}_2$ solution is $0.50\text{ M}$ .