Chapter 11, Problem 79QP

(a)

Interpretation Introduction

Interpretation:

The volume of $0.1000\text{M}\text{NaOH}$ required to neutralize $10.00\text{mL}$ of $0.1000\text{M}\text{HCl}$ is to be calculated.

Explanation of Solution

The balanced chemical equation is,

$\text{HCl}(aq)+\text{NaOH}(aq)\to \text{NaCl}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

The moles of $\text{HCl}$ present in $10.00\text{mL}$ of $0.1000\text{M}\text{HCl}$ after converting milliliter to liter unit is calculated as follows,

$\begin{array}{c}\text{Moles HCl}=10.00\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times 0.1000{\text{mol L}}^{-1}\\ =\text{0}\text{.001}\text{mol}\end{array}$

From the balanced chemical reaction, it is clear that $1\text{mol}$ of $\text{HCl}$ reacts with $1\text{mol}$ of $\text{NaOH}$ . The mol of $\text{NaOH}$ is calculated using the mol ratio as shown below.

$\begin{array}{c}\text{Moles NaOH}=\text{0}\text{.001}\text{mol}\text{HCl}\times \frac{1\text{mol NaOH}}{1\text{mol HCl}}\\ =0.001\text{mol NaOH}\end{array}$

The relationship between molarity, moles of solute, and volume of the solution is,

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in liter}}\mathrm{......}\left(1\right)$

Equation $\left(1\right)$ can be rearranged to the following equation.

$\text{Volume in liter}=\frac{\text{Moles of solute}}{\text{Molarity}}\mathrm{......}\left(2\right)$

The volume of $\text{NaOH}$ required is,

$\begin{array}{c}\text{Volume of NaOH in liter}=\frac{0.001\text{mol}}{\text{0}\text{.1000}{\text{mol L}}^{-1}}\\ =0.01\text{L}\end{array}$

The required volume is in litera and is converted into milliliters.

$\begin{array}{c}\text{Volumeof NaOH in milliliters}=0.01\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =10\text{mL}\end{array}$

Hence, $10\text{mL}$ of $0.1000\text{M}\text{NaOH}$ is required to neutralize $10.00\text{mL}$ of $0.1000\text{M}\text{HCl}$ .

(b)

Interpretation Introduction

Interpretation:

The volume of $0.1000\text{M}\text{NaOH}$ required to neutralize $15.00\text{mL}$ of $0.3500\text{M}{\text{HNO}}_{\text{3}}$ is to be calculated.

Explanation of Solution

The balanced chemical equation is,

${\text{HNO}}_{\text{3}}(aq)+\text{NaOH}(aq)\to {\text{NaNO}}_{\text{3}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

The moles of ${\text{HNO}}_3$ present in $15.00\text{mL}$ of $0.3500\text{M}{\text{HNO}}_3$ after converting milliliter to liter is calculated as follows,

$\begin{array}{c}\text{Moles HCl}=15.00\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times 0.3500{\text{mol L}}^{-1}\\ =\text{0}\text{.00525}\text{mol}\end{array}$

From the balanced chemical reaction, it is clear that $1\text{mol}$ of ${\text{HNO}}_3$ reacts with $1\text{mol}$ of $\text{NaOH}$ . The mol of $\text{NaOH}$ is calculated using the mol ratio as shown below.

$\begin{array}{c}\text{Mole of NaOH}=\text{0}\text{.00525}{\text{mol HNO}}_3\times \frac{1\text{mol NaOH}}{1{\text{mol HNO}}_3}\\ =0.00525\text{mol NaOH}\end{array}$

The relationship between molarity, moles of solute, and the volume of the solution is,

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\mathrm{......}\left(1\right)$

Equation $\left(1\right)$ can be rearranged to the following equation.

$\text{Volume in liter}=\frac{\text{Moles of solute}}{\text{Molarity}}\mathrm{......}\left(2\right)$

The volume of $\text{NaOH}$ required is as given below.

$\begin{array}{c}\text{Volume of NaOH in liter}=\frac{0.00525\text{mol}}{\text{0}\text{.1000}{\text{mol L}}^{-1}}\\ =0.0525\text{L}\end{array}$

The required volume in liter unit is very small and is converted into milliliters.

$\begin{array}{c}\text{Volume of NaOH in milliliter}=0.0525\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =52.5\text{mL}\end{array}$

Hence, $52.5\text{mL}$ of $0.1000\text{M}\text{NaOH}$ required to neutralize $15.00\text{mL}$ of $0.3500\text{M}{\text{HNO}}_3$ .

(c)

Interpretation Introduction

Interpretation:

The volume of $0.1000\text{M}\text{NaOH}$ required to neutralize $25.00\text{mL}$ of $0.0500\text{ M}{\text{H}}_3{\text{PO}}_4$ is to be calculated.

Explanation of Solution

The balanced chemical equation is,

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+3\text{NaOH}(aq)\to {\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}(aq)+3{\text{H}}_{\text{2}}\text{O}(l)$

The moles of ${\text{H}}_3{\text{PO}}_4$ present in $25.00\text{mL}$ of $0.0500\text{ M}{\text{H}}_3{\text{PO}}_4$ after converting milliliter into liter unit is calculated as follows,

$\begin{array}{c}{\text{Moles of H}}_3{\text{PO}}_4=25.00\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times 0.0500{\text{mol L}}^{-1}\\ =\text{0}\text{.00125}\text{mol}\end{array}$

From the balanced chemical reaction, it is clear that $1\text{mol}$ of ${\text{H}}_3{\text{PO}}_4$ reacts with $3\text{moles}$ of $\text{NaOH}$ . The moles of $\text{NaOH}$ is calculated using the mol ratio as shown below.

$\begin{array}{c}\text{Moles of NaOH}=\text{0}\text{.00125}\text{mol}\text{H3PO4}\times \frac{3\text{mol NaOH}}{1{\text{mol H}}_3{\text{PO}}_4}\\ =0.00375\text{mol NaOH}\end{array}$

The relationship between molarity, moles of solute, and volume of the solution is,

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in litre}}\mathrm{......}\left(1\right)$

Equation $\left(1\right)$ can be rearranged to the following equation.

$\text{Volume in liter}=\frac{\text{Moles of solute}}{\text{Molarity}}$

The volume of $\text{NaOH}$ required is as given below.

$\begin{array}{c}\text{Volumeof NaOH in liter}=\frac{0.00375\text{mol}}{\text{0}\text{.1000}{\text{mol L}}^{-1}}\\ =0.0375\text{L}\end{array}$

The required volume in liter unit is very small and is converted into milliliter.

$\begin{array}{c}\text{Volume of NaOH in milliliter}=0.0375\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =37.5\text{mL}\end{array}$

Hence, $37.5\text{mL}$ of $0.1000\text{M}\text{NaOH}$ required to neutralize $25.00\text{mL}$ of $0.0500\text{ M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ .