Chapter 11, Problem 80QP

(a)

Interpretation Introduction

Interpretation:

The volume of $0.5000\text{ M}\text{HCl}$ required to neutralize $35.00\text{mL}$ of $0.0500\text{ M}\text{NaOH}$ 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 $\text{NaOH}$ present in $35.00\text{mL}$ of $0.0500\text{}\text{ M}\text{NaOH}$ after converting millilitre to litre unit is calculated as follows,

$\begin{array}{c}\text{Moles of NaOH}=35.00\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times 0.0500\text{mol/L}\\ =\text{0}\text{.00175}\text{mol}\end{array}$

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

$\begin{array}{c}\text{Moles of HCl}=\text{0}\text{.00175}\text{mol}\text{NaOH}\times \frac{1\text{mol HCl}}{1\text{mol NaOH}}\\ =0.00175\text{mol HCl}\end{array}$

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.

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

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

$\begin{array}{c}\text{Volume of HCl in litre}=\frac{0.00175\text{mol}}{\text{0}\text{.5000}{\text{mol L}}^{-1}}\\ =0.0035\text{L}\end{array}$

The required volume in liter unit is very small. Thus, it must be converted to milliliter unit.

$\begin{array}{c}\text{Volumeof NaOH in millilitre}=0.0035\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =3.50\text{mL}\end{array}$

Hence, $3.50\text{mL}$ of $0.0500\text{ M}\text{NaOH}$ is required to neutralize $35.0\text{mL}$ of $0.5000\text{}\text{ M}\text{HCl}$ .

(b)

Interpretation Introduction

Interpretation:

The volume of $0.5000\text{}\text{ M}\text{HCl}$ required to neutralize $10.00\text{mL}$ of $0.2000\text{}\text{ M}{\text{Ba(OH)}}_{\text{2}}$ is to be calculated.

Explanation of Solution

The balanced chemical equation is,

${\text{Ba(OH)}}_2(aq)+2\text{HCl}(aq)\to {\text{BaCl}}_2(aq)+2{\text{H}}_2\text{O}(l)$

The moles of ${\text{Ba(OH)}}_2$ present in $15.00\text{mL}$ of $0.2000\text{ M}{\text{Ba(OH)}}_2$ after converting milliliter into liter unit is calculated as follows,

$\begin{array}{c}{\text{Moles of Ba(OH)}}_2=15.00\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times 0.2000{\text{mol L}}^{-1}\\ =\text{0}\text{.003}\text{mol}\end{array}$

From the balanced chemical reaction, it is clear that $2\text{moles}$ of $\text{HCl}$ react with $1\text{mol}$ of ${\text{Ba(OH)}}_2$ . The moles of $\text{HCl}$ is calculated using the mol ratio as shown below.

$\begin{array}{c}\text{Moles of HCl}=\text{0}\text{.003}\text{mol Ba(OH)2}\times \frac{2\text{mol HCl}}{1{\text{mol Ba(OH)}}_2}\\ =0.006\text{mol HCl}\end{array}$

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.

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

The volume of $\text{HCl}$ required is as shown below.

$\begin{array}{c}\text{Volumeof HCl in litre}=\frac{0.006\text{mol}}{\text{0}\text{.5000}{\text{mol L}}^{-1}}\\ =0.012\text{L}\end{array}$

The required volume in liter unit is very small. Thus, it must be converted to milliliter.

$\begin{array}{c}\text{Volumeof HCl in millilitre}=0.012\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =12.0\text{mL}\end{array}$

Hence, $12.0\text{mL}$ of $0.5000\text{}\text{ M}\text{HCl}$ required to neutralize $15.00\text{mL}$ of $0.2000\text{ M}{\text{Ba(OH)}}_2$ .

(c)

Interpretation Introduction

Interpretation:

The volume of $0.5000\text{}\text{ M}\text{HCl}$ required to neutralize $15.00\text{mL}$ of $0.2500\text{M}{\text{NH}}_3$ is to be calculated.

Explanation of Solution

The balanced chemical equation is,

$\text{HCl}(aq)+{\text{NH}}_3(aq)\to {\text{NH}}_4\text{Cl}(aq)$

The moles of ${\text{NH}}_3$ present in $15.00\text{mL}$ of $0.2500\text{ M}{\text{NH}}_3$ after converting milliliter to liter unit is calculated as follows,

$\begin{array}{c}\text{Moles NH3}=15.00\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times 0.2500{\text{mol L}}^{-1}\\ =\text{0}\text{.00375}\text{mol}\end{array}$

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

$\begin{array}{c}\text{Moles HCl}=\text{0}\text{.00375}{\text{mol NH}}_3\times \frac{1\text{mol HCl}}{1{\text{mol NH}}_3}\\ =0.00375\text{mol HCl}\end{array}$

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.

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

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

$\begin{array}{c}\text{Volumeof HCl in litre}=\frac{0.00375\text{mol}}{\text{0}\text{.5000}{\text{mol L}}^{-1}}\\ =0.0075\text{L}\end{array}$

The required volume in liter unit is very small. Thus, it must be converted into milliliters.

$\begin{array}{c}\text{Volumeof HCl in mollilitre}=0.0075\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =7.5\text{mL}\end{array}$

Hence, $7.5\text{mL}$ of $0.5000\text{ M}\text{HCl}$ required to neutralize $15.00\text{mL}$ of $0.2500\text{M}{\text{NH}}_{\text{3}}$ .