Chapter 14, Problem 92AE

Interpretation Introduction

Interpretation: The titration of Nitric acid with different volumes of $\text{NaOH}$ is given. The $\text{pH}$ of each solution is to be calculated and the graph between $\text{pH}$ and volume of base added is to be plotted.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

Answer to Problem 92AE

Answer

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.0}$.

The value of $\text{pH}$ of solution when $4.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.14}$.

The value of $\text{pH}$ of solution when $8.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.28}$.

The value of $\text{pH}$ of solution when $12.5\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.48}$.

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.95}$.

The value of $\text{pH}$ of solution when $24.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{2.69}$.

The value of $\text{pH}$ of solution when $24.5\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{3.0}$.

The value of $\text{pH}$ of solution when $24.9\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{3.69}$.

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{7.0}$.

The value of $\text{pH}$ of solution when $25.1\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{10.35}$.

The value of $\text{pH}$ of solution when $26.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.31}$.

The value of $\text{pH}$ of solution when $28.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.78}$.

The value of $\text{pH}$ of solution when $30.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.96}$.

The graph between $\text{pH}$ and volume of base added is shown in Figure 1.

Explanation of Solution

Explanation

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.0}$.

Given:

The concentration of ${\text{HNO}}_{\text{3}}$ acid is $0.100\text{M}$

The concentration of $\text{NaOH}$ is $0.100\text{M}$.

The volume of ${\text{HNO}}_{\text{3}}$ is $25.0\text{mL}$.

The volume of $\text{NaOH}$ is $0.0\text{mL}$.

When no base is added, then solution contains only the strong acid ${\text{HNO}}_{\text{3}}$. Therefore, $\text{pH}$ is given by concentration of ${\text{H}}^{\text{+}}$ only.

The $\text{pH}$ of a solution is shown below.

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$ (1)

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of ions present in a solution.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.100\right)\\ =\underset{\_}{1.0}\end{array}$

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.0}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.0724\text{M}$.

Given

The volume of $\text{NaOH}$ is $4.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $4.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{4}\text{.0}\text{mL}=4.0\times 0.001\text{L}\\ =\text{0}\text{.004}\text{L}\end{array}$

The concentration of any species is given as,

$\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$ (2)

Rearrange the above equation to obtain the value of number of moles.

$\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}$ (3)

Substitute the value of concentration and volume of ${\text{HNO}}_{\text{3}}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.025\text{L}\\ =\text{0}\text{.0025}\text{moles}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.004\text{L}\\ =\text{0}\text{.0004}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.0004& & \\ \text{Change }& 0.0025-0.0004& & -0.0004& & \\ \text{Final}\text{moles}& 0.0021& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.004\text{L}\\ =\text{0}\text{.029}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0021\text{moles}}{0.029\text{}\text{L}}\\ =0.0724\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $4.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.14}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.0724\right)\\ =\underset{\_}{1.14}\end{array}$

The value of $\text{pH}$ of solution when $4.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.14}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.0515\text{M}$.

Given

The volume of $\text{NaOH}$ is $8.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $8.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{8}\text{.0}\text{mL}=8.0\times 0.001\text{L}\\ =\text{0}\text{.008}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.008\text{L}\\ =\text{0}\text{.0008}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.0008& & \\ \text{Change }& 0.0025-0.0008& & -0.0008& & \\ \text{Final}\text{moles}& 0.0017& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.008\text{L}\\ =\text{0}\text{.033}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0017\text{moles}}{0.033\text{}\text{L}}\\ =0.0515\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $8.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.28}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.0515\right)\\ =\underset{\_}{1.28}\end{array}$

The value of $\text{pH}$ of solution when $8.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.28}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.033\text{M}$.

Given

The volume of $\text{NaOH}$ is $12.5\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $12.5\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{12}\text{.5}\text{mL}=12.5\times 0.001\text{L}\\ =\text{0}\text{.0125}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.0125\text{L}\\ =\text{0}\text{.00125}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.00125& & \\ \text{Change }& 0.0025-0.00125& & -0.00125& & \\ \text{Final}\text{moles}& 0.00125& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.0125\text{L}\\ =\text{0}\text{.0375}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00125\text{moles}}{0.0375\text{}\text{L}}\\ =0.033\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $12.5\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.48}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.033\right)\\ =\underset{\_}{1.48}\end{array}$

The value of $\text{pH}$ of solution when $12.5\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.48}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.011\text{M}$.

The volume of $\text{NaOH}$ is $20.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $20.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{20}\text{.0}\text{mL}=20.0\times 0.001\text{L}\\ =\text{0}\text{.02}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.02\text{L}\\ =\text{0}\text{.002}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.002& & \\ \text{Change }& 0.0025-0.002& & -0.002& & \\ \text{Final}\text{moles}& 0.0005& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.02\text{L}\\ =\text{0}\text{.045}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0005\text{moles}}{0.045\text{}\text{L}}\\ =0.011\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.95}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.011\right)\\ =\underset{\_}{1.95}\end{array}$

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{1.95}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.002\text{M}$.

Given

The volume of $\text{NaOH}$ is $24.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $24.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{24}\text{.0}\text{mL}=24.0\times 0.001\text{L}\\ =\text{0}\text{.024}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.024\text{L}\\ =\text{0}\text{.0024}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.0024& & \\ \text{Change }& 0.0025-0.0024& & -0.0024& & \\ \text{Final}\text{moles}& 0.0001& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.024\text{L}\\ =\text{0}\text{.049}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0001\text{moles}}{0.049\text{}\text{L}}\\ =0.002\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $24.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{2.69}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.002\right)\\ =\underset{\_}{2.69}\end{array}$

The value of $\text{pH}$ of solution when $24.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{2.69}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.001\text{M}$.

Given

The volume of $\text{NaOH}$ is $24.5\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $24.5\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{24}\text{.5}\text{mL}=24.5\times 0.001\text{L}\\ =\text{0}\text{.0245}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.0245\text{L}\\ =\text{0}\text{.00245}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.00245& & \\ \text{Change }& 0.0025-0.00245& & -0.00245& & \\ \text{Final}\text{moles}& 0.00005& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.0245\text{L}\\ =\text{0}\text{.0495}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00005\text{moles}}{0.0495\text{}\text{L}}\\ =0.001\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $24.5\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{3.0}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.001\right)\\ =\underset{\_}{3.0}\end{array}$

The value of $\text{pH}$ of solution when $24.5\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{3.0}$.

Explanation

The concentration of ${\text{H}}^{+}$ is $0.0002\text{M}$.

Given

The volume of $\text{NaOH}$ is $24.9\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $24.9\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{24}\text{.9}\text{mL}=24.9\times 0.001\text{L}\\ =\text{0}\text{.0249}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.0249\text{L}\\ =\text{0}\text{.00249}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.00249& & \\ \text{Change }& 0.0025-0.00249& & -0.00249& & \\ \text{Final}\text{moles}& 0.00001& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.0249\text{L}\\ =\text{0}\text{.0499}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{HNO}}_{\text{3}}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00001\text{moles}}{0.0499\text{}\text{L}}\\ =0.0002\text{M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $24.9\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{3.69}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the equation (1).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.0002\right)\\ =\underset{\_}{3.69}\end{array}$

The value of $\text{pH}$ of solution when $24.9\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{3.69}$.

Explanation

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{7.0}$.

Given

The volume of $\text{NaOH}$ is $25.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $25.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{25}\text{.0}\text{mL}=25.0\times 0.001\text{L}\\ =\text{0}\text{.025}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.025\text{L}\\ =\text{0}\text{.0025}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.0025& & \\ \text{Change }& 0.0025-0.0025& & -0.0025& & \\ \text{Final}\text{moles}& 0.0& & 0& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.025\text{L}\\ =\text{0}\text{.05}\text{L}\end{array}$

As all the moles have been neutralized, therefore the value of $\text{pH}$ is $\underset{\_}{7.0}$.

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{7.0}$.

Explanation

The concentration of ${\text{OH}}^{-}$ is $0.00022\text{M}$.

Given

The volume of $\text{NaOH}$ is $25.1\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $25.1\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{25}\text{.1}\text{mL}=25.1\times 0.001\text{L}\\ =\text{0}\text{.0251}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.0251\text{L}\\ =\text{0}\text{.00251}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.00251& & \\ \text{Change }& 0.0025-0.0025& & 0.00251-0.0025& & \\ \text{Final}\text{moles}& 0& & 0.00001& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.0251\text{L}\\ =\text{0}\text{.0451}\text{L}\end{array}$

Substitute the value of number of moles of $\text{NaOH}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00001\text{moles}}{0.0451\text{}\text{L}}\\ =0.00022\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

Explanation

The value of $\text{pOH}$ of solution when $25.1\text{}\text{mL}$ $\text{NaOH}$ has been added is. $3.65$.

The $\text{pOH}$ of the solution is shown below.

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ (4)

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of Hydroxide ions.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.00022\right)\\ =3.65\end{array}$

Explanation

The value of $\text{pH}$ of solution when $25.1\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{10.35}$.

The relationship between $\text{pOH}$ is given as,

$\text{pH}+\text{pOH}=14$

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+3.65=14\\ \text{pH}=\underset{\_}{10.35}\end{array}$

The value of $\text{pH}$ of solution when $25.1\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{10.35}$.

Explanation

The concentration of ${\text{OH}}^{-}$ is $0.002\text{M}$.

Given

The volume of $\text{NaOH}$ is $26.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $26.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{26}\text{.0}\text{mL}=26.0\times 0.001\text{L}\\ =\text{0}\text{.026}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.026\text{L}\\ =\text{0}\text{.0026}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.0026& & \\ \text{Change }& 0.0025-0.0025& & 0.0026-0.0025& & \\ \text{Final}\text{moles}& 0& & 0.0001& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.026\text{L}\\ =\text{0}\text{.051}\text{L}\end{array}$

Substitute the value of number of moles of $\text{NaOH}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0001\text{moles}}{0.051\text{}\text{L}}\\ =0.002\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

Explanation

The value of $\text{pOH}$ of solution when $26.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $2.69$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the equation (4).

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.002\right)\\ =2.69\end{array}$

The value of $\text{pOH}$ of solution when $26.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $2.69$.

Explanation

The value of $\text{pH}$ of solution when $26.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.31}$.

The relationship between $\text{pOH}$ is given as,

$\text{pH}+\text{pOH}=14$ (5)

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+2.69=14\\ \text{pH}=\underset{\_}{11.31}\end{array}$

The value of $\text{pH}$ of solution when $26.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.31}$.

Explanation

The concentration of ${\text{OH}}^{-}$ is $0.006\text{M}$.

Given

The volume of $\text{NaOH}$ is $28.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $28.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{28}\text{.0}\text{mL}=28.0\times 0.001\text{L}\\ =\text{0}\text{.028}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.026\text{L}\\ =\text{0}\text{.0026}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.0028& & \\ \text{Change }& 0.0025-0.0025& & 0.0028-0.0025& & \\ \text{Final}\text{moles}& 0& & 0.0003& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.028\text{L}\\ =\text{0}\text{.053}\text{L}\end{array}$

Substitute the value of number of moles of $\text{NaOH}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0003\text{moles}}{0.053\text{}\text{L}}\\ =0.006\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

Explanation

The value of $\text{pOH}$ of solution when $28.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $2.22$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the equation (4).

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.006\right)\\ =2.22\end{array}$

The value of $\text{pOH}$ of solution when $28.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $2.22$.

Explanation

The value of $\text{pH}$ of solution when $25.1\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.78}$.

Substitute the value of $\text{pOH}$ in the above equation (5).

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+2.22=14\\ \text{pH}=\underset{\_}{11.78}\end{array}$

The value of $\text{pH}$ of solution when $28.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.78}$.

Explanation

The concentration of ${\text{OH}}^{-}$ is $0.009\text{M}$.

Given

The volume of $\text{NaOH}$ is $30.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $30.0\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{30}\text{.0}\text{mL}=30.0\times 0.001\text{L}\\ =\text{0}\text{.03}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{NaOH}$ in equation (3) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.03\text{L}\\ =\text{0}\text{.003}\text{moles}\end{array}$

Make the table for the reaction between ${\text{HNO}}_{\text{3}}$ and $\text{NaOH}$.

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \to & {\text{H}}_2\text{O}\\ \text{Initial}\text{moles }& 0.0025& & 0.003& & \\ \text{Change }& 0.0025-0.0025& & 0.003-0.0025& & \\ \text{Final}\text{moles}& 0& & 0.0005& & \end{array}$

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{HNO}}_3+\text{Volume}\text{of}\text{NaOH}\\ =\text{0}\text{.025}\text{L}+0.03\text{L}\\ =\text{0}\text{.055}\text{L}\end{array}$

Substitute the value of number of moles of $\text{NaOH}$ and final volume of solution in equation (2).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0005\text{moles}}{0.055\text{}\text{L}}\\ =0.009\text{M}\end{array}$

It is the concentration of ${\text{OH}}^{-}$.

Explanation

The value of $\text{pOH}$ of solution when $30.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $2.04$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the equation (4).

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.009\right)\\ =2.04\end{array}$

The value of $\text{pOH}$ of solution when $30.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $2.04$.

Explanation

The value of $\text{pH}$ of solution when $30.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.96}$.

Substitute the value of $\text{pOH}$ in the above equation (5).

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+2.04=14\\ \text{pH}=\underset{\_}{11.96}\end{array}$

The value of $\text{pH}$ of solution when $30.0\text{}\text{mL}$ $\text{NaOH}$ has been added is. $\underset{\_}{11.96}$.

Explanation

The graph plotted between $\text{pH}$ and volume of base added is shown below.

The values of $\text{pH}$ obtained are shown in the below table.

$\text{pH}$	Volume of $\text{NaOH}$ in $\text{mL}$
$1.0$	$0.0$
$1.14$	$4.0$
$1.28$	$8.0$
$1.48$	$12.5$
$1.95$	$20.0$
$2.69$	$24.0$
$3.0$	$24.5$
$3.69$	$24.9$
$7.0$	$25.0$
$10.35$	$25.1$
$11.31$	$26.0$
$11.78$	$28.0$
$11.96$	$30.0$

Table 1

The graph between $\text{pH}$ and volume of base added is shown below.

Figure 1

Conclusion

Conclusion

The amount of species present in the solution during titration depends on the volume of titrant added in the solution and this further defines the value of $\text{pH}$ of the solution.