Chapter 15, Problem 69E

Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M NH₃ (K_b = 1.8 × 10⁻⁵) with 0.100 M HCl.

Interpretation Introduction

Interpretation: The titration of ${\text{NH}}_3$ with different volumes of $\text{HCl}$ is given. The $\text{pH}$ of each solution is to be calculated and then graph of $\text{pH}$ versus volume of titrant 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.

To determine: The $\text{pH}$ of each solution is to be calculated and graph of $\text{pH}$ versus volume of titrant added.

Answer to Problem 69E

Answer

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{11.11}$.

The value of $\text{pH}$ of solution when $4.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{9.85}$.

The value of $\text{pH}$ of solution when $8.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{9.61}$.

The value of $\text{pH}$ of solution when $12.5\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{9.26}$.

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{8.66}$.

The value of $\text{pH}$ of solution when $24.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{7.86}$.

The value of $\text{pH}$ of solution when $24.5\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{7.56}$.

The value of $\text{pH}$ of solution when $24.9\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{6.86}$.

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{5.25}$.

The value of $\text{pH}$ of solution when $25.1\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{3.69}$

The value of $\text{pH}$ of solution when $26.0\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{2.69}$

The value of $\text{pH}$ of solution when $28.0\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{2.22}$

The value of $\text{pH}$ of solution when $30.0\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{2.04}$

The graph between $\text{pH}$ and volume of $\text{HCl}$ added is shown in Figure 2.

Explanation of Solution

Explanation

Given

The concentration of ${\text{NH}}_3$ is $0.100\text{M}$

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

The volume of ${\text{NH}}_3$ is $25.0\text{mL}$.

The volume of $\text{HCl}$ is $0.0\text{mL}$, $4.0\text{mL}$, $8.0\text{mL}$, $12.5\text{mL}$, $20.0\text{mL}$, $24.0\text{mL}$, $24.5\text{mL}$, $24.9\text{mL}$, $25.0\text{mL}$, $25.1\text{mL}$, $26.0\text{mL}$, $28.0\text{mL}$, $30.0\text{mL}$

The value of $K_{\text{b}}$ of ${\text{NH}}_3$ is $1.8\times {10}^{-5}$ .

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

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

Hence the conversion of $100\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{100}\text{mL}=100\times 0.001\text{L}\\ =\text{0}\text{.100}\text{L}\end{array}$

The reaction is represented as,

${\text{NH}}_3+{\text{H}}_2\text{O}\rightleftharpoons {\text{NH}}_4{}^{+}+{\text{OH}}^{-}$

Make the ICE table for the reaction between ${\text{H}}_2{\text{NNH}}_2$ and ${\text{H}}_{\text{2}}\text{O}$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_{\text{2}}\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& \text{OH}-\\ \text{Initial moles:}& 0.100& & & & 0& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.100-x& & & & x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.100-x\right)$ is taken as $0.100$.

Simplify the above equation,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.100\right)\text{M}}\\ x=0.0013\text{}\text{M}\end{array}$

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

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

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ (1)

Where,

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.0013\right)\\ =2.89\end{array}$

The value of $\text{pOH}$ of the solution is $2.89$.

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

$\text{pH}+\text{pOH}=14$ (2)

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+2.89=14\\ \text{pH}=\underset{\_}{11.11}\end{array}$

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{11.11}$.

The volume of $\text{HCl}$ 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,

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

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

$\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}$ (4)

Substitute the value of concentration and volume of $\text{HCl}$ in equation (4) as,

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

Substitute the value of concentration and volume of ${\text{NH}}_3$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.0004& & 0\\ \text{Change in moles }& -0.0004& & -0.0004& & +0.0004\\ \text{Equilibrium}\text{moles}& 0.0021& & 0.0004& & 0.0004\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0021\text{moles}}{0.029\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0004\text{moles}}{0.029\text{}\text{L}}\\ =0.013\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.05& & & & 0.013& & \\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.05-x& & & & 0.013+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.013+x\right)\left(x\right)}{\left(0.05-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.05-x\right)$ is taken as $\left(0.05\right)$ and $\left(0.013+x\right)$ is taken as $0.013$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.013+x\right)\left(x\right)}{\left(0.05-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.013\right)\left(x\right)}{\left(0.05\right)\text{M}}\\ x=7.0\times {10}^{-5}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(7.0\times {10}^{-5}\right)\\ =4.15\end{array}$

The value of $\text{pOH}$ of the solution is $4.15$.

Substitute the value of $\text{pOH}$ in the equation (2)

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{4}\text{.15}=14\\ \text{pH}=\underset{\_}{9.85}\end{array}$

The value of $\text{pH}$ of solution when $4.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{9.85}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.0008& & 0\\ \text{Change in moles }& -0.0004& & -0.0008& & +0.0008\\ \text{Equilibrium}\text{moles}& 0.0017& & 0.0008& & 0.0008\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0008\text{moles}}{0.033\text{}\text{L}}\\ =0.0242\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.0515& & & & 0.0242& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.0515-x& & & & 0.0242+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.0242+x\right)\left(x\right)}{\left(0.0515-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.0515-x\right)$ is taken as $\left(0.0515\right)$ and $\left(0.0242+x\right)$ is taken as $0.0242$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.0242+x\right)\left(x\right)}{\left(0.0515-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.0242\right)\left(x\right)}{\left(0.0515\right)\text{M}}\\ x=4.0\times {10}^{-5}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(4.0\times {10}^{-5}\right)\\ =4.39\end{array}$

The value of $\text{pOH}$ of the solution is $4.39$.

Substitute the value of $\text{pOH}$ in the equation (2)

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{4}\text{.39}=14\\ \text{pH}=\underset{\_}{9.61}\end{array}$

The value of $\text{pH}$ of solution when $8.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{9.61}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.00125& & 0\\ \text{Change in moles }& -0.00125& & -0.00125& & +0.00125\\ \text{Equilibrium}\text{moles}& 0.00125& & 0.00125& & 0.00125\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

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

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.033& & & & 0.033& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.033-x& & & & 0.033+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.033+x\right)\left(x\right)}{\left(0.033-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.033-x\right)$ is taken as $\left(0.033\right)$ and $\left(0.033+x\right)$ is taken as $0.033$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.033+x\right)\left(x\right)}{\left(0.033-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.033\right)\left(x\right)}{\left(0.033\right)\text{M}}\\ x=1.8\times {10}^{-5}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(1.8\times {10}^{-5}\right)\\ =4.74\end{array}$

The value of $\text{pOH}$ of the solution is $4.74$.

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{4}\text{.74}=14\\ \text{pH}=\underset{\_}{9.26}\end{array}$

The value of $\text{pH}$ of solution when $12.5\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{9.26}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.002& & 0\\ \text{Change in moles }& -0.002& & -0.002& & +0.002\\ \text{Equilibrium}\text{moles}& 0.0005& & 0.002& & 0.002\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.002\text{moles}}{0.045\text{}\text{L}}\\ =0.044\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.011& & & & 0.044& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.011-x& & & & 0.044+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.044+x\right)\left(x\right)}{\left(0.011-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.011-x\right)$ is taken as $\left(0.011\right)$ and $\left(0.044+x\right)$ is taken as $0.044$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.044+x\right)\left(x\right)}{\left(0.011-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.044\right)\left(x\right)}{\left(0.011\right)\text{M}}\\ x=4.5\times {10}^{-6}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(4.5\times {10}^{-6}\right)\\ =5.34\end{array}$

The value of $\text{pOH}$ of the solution is $5.34$.

Substitute the value of $\text{pOH}$ in the equation (2).

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+5.34=14\\ \text{pH}=\underset{\_}{8.66}\end{array}$

The value of $\text{pH}$ of solution when $20.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{8.66}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.0024& & 0\\ \text{Change in moles }& -0.0024& & -0.0024& & +0.0024\\ \text{Equilibrium}\text{moles}& 0.0001& & 0.0024& & 0.0024\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0024\text{moles}}{0.049\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.002& & & & 0.05& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.002-x& & & & 0.05+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.002-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.002-x\right)$ is taken as $\left(0.002\right)$ and $\left(0.05+x\right)$ is taken as $0.05$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.002-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05\right)\left(x\right)}{\left(0.002\right)\text{M}}\\ x=7.2\times {10}^{-7}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(7.2\times {10}^{-7}\right)\\ =6.14\end{array}$

The value of $\text{pOH}$ of the solution is $6.14$.

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

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+6.14=14\\ \text{pH}=\underset{\_}{7.86}\end{array}$

The value of $\text{pH}$ of solution when $24.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{7.86}$.

The volume of $\text{HCl}$ 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,

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

Substitute the value of concentration and volume of $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.00245& & 0\\ \text{Change in moles }& -0.00245& & -0.00245& & +0.00245\\ \text{Equilibrium}\text{moles}& 0.00005& & 0.00245& & 0.00245\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00245\text{moles}}{0.0495\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.001& & & & 0.05& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.001-x& & & & 0.05+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.001-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.001-x\right)$ is taken as $\left(0.001\right)$ and $\left(0.05+x\right)$ is taken as $0.05$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.001-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05\right)\left(x\right)}{\left(0.001\right)\text{M}}\\ x=3.6\times {10}^{-7}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(3.6\times {10}^{-7}\right)\\ =6.44\end{array}$

The value of $\text{pOH}$ of the solution is $6.44$.

Substitute the value of $\text{pOH}$ in the equation (2).

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+6.44=14\\ \text{pH}=\underset{\_}{7.56}\end{array}$

The value of $\text{pH}$ of solution when $24.5\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{7.56}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.00249& & 0\\ \text{Change in moles }& -0.00249& & -0.00249& & +0.00249\\ \text{Equilibrium}\text{moles}& 0.00001& & 0.00249& & 0.00249\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_3$ and final volume of solution in equation (1).

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00249\text{moles}}{0.0499\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{NH}}_3$.

$\begin{array}{cccccccc}& {\text{NH}}_3& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4{}^{+}& +& {\text{OH}}^{-}\\ \text{Initial moles:}& 0.0002& & & & 0.05& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.0002-x& & & & 0.05+x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_3\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.0002-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.0002-x\right)$ is taken as $\left(0.0002\right)$ and $\left(0.05+x\right)$ is taken as $0.05$.

Simplify the above equation,

$\begin{array}{c}\text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.0002-x\right)\text{M}}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05\right)\left(x\right)}{\left(0.0002\right)\text{M}}\\ x=7.2\times {10}^{-8}\text{}\text{M}\end{array}$

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(7.2\times {10}^{-8}\right)\\ =7.14\end{array}$

The value of $\text{pOH}$ of the solution is $7.14$.

Substitute the value of $\text{pOH}$ in the equation (2).

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+7.14=14\\ \text{pH}=\underset{\_}{6.86}\end{array}$

The value of $\text{pH}$ of solution when $24.9\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{6.86}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.0025& & 0\\ \text{Change in moles }& -0.0025& & -0.0025& & +0.0025\\ \text{Equilibrium}\text{moles}& 0.0& & 0.0& & 0.0025\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of ${\text{NH}}_4{}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.0025\text{moles}}{0.05\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Make the ICE table for the reaction of ${\text{NH}}_4{}^{+}$ with water.

$\begin{array}{cccccccc}& {\text{NH}}_4{}^{+}& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_3& +& {\text{H}}_3{\text{O}}^{\text{+}}\\ \text{Initial moles:}& 0.05& & & & 0& & 0\\ \text{Change}:& -x& & & & +x& & x\\ \text{Final}\text{moles}:& 0.05-x& & & & x& & x\end{array}$

The relationship between $K_{\text{a}}$ and $K_{\text{b}}$ is given as,

$K_{\text{a}}\times K_{\text{b}}={10}^{-14}$

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

$\begin{array}{c}{K}_{\text{a}}\times K_{\text{b}}={10}^{-14}\\ K_{\text{a}}\times 1.8\times {10}^{-5}={10}^{-14}\\ K_{\text{a}}=5.6\times {10}^{-10}\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{NH}}_3\right]\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]}{\left[{\text{NH}}_4{}^{+}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{NH}}_3\right]\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]}{\left[{\text{NH}}_4{}^{+}\right]}\\ \text{5}\text{.6}\times {\text{10}}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.05-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{a}}$ is very small, hence, $\left(0.05-x\right)$ is taken as $\left(0.05\right)$.

Simplify the above equation,

$\begin{array}{c}\text{5}\text{.6}\times {\text{10}}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.05-x\right)\text{M}}\\ \text{5}\text{.6}\times {\text{10}}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.05\right)\text{M}}\\ x=5.5\times {10}^{-6}\text{}\text{M}\end{array}$

It is the concentration of ${\text{H}}_3{\text{O}}^{\text{+}}$.

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

$\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{\text{+}}\right]$ (5)

Where,

• $\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]$ is the concentration of Hydronium ions in the solution.

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{\text{+}}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}_3{\text{O}}^{\text{+}}\right]\\ =-\log \left(5.5\times {10}^{-6}\right)\\ =\underset{\_}{5.25}\end{array}$

The value of $\text{pH}$ of solution when $25.0\text{}\text{mL}$ $\text{HCl}$ has been added is. $\underset{\_}{5.25}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.00251& & 0\\ \text{Change in moles }& -0.0025& & -0.0025& & +0.0025\\ \text{Equilibrium}\text{moles}& 0.0& & 0.00001& & 0.0025\end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{NH}}_3+\text{Volume}\text{of}\text{HCl}\\ =\text{0}\text{.025}\text{L}+0.0251\text{L}\\ =\text{0}\text{.0501}\text{L}\end{array}$

Substitute the value of number of moles of $\text{HCl}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.00001\text{moles}}{0.0501\text{}\text{L}}\\ =0.0002\text{M}\end{array}$

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

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the equation (5).

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

The value of $\text{pH}$ of solution when $25.1\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{3.69}$.

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.0026& & 0\\ \text{Change in moles }& -0.0025& & -0.0025& & +0.0025\\ \text{Equilibrium}\text{moles}& 0.0& & 0.0001& & 0.0025\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of $\text{HCl}$ and final volume of solution in equation (1).

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

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

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

The value of $\text{pH}$ of solution when $26.0\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{2.69}$

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.028\text{L}\\ =\text{0}\text{.0028}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.0028& & 0\\ \text{Change in moles }& -0.0025& & -0.0025& & +0.0025\\ \text{Equilibrium}\text{moles}& 0.0& & 0.0003& & 0.0025\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of $\text{HCl}$ and final volume of solution in equation (1).

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

The concentration of ${\text{H}}^{\text{+}}$ is $0.006\text{M}$.

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the equation (5).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ =-\log \left(0.006\right)\\ =\underset{\_}{2.22}\end{array}$

The value of $\text{pH}$ of solution when $28.0\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{2.22}$

The volume of $\text{HCl}$ 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 $\text{HCl}$ in equation (4) 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 ICE table for the reaction between ${\text{NH}}_3$ and $\text{HCl}$.

$\begin{array}{cccccc}& {\text{NH}}_3& +& \text{HCl}& \stackrel{}{\rightleftharpoons }& {\text{NH}}_4\text{Cl}\\ \text{Initial}\text{moles }& 0.0025& & 0.003& & 0\\ \text{Change in moles }& -0.0025& & -0.0025& & +0.0025\\ \text{Equilibrium}\text{moles}& 0.0& & 0.0005& & 0.0025\end{array}$

The above reaction shows the presence of equilibrium in the solution.

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

Substitute the value of number of moles of $\text{HCl}$ and final volume of solution in equation (1).

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

The concentration of ${\text{H}}^{\text{+}}$ is $0.009\text{M}$.

Substitute the value of $\left[{\text{H}}^{\text{+}}\right]$ in the equation (5).

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ =-\log \left(0.009\right)\\ =\underset{\_}{2.04}\end{array}$

The value of $\text{pH}$ of solution when $30.0\text{}\text{mL}$ $\text{HCl}$ has been added is $\underset{\_}{2.04}$

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

Volume of $\text{HCl}$ in $\text{mL}$	$\text{pH}$
$0.0$	$11.11$
$4.0$	$9.85$
$8.0$	$9.61$
$12.5$	$9.26$
$20.0$	$8.66$
$24.0$	$7.86$
$24.5$	$7.56$
$24.9$	$6.86$
$25.0$	$5.25$
$25.1$	$3.69$
$26.0$	$2.69$
$28.0$	$2.22$
$30.0$	$2.04$

Figure 1

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

Figure 2

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