Chapter 17, Problem 36QP

A 10.0-ml solution of 0.300 M ${\text{NH}}_{\text{3}}$ is titrated with a 0.100 M $\text{HCl}$ solution. Calculate the pH after the following additions of the $\text{HCl}$ solution: (a) 0.0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 30.0 mL, (e) 40.0 mL.

Interpretation Introduction

Interpretation:

The values of the $\text{pH}$ of the solution, after the addition of different volumes of $\text{HCl}$, are to be determined.

Concept Introduction:

Acid–base titration is a technique to analyze the unknown concentration of the acid or base through the known concentration of the acid and base.

Equivalence point is the point in the acid–base titration in the chemical reaction where the number of moles of the titrant and the unknown concentration of the analyte are equal, which is used to identify the unknown concentration of the analyte.

Answer to Problem 36QP

Solution:

(a)

$\text{pH}=11.36$

(b)

$\text{pH}=9.55$

(c)

$\text{pH}=8.95$

(d)

$\text{pH}=5.19$

(e)

$\text{pH}=1.70$

Explanation of Solution

a) $0.0\text{mL}$ solution

Thereaction between ${\text{NH}}_{\text{3}}$ and $\text{HCl}$ is as follows:

${\text{NH}}_3\left(aq\right)\text{+HCl}\left(aq\right)\rightleftharpoons {\text{NH}}_4\text{Cl}\left(aq\right)$.

From the above equation, one mole of ${\text{NH}}_3$ reacts with one mole of $\text{HCl}$.

Consider $x$ to be the degree of dissociation.

The initial change equilibrium for the ionisation of ${\text{NH}}_3$ is as follows:

$\begin{array}{l}{\text{NH}}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_4^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{M}\right)& 0.300& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.300-x& -& x& x\end{array}\end{array}$

$K_b{\text{ of NH}}_3=1.8\times {10}^{-5}$.

The equilibrium expression for the reaction is represented below.

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

Here, $\left[{\text{OH}}^{-}\right]$ is the concentration of the hydroxide ion, $\left[{\text{NH}}_4^{+}\right]$ is the concentration of the ammonium cation, $\left[{\text{NH}}_3\right]$ is the concentration of ammonia, and $K_b$ is the base dissociation constant.

Substitute the values of $\left[{\text{OH}}^{-}\right]$, $\left[{\text{NH}}_4^{+}\right]$, $\left[{\text{NH}}_3\right]$, and $K_{\text{b}}$ in the expression.

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[0.300-x\right]}\\ =\frac{\left(x\right)\left(x\right)}{\left(0.300\right)}\\ 1.8\times {10}^{-5}\times 0.300=x^2\end{array}$

The value of $x$ is very small as compared to $0.300$. Hence, it can be neglected.

$\begin{array}{l}x=\sqrt{1.8\times {10}^{-5}\times 0.300}\\ x=2.3\times {10}^{-3}\end{array}$

The concentration of the hydroxide ion is calculated as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=x\\ =\text{ 2}.3\times {10}^{-3}\end{array}$

The value of $\text{pOH}$ is calculated by the expression as follows:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$.

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

$\begin{array}{l}\text{pOH}=-\log \left(2.3\times {10}^{-3}\right)\\ \text{pOH}=\mathrm{2.64.}\end{array}$

The value of $\text{pH}$ is calculated by the expression as follows:

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\text{.}\end{array}$

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

$\begin{array}{l}\text{pH}=14-2.64\\ =\mathrm{11.36.}\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $11.36$.

b) $\text{10 mL solution}$

The number of moles of ${\text{NH}}_3$ initially present in $\text{10 mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of ${\text{NH}}_3$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{NH}}_3=\text{ 10 mL×}\frac{\text{0}{\text{.300 mol NH}}_3}{{\text{1000 mL NH}}_3\text{ solution}}\\ =0.300\text{ mol}\\ =\text{3}\times {\text{10}}^{-3}\text{mol}\text{.}\end{array}$

The number of moles of $\text{HCl}$ initially present in $\text{10}\text{mL}$ of thesolution is calculated as follows:

$\text{Number of moles of HCl in 10 mL}$ of the solution is as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of $\text{HCl}$ in the above expression.

$\begin{array}{l}\text{Moles}\text{of}\text{HCl}=10\text{mL}\times \frac{0.100\text{mol HCl}}{1000\text{mL HCl solution}}\\ =1.00\times {10}^{-3}\text{mol}\text{.}\end{array}$

On mixing the two solutions, the molarity of the solution changes. The number of moles will remain the same.

The changes in the number of moles of ${\text{NH}}_3$ and $\text{HCl}$ at equilibrium are as follows:

$\begin{array}{l}{\text{NH}}_3\left(aq\right)+\text{HCl}\left(aq\right)\rightleftharpoons {\text{NH}}_4\text{Cl}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 3.00\times {10}^{-3}& \text{}1.00\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -1.00\times {10}^{-3}& -1.00\times {10}^{-3}& +1.00\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 2.0\times {10}^{-3}& 0& 1.00\times {10}^{-3}& -\end{array}\end{array}$

At this stage, this is a buffer solution.

Determine the pH of the buffer solution by the Henderson–Hasselbalch equation.

This is a basic buffer.

$K_b\text{of}{\text{NH}}_{\text{3}}=5.6\times {10}^{-10}$

$\begin{array}{l}\text{pH}=p{K}_a+\log \frac{[\text{Weak}\text{Base}]}{[\text{Conjugate Base}]}\\ =p{K}_a+\log \frac{[{\text{NH}}_3]}{[{\text{NH}}_4^{+}]}.\end{array}$

Here, $\left[{\text{NH}}_4^{+}\right]$ is the concentration of the ammonium cation, $\left[{\text{NH}}_3\right]$ is the concentration of ammonia, and $K_b$ is the base dissociation constant.

Substitute the values of $\left[{\text{NH}}_4^{+}\right]$, $\left[{\text{NH}}_3\right]$, and $K_b$ in the expression.

$\begin{array}{l}\text{pH}=-\log \left(5.6\times {10}^{-10}\right)+\log \frac{\left[2.0\times {10}^{-3}\right]}{\left[1.0\times {10}^{-3}\right]}\\ =\mathrm{9.55.}\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $9.55$.

c) $\text{20}\text{mL solution}$.

The number of moles of ${\text{NH}}_3$ initially present in $\text{10}\text{mL}$ of thesolution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of ${\text{NH}}_3$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{NH}}_3\text{= 20 mL}\times \frac{\text{0}\text{.100}{\text{mol NH}}_3}{\text{1000 mL HCl solution}}\\ =0.200\text{ mol}\\ \text{=}2\times {\text{10}}^{-3}\text{mol}\text{.}\end{array}$

The number of moles of $\text{HCl}$ present in $\text{20}\text{mL}$ of the solution is calculated as follows:

$\text{Number of moles of HCl in 20}\text{mL}$ ofsolutions is as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of $\text{HCl}$ in the above expression.

$\begin{array}{l}\text{Moles}\text{of}\text{HCl}=\text{ 2}0\text{mL}\times \frac{0.100\text{mol HCl}}{1000\text{mL HCl solution}}\\ =\text{ 2}.00\times {10}^{-3}\text{mol}\text{.}\end{array}$

On mixing the two solutions, the molarity of the solution changes. The number of moles will remain the same.

The changes in the number of moles of ${\text{NH}}_3$ and $\text{HCl}$ at equilibrium are as follows:

$\begin{array}{l}{\text{NH}}_3\left(aq\right)+\text{HCl}\left(aq\right)\rightleftharpoons {\text{NH}}_4\text{Cl}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 3.00\times {10}^{-3}& \text{}2.00\times {10}^{-3}& 0& \\ \text{Change}\left(\text{mole}\right)& -2.00\times {10}^{-3}& -2.00\times {10}^{-3}& +2.00\times {10}^{-3}& \\ \text{Equilibrium}\left(\text{mole}\right)& 1.0\times {10}^{-3}& 0& 2.00\times {10}^{-3}& \end{array}\end{array}$

At this stage, this is a buffer solution.

Determine the pH of the buffer solution by the Henderson–Hasselbalch equation.

This is a basic buffer.

$K_b\text{of}{\text{NH}}_{\text{3}}=5.6\times {10}^{-10}$.

$\begin{array}{l}\text{pH}=p{K}_a+\log \frac{[\text{Weak}\text{Base}]}{[\text{Conjugate Base}]}\\ =p{K}_a+\log \frac{[{\text{NH}}_3]}{[{\text{NH}}_4^{+}]}.\end{array}$

Here, $\left[{\text{NH}}_4^{+}\right]$ is the concentration of theammonium cation, $\left[{\text{NH}}_3\right]$ is the concentration of ammonia, and $K_b$ is the base dissociation constant.

${\text{K}}_b{\text{ of NH}}_3=1.8\times {10}^{-5}$.

Substitute the values of $\left[{\text{NH}}_4^{+}\right]$, $\left[{\text{NH}}_3\right]$, and $K_b$ in the expression.

$\begin{array}{l}\text{pH}=-\log \left(5.6\times {10}^{-10}\right)+\log \frac{\left[1.0\times {10}^{-3}\right]}{\left[2.0\times {10}^{-3}\right]}\\ =\mathrm{8.95.}\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $8.95$.

d) $\text{30 mL solution}$.

The number of moles of ${\text{NH}}_3$ initially present in $\text{10}\text{mL}$ of thesolution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of ${\text{NH}}_3$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{NH}}_3=\text{30 mL}\times \frac{\text{0}\text{.300}{\text{mol NH}}_3}{{\text{1000 mL NH}}_3\text{ solution}}\\ =0.300\text{mol}\\ =\text{3}\times {\text{10}}^{-3}\text{mol}\text{.}\end{array}$

The number of moles of $\text{HCl}$ present in $\text{30}\text{mL}$ of thesolution is calculated as follows:

$\text{Number of moles of HCl in 30}\text{mL}$ of thesolution is as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of $\text{HCl}$ in the above expression:

$\begin{array}{l}\text{Moles}\text{of}\text{HCl}=\text{ 3}0\text{mL}\times \frac{0.100\text{mol HCl}}{1000\text{mL HCl solution}}\\ =\text{ 3}.00\times {10}^{-3}\text{mol}\text{.}\end{array}$

On mixing the two solutions, the molarity of the solution changes. The number of moles will remain the same.

The changes in the number of moles of ${\text{NH}}_3$ and $\text{HCl}$ at equilibrium are as follows:

$\begin{array}{l}{\text{NH}}_3\left(aq\right)+\text{HCl}\left(aq\right)\rightleftharpoons {\text{NH}}_4\text{Cl}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 3.00\times {10}^{-3}& \text{}3.00\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -3.00\times {10}^{-3}& -3.00\times {10}^{-3}& +3.00\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 0& 0& 3.00\times {10}^{-3}& -\end{array}\end{array}$

At this stage, an equivalence point of thetitration is reached. $3.0\times {10}^{-3}$ moles of ${\text{NH}}_3$ and $3.0\times {10}^{-3}$ moles of $\text{HCl}$ produce $3.0\times {10}^{-3}$ moles of ${\text{NH}}_4\text{Cl}$. At the equivalence point, only salt is present, which is the conjugate acid of ${\text{NH}}_3$.

Calculate the molarity of ${\text{NH}}_4^{+}$.

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$.

$\begin{array}{c}\text{Total}\text{Volume}=10\text{ mL}+30\text{ mL}\\ =\text{40}\text{.0 mL}\\ =\text{0}\text{.0400}\text{L}\text{.}\end{array}$

Substitute the values of the number of moles and the volume ${\text{NH}}_4^{+}$ in the above expression.

$\begin{array}{l}\text{Molarity}=\frac{3.0\times {10}^{-3}\text{mol}}{0.0400\text{L}}\\ =0.0750\text{M}\text{.}\end{array}$

This is weak acid. ${\text{NH}}_4^{+}$ undergoes hydrolysis.

The ICE table for the ionization of ${\text{NH}}_3$ is as follows:

Let $x$ be the degree of dissociation.

$\begin{array}{l}{\text{NH}}_4^{+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_3\left(\text{aq}\right)+{\text{H}}_3{\text{O}}^{+}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{M}\right)& 0.0750& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.0750-x& -& x& x\end{array}\end{array}$

The equilibrium expression for a reaction is written as:

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

Here, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of the hydronium ion, $\left[{\text{NH}}_4^{+}\right]$ is the concentration of the ammonium cation, $\left[{\text{NH}}_3\right]$ is the concentration of ammonia, and $K_a$ is the acid dissociation constant.

${\text{K}}_a{\text{ of NH}}_3=5.6\times {10}^{-10}$.

Substitute the values of $\left[{\text{OH}}^{-}\right]$, $\left[{\text{NH}}_4^{+}\right]$, $\left[{\text{NH}}_3\right]$, and $K_b$ in the expression.

$\begin{array}{l}1.8\times {10}^{-5}=\frac{\left[\text{x}\right]\left[\text{x}\right]}{\left[\text{0}\text{.0750-x}\right]}\\ =\frac{\left(x\right)\left(x\right)}{\left(0.0750\right)}\\ 1.8\times {10}^{-5}\times 0.0750=x^2\end{array}$

The value of $x$ is very small as compared to $0.0750$. Hence, it can be neglected.

$\begin{array}{l}x=\sqrt{1.8\times {10}^{-5}\times 0.0750}\\ x=6.5\times {10}^{-6}\end{array}$

Concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]=x=\text{ 6}.5\times {10}^{-6}$.

The value of $\text{pH}$ is calculated by using the expression as follows:

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

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

$\begin{array}{l}\text{pH}=-\log \left(6.5\times {10}^{-6}\right)\\ =\mathrm{5.19.}\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $5.19$.

e) $\text{40}\text{mL solution}$.

The number of moles of ${\text{NH}}_3$ initially present in $\text{10}\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of ${\text{NH}}_3$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{NH}}_3=\text{ 30 mL}\times \frac{\text{0}\text{.300}{\text{mol NH}}_3}{{\text{1000 mL CH}}_{\text{3}}\text{COOH solution}}\\ =0.300\text{mol}\\ =\text{3}\times {\text{10}}^{-3}\text{mol}\text{.}\end{array}$

The number of moles of $\text{HCl}$ present in $\text{30}\text{mL}$ of thesolution is calculated as follows:

$\text{Number of moles of HCl in 40}\text{mL}$ ofsolutions is as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$.

Substitute the values of the concentration and the volume of $\text{HCl}$ in the above expression:

$\begin{array}{l}\text{Moles}\text{of}\text{HCl}=\text{ 4}0\text{mL}\times \frac{0.100\text{mol HCl}}{1000\text{mL HCl solution}}\\ =\text{ 4}.00\times {10}^{-3}\text{mol}\text{.}\end{array}$

On mixing the two solutions, the molarity of the solution changes. The number of moles will remain the same.

The changes in the number of moles of ${\text{NH}}_3$ and $\text{HCl}$ at equilibrium are as follows:

$\begin{array}{l}{\text{NH}}_3\left(aq\right)+\text{HCl}\left(aq\right)\rightleftharpoons {\text{NH}}_4\text{Cl}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 3.00\times {10}^{-3}& \text{}4.00\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -3.00\times {10}^{-3}& -3.00\times {10}^{-3}& +3.00\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 0& 1.00\times {10}^{-3}& 3.00\times {10}^{-3}& -\end{array}\end{array}$

Calculate the molarity of $\text{HCl}$ as follows:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$.

$\begin{array}{c}\text{Total}\text{Volume}=10\text{ mL}+40\text{ mL}\\ =\text{50}\text{.0 mL}\\ =\text{0}\text{.0500}\text{L}\text{.}\end{array}$

Substitute the values of the number of moles and the volume of ${\text{NH}}_4^{+}$ in the above expression.

$\begin{array}{l}\text{Molarity}=\frac{1.0\times {10}^{-3}\text{mol}}{0.0500\text{L}}\\ =0.020\text{M}\text{.}\end{array}$

$\text{HCl}$ is a strong acid. It dissociates completely into water.

Concentration of $\left[{\text{H}}^{+}\right]=0.020\text{M}$.

The $\text{pH}$ is calculated by using the expression as follows:

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

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

$\begin{array}{l}\text{pH}=-\log \left(0.0200\right)\\ =\mathrm{1.70.}\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $1.70$.