Question

Consider the titration of 100.0 mL of 0.200 M ${\mathbf{HONH}}_{\mathbf{2}}$by 0.100 M HCl.$\left(K_b\mathrm{for}{\mathrm{HONH}}_2=1.1\times {10}^{-8}.\right)$

1. Calculate the pH after 0.0 mL of HCl has been added.
2. Calculate the pH after 25.0 mL of HCl has been added.
3. Calculate the pH after 70.0 mL of HCl has been added.
4. Calculate the pH at the equivalence point.
5. Calculate the pH after 300 mL of HCl has been added.
6. At what volume of HCl added does the$\mathbf{pH}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{04}$ ?

Short answer

1. The pH of the initial solution is 9.67.
2. The pH of the solution is 6.88.
3. The pH of the solution is 6.31.
4. The pH of the solution is 3.60.
5. The pH of the solution is 1.60.
6. The volume of HCl for the pH 6.04is 100 mL.

Step-by-step solution

Definition of pH

A quantitative measure of the acidity or basicity of aqueous or other liquid solutions is defined as pH.

Subpart (a)

The ICE table for the amount of lactate ion in the solution after the acid dissociation can be determined

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

$\begin{array}{l}{\mathrm{K}}_{\mathrm{b}}=\frac{\left[{{\mathrm{HONH}}_3}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{HONH}}_2\right]}\\ 1.1\times {10}^{-8}=\frac{{\mathrm{x}}^2}{0.2}\\ \mathrm{x}=4.7\times 10{}^{-5}\end{array}$

The pH of the initial solution can be determined.

$\begin{array}{l}\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]\\ \mathrm{pOH}=4.33\\ \mathrm{pH}=14-4.33\\ \mathrm{pH}=9.67\end{array}$

The pH of the initial solution is 9.67.

Subpart (b)

To determine the pH of the solution with the help of the Henderson-Hesselbalch equation, the ICE table after the addition of the base is created.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is

$\begin{array}{l}{\mathrm{pK}}_{\mathrm{b}}=7.96\\ \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{conjugate}\mathrm{base}\right]}{\left[\mathrm{weak}\mathrm{base}\right]}\\ \mathrm{pOH}=7.12\\ \mathrm{pH}=14-7.12\\ \mathrm{pH}=6.88\end{array}$

The pH of the solution is 6.88.

Subpart (c)

To determine the pH of the solution with the help of the Henderson-Hesselbalch equation, the ICE table after the addition of the base is created.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

$\begin{array}{l}{\mathrm{pK}}_{\mathrm{b}}=7.96\\ \mathrm{pOH}=7.96+\log \frac{\left[\mathrm{conjugate}\mathrm{base}\right]}{\left[\mathrm{weak}\mathrm{base}\right]}\\ \mathrm{pOH}=7.69\end{array}$

$\begin{array}{l}\mathrm{pH}=14-7.69\\ \mathrm{pH}=6.31\end{array}$

The pH of the solution is 6.31.

Subpart (d)

To determine the pH of the solution with the help of the Henderson-Hesselbalch equation, the ICE table after the addition of the base is created.

Reaction	${\mathrm{HONH}}_{3^{+}}$	${\mathrm{HONH}}_2$	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

$\begin{array}{l}\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ \mathrm{pH}=3.60\end{array}$

The pH of the solution is 3.60

Subpart (e)

The amount of acid added is 0.03 mol

The excess of acid in the solution is 0.01 mol

The amount of hydrogen ions is 0.01 mol

The concentration of hydrogen ions is 0.025 mol/L

$\begin{array}{l}\mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ \mathrm{pH}=-\log \left[0.025\right]\\ \mathrm{pH}=1.60\end{array}$

The pH of the solution is 1.60.

Subpart (f)

For the given pH value, pOH of the solution is

$\begin{array}{l}\mathrm{pOH}=14-\mathrm{pH}\\ \mathrm{pOH}=7.96\\ \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\\ 7.96=7.96+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}\end{array}$

The number of moles of salt and base will be the same that is${\mathrm{NH}}_3{\mathrm{OH}}^{+}$ and${\mathrm{NH}}_2\mathrm{OH}$ respectively.

For the equivalence point, the volume of HCl required was 200 mL, A halfway equivalence point the volume will be half to it. So,the volume of HCl is 100 mL.