Chapter 17, Problem 17.36QP

A 10.0-mL solution of 0.300 M NH₃ is titratee with a 0.100 M HCl solution. Calculate the pH after the following additions of the 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 $\text{pH}$ of titration ${\text{NH}}_{\text{3}}\text{Vs}\text{HCl}$ on adding various amount of $\text{HCl}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when no $\text{HCl}$ is added.

Answer to Problem 17.36QP

pH = $11.36$

Explanation of Solution

Concentration of ${\text{NH}}_{\text{3}}$ is $\text{0}\text{.30}0\text{M}$

Concentration of $\text{HCl}$ is $\text{0}\text{.100}\text{M}$

$\begin{array}{l}{\text{K}}_{\text{b}}\text{value for ammonia is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}{}^{\text{+}}{\text{ ][OH}}^{\text{-}}\text{ ]}}{{\text{[NH}}_{\text{3}}\text{]}}\\ \text{1}\text{.8 ×}{\text{10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.300-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [OH}}^{\text{-}}\text{] = 2}\text{.3 ×}{\text{10}}^{\text{-3}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(2}\text{.3 ×}{\text{10}}^{\text{-3}}\text{)}\\ \text{pOH}\text{=}\text{2}\text{.64}\\ \text{pH = 14}\text{.00 - 2}\text{.64 }\\ \text{= 11}\text{.36}\end{array}$

When there is no $\text{HCl}$ added, $\text{pH}$ of solution can be determined by calculating the ammonia $\text{pH}$.  The $\text{pH}$ of ammonia in water can be determined from base dissociation constant.  From the concentration of hydroxide ion, the $\text{pH}$ is calculated by taking negative log of hydroxide ion.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration ${\text{NH}}_{\text{3}}\text{Vs}\text{HCl}$ on adding various amount of $\text{HCl}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $10\text{mL}$ $\text{HCl}$ is added.

Answer to Problem 17.36QP

pH = $9.55$

Explanation of Solution

Find the number of moles of ${\text{NH}}_{\text{3}}\text{and}\text{HCl}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}{\text{NH}}_{\text{3}}\text{ in 10 mL}\text{=}10.\text{0 mL}\text{×}\frac{\text{0}\text{.300 mol}}{\text{1000}\text{mL}}\\ \text{=}3.\text{0}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{HCl in 10 mL}\text{=}10.\text{0 mL}\text{×}\frac{\text{0}\text{.100 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

The number of moles of ammonia and $\text{HCl}$ can be calculated using volume and given concentration of the ammonia and $\text{HCl}$.

Find the $\text{pH}$ when $10\text{mL}$ $\text{HCl}$ is added

$\begin{array}{l}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}\text{+ }{\text{HCl}}_{\text{(aq)}}\to {\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{3}\text{.00 ×}{\text{10}}^{\text{-3}}\text{1}\text{.0 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change in concentration (M):}\text{-1}\text{.0 ×}{\text{10}}^{\text{-3}}\text{-1}\text{.0 ×}{\text{10}}^{\text{-3}}\text{+1}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \text{Final}\text{concentration (M): }\text{2}\text{.00 ×}{\text{10}}^{\text{-3}}\text{0}\text{+1}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \text{pH}\text{=}\text{-log(5}\text{.6 ×}{\text{10}}^{\text{-10}}\text{)}\text{+}\text{log}\frac{\text{2}\text{.0 ×}{\text{10}}^{\text{-3}}}{\text{1}\text{.0 ×}{\text{10}}^{\text{-3}}}\\ \text{pH}\text{=}\text{9}\text{.55}\end{array}$

When $\text{HCl}$ adds to ammonia, the volume of solution rises which varies the concentration of solution but the moles remain same.  The number of moles after addition of $\text{HCl}$ changes and is given in the table.  The Henderson-Hasselbalch equation can be used to calculate the $\text{pH}$ because the solution acts as buffer system.  From the equation and using acid dissociation constant, the $\text{pH}$ is calculated.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration ${\text{NH}}_{\text{3}}\text{Vs}\text{HCl}$ on adding various amount of $\text{HCl}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $20\text{mL}$ $\text{KOH}$ is added

Answer to Problem 17.36QP

pH = $8.95$

Explanation of Solution

To find the number of moles of ${\text{NH}}_{\text{3}}\text{and}\text{HCl}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}{\text{NH}}_{\text{3}}\text{ in 10 mL}\text{=}\text{10}\text{.0 mL}\text{×}\frac{\text{0}\text{.300 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{3}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{HCl in 20 mL}\text{=}\text{20}\text{.0 mL}\text{×}\frac{\text{0}\text{.100 mol}}{\text{1000}\text{mL}}\\ \text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{mol}\end{array}$

The number of moles of acetic acid can be calculated using volume and given concentration of the $\text{KOH}$.

Find the $\text{pH}$ when $20\text{mL}$ $\text{KOH}$ is added.

$\begin{array}{l}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}\text{+ }{\text{HCl}}_{\text{(aq)}}\to {\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{3}\text{.00 ×}{\text{10}}^{\text{-3}}2.\text{0 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change in concentration (M):}\text{-2}\text{.0 ×}{\text{10}}^{\text{-3}}\text{-2}\text{.0 ×}{\text{10}}^{\text{-3}}\text{+2}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \text{Final}\text{concentration (M): }1.\text{00 ×}{\text{10}}^{\text{-3}}\text{0}\text{+2}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \\ \text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}\\ \text{pH}\text{=}\text{-log(5}\text{.6 ×}{\text{10}}^{\text{-10}}\text{)}\text{+}\text{log}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-3}}}{\text{2}\text{.0 ×}{\text{10}}^{\text{-3}}}\\ \text{pH}\text{=}8.9\text{5}\end{array}$

When $\text{HCl}$ adds to ammonia, the volume of solution rises which varies the concentration of solution but the moles remain same.  The number of moles after addition of $\text{HCl}$ changes and is given in the table.  The Henderson-Hasselbalch equation can be used to calculate the $\text{pH}$ because the solution acts as buffer system.  From the equation and using acid dissociation constant, the $\text{pH}$ is calculated.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration ${\text{NH}}_{\text{3}}\text{Vs}\text{HCl}$ on adding various amount of $\text{HCl}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $30\text{mL}$ $\text{HCl}$ is added.

Answer to Problem 17.36QP

pH = $5.19$

Explanation of Solution

$\begin{array}{l}{\text{M(NH}}_{\text{4}}{}^{\text{+}}\text{)}\text{=}\frac{\text{3}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{mol}}{\text{0}\text{.0400}\text{L}}\text{=}\text{0}\text{.0750}\text{M}\\ {\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}\text{+ }{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+ }{\text{NH}}_{\text{3}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{.0750 }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-x}\text{+x}\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{0}\text{.0750-x}\text{x}\text{x}\\ \\ {\text{K}}_{\text{b}}\text{ value for ammonia is 1}\text{.8 ×}{\text{10}}^{\text{-5}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\text{=}\frac{\text{1}\text{.0 ×}{\text{10}}^{\text{-14}}}{\text{1}\text{.8 ×}{\text{10}}^{\text{-5}}}\text{=}\text{5}\text{.6×}{\text{10}}^{\text{-10}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][NH}}_{\text{3}}\text{]}}{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}}\\ \text{5}\text{.6×}{\text{10}}^{\text{-10}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0750-x}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ {\text{x = [H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = 6}\text{.5 ×}{\text{10}}^{\text{-6}}\text{M}\\ \text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{=}\text{-log(6}\text{.09 ×}{\text{10}}^{\text{-6}}\text{)}\\ \text{pH}\text{=}\text{5}\text{.19}\end{array}$

By adding $30\text{mL}\text{HCl}$, we reached equivalence point and equal number of moles of acetic acid $\text{3}\text{.00 ×}{\text{10}}^{\text{-3}}\text{mol}$ reacts with $\text{3}\text{.00 ×}{\text{10}}^{\text{-3}}\text{mol}$ of $\text{HCl}$.  At equivalence point $\text{3}\text{.00 ×}{\text{10}}^{\text{-3}}\text{mol}$ of ammonium chloride present in solution.  The concentration of hydrogen ion can be calculated from the final concentration of hydrogen ion in equilibrium table using base dissociation constant.  The $\text{pH}$ is determined by taking negative logarithm of  hydrogen ion concentration.

Interpretation Introduction

Interpretation:

The $\text{pH}$ of titration ${\text{NH}}_{\text{3}}\text{Vs}\text{HCl}$ on adding various amount of $\text{HCl}$ has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$  in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.
• $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$
• The point at which amount of standard solution and analyte becomes equal and neutralisation happens in titration is called equivalence point.
• $\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$ is Henderson-Hasselbalch equation
• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.

To find: the $\text{pH}$ when $40\text{mL}$ $\text{HCl}$ is added.

Answer to Problem 17.36QP

pH = $1.70$

Explanation of Solution

Find the number of moles of $\text{HCl}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{HCl in 40 mL}\text{=}40.\text{0 mL}\text{×}\frac{\text{0}\text{.100 mol}}{\text{1000}\text{mL}}\\ \text{=}4.\text{00}\text{×}{\text{10}}^{\text{-3}}\text{mol}\\ \end{array}$

The number of moles of $\text{HCl}$ can be calculated using volume and given concentration of the $\text{HCl}$.

Find the $\text{pH}$ when $40\text{mL}$ $\text{HCl}$ is added

$\begin{array}{l}{\text{NH}}_{\text{3}}{}_{\text{(aq)}}\text{+ }{\text{HCl}}_{\text{(aq)}}\to {\text{NH}}_{\text{4}}{\text{Cl}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{3}\text{.00 ×}{\text{10}}^{\text{-3}}\text{4}\text{.0 ×}{\text{10}}^{\text{-3}}\text{0}\\ \text{Change in concentration (M):}\text{-3}\text{.0 ×}{\text{10}}^{\text{-3}}\text{-3}\text{.0 ×}{\text{10}}^{\text{-3}}\text{+3}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \text{Final}\text{concentration (M): }\text{0}\text{1}\text{.00 ×}{\text{10}}^{\text{-3}}\text{+3}\text{.0 ×}{\text{10}}^{\text{-3}}\\ \\ \text{The}\text{molarity}\text{of}\text{HCl}\text{in}\text{0}\text{.0500L}\text{after}\text{reacting}\\ \text{M(HCl)}\text{=}\frac{\text{1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{mol}}{\text{0}\text{.0500 L}}\text{=}\text{0}\text{.0200 M}\\ \text{pH =}\text{-log(0}\text{.0200) = 1}\text{.70}\end{array}$

The number of moles after addition of $\text{HCl}$ changes and is given in the table.  By calculating the strength of $\text{HCl}$, the $\text{pH}$ of the solution determined.