Question

1) A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 16.0 mL of KOH.

Express your answer numerically.   pH=_______ 

2) A 75.0-mL volume of 0.200 M NH3 (Kb = 1.8 x10-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 13.0 mL of HNO3.

Express your answer numerically.   pH=_______ 

3) A 52.0-mL volume of 0.35 M CH3COOH (Ka = 1.8 x10-5 ) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 17.0 mL of NaOH.

Express your answer numerically.   pH=_______ 

Answer 1

Concepts and reason

The concept used to solve this question is to calculate the pH of the solution during the acid-base titration in three different cases. 1) Titration of strong acid HBr with strong base KOH solution, 2) Titration of strong acid ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ with weak base ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ solution, 3) Titration of weak acid, ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ with strong base, NaOH solution

To calculate the pH of the solution, the nature of the solution should be known whether it is acidic or basic in case of titration of strong acid with strong base.

When either weak acid or weak base is used in the titration then the pH of the solution calculated depending on the nature of species present in the solution.

Fundamentals

Titration is a common analytical technique which is used to measure the amounts of compounds in solution. The glassware called a buret which holds one of the reactants, called the titrant, and well-situated adds it into a reaction vessel which contains the second reactant.

Either acid or base can be a titrant in the titration process. The calculation of pH depends on the nature of acid or base.

If both acid and base used in the titration are strong then it can be calculated by knowing the nature of the solution during the titration. If the solution is acidic the concentration of ${{\rm{H}}^ + }$ affects the pH of the solution and if the solution is basic then the concentration of ${\rm{O}}{{\rm{H}}^ - }$ ion affects the pH of the solution.

The weak acid or weak base is partially ionized. Thus, when either weak acid or weak base is used in the titration then the pH of the solution calculated depending on the nature of species present in the solution. Before the equivalence point, the pH depends on the concentration of weak acid and conjugate base (or weak base and conjugate acid) in the solution. At the equivalence point, the pH depends on the concentration of conjugate base (or conjugate acid) in the solution. Beyond the equivalence point, the pH depends on the concentration of ${\rm{O}}{{\rm{H}}^ - }$ (or H+) in the solution

1)

The given acid is HBr and base is KOH so it is a titration of strong acid with strong base.

Given,

The concentration of HBr solution = 0.15 M.

The volume of HBr solution = 50.0 mL = 0.05 L.

The concentration of KOH solution = 0.25 M.

The volume of KOH solution added = 16 mL = 0.016.

$\begin{array}{c}\\{\rm{The number of moles of HBr solution = concentration }} \times {\rm{ volume in litre}}\\\\{\rm{ = 0}}{\rm{.15 moles /L }} \times {\rm{ 0}}{\rm{.05 L}}\\\\{\rm{ = 0}}{\rm{.0075 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The number of moles of KOH solution = concentration }} \times {\rm{ volume in litre}}\\\\{\rm{ = 0}}{\rm{.25 moles /L }} \times {\rm{ 0}}{\rm{.016 L}}\\\\{\rm{ = 0}}{\rm{.004 moles}}\\\end{array}$

The chemical reaction of KOH with HBr is an acid-base reaction, which is shown below.

${\rm{ KOH + HBr }} \to {\rm{ KBr + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Hence, 1 mole of KOH solution neutralizes 1 mole of HBr solution.

Therefore, 0.004 moles of KOH solution neutralizes 0.004 moles of HBr solution. Hence there will be excess of HBr present in the solution so the solution is acidic.

$\begin{array}{c}\\{\rm{The number moles of excess HBr in the solution = 0}}{\rm{.0075 moles - 0}}{\rm{.004 moles}}\\\\ = {\rm{ 0}}{\rm{.0035 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The total volume of solution = 50 mL + 16 mL}}\\\\{\rm{ = 66 mL}}\\\\{\rm{ = 0}}{\rm{.066 L}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of excess HBr solution = }}\frac{{{\rm{Number of moles}}}}{{{\rm{Total volume in litres}}}}\\\\ = \frac{{{\rm{0}}{\rm{.0035 moles}}}}{{{\rm{0}}{\rm{.066 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.053 M}}\\\end{array}$

Hence, ${\rm{the concentration of H + ion, [H + ] = 0}}{\rm{.053 M}}$ .

The pH of the solution can be calculated by the following formula.

${\rm{pH = - log[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}$

The pH of the given solution is calculated as follows:

$\begin{array}{c}\\{\rm{pH = - log(0}}{\rm{.053)}}\\\\{\rm{ = 1}}{\rm{.27}}\\\end{array}$

Therefore, the pH of the given solution is 1.27.

2)

Given,

The concentration of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ solution is 0.2 M.

The volume of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ solution is 75.0 mL = 0.075 L.

The concentration of ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ solution is 0.5 M.

The volume of ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ solution added is 13 mL = 0.013 L.

$\begin{array}{c}\\{\rm{The number of moles of N}}{{\rm{H}}_{\rm{3}}}{\rm{ solution = concentration }} \times {\rm{ volume in litre}}\\\\{\rm{ = 0}}{\rm{.2 moles /L }} \times {\rm{ 0}}{\rm{.075 L}}\\\\{\rm{ = 0}}{\rm{.015 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The number of moles of HN}}{{\rm{O}}_{\rm{3}}}{\rm{ solution = concentration }} \times {\rm{ volume in litre}}\\\\{\rm{ = 0}}{\rm{.5 moles /L }} \times {\rm{ 0}}{\rm{.013 L}}\\\\{\rm{ = 0}}{\rm{.0065 moles}}\\\end{array}$

The titration of ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ with ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ is a strong acid-weak base titration and ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ is completely dissociated and ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ is partially dissociated.

The neutralization reaction is written as follows:

${\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{(aq) + HN}}{{\rm{O}}_{\rm{3}}}{\rm{(aq) }} \to {\rm{ N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{N}}{{\rm{O}}_{\rm{3}}}(aq)$

The net ionic equation is as follows:

${\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{(aq) + }}{{\rm{H}}^{\rm{ + }}}{\rm{(aq) }} \to {\rm{ N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}(aq)$

From the reaction, 1 mole of ${{\rm{H}}^{\rm{ + }}}$ ions neutralizes 1 mole of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ solution to give 1 mole of conjugate acid of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ .

Therefore, 0.0065 moles of ${\rm{HN}}{{\rm{O}}_{\rm{3}}}$ solution neutralizes 0.0065 moles of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ solution to give 0.0065 moles of ${\rm{N}}{{\rm{H}}_{\rm{4}}}^ +$ .

Hence, there will be excess of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ solution present in the solution.

$\begin{array}{c}\\{\rm{The number moles of excess N}}{{\rm{H}}_{\rm{3}}}{\rm{ in the solution = 0}}{\rm{.015 moles - 0}}{\rm{.0065 moles}}\\\\ = {\rm{ 0}}{\rm{.0085 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The total volume of solution = 75 mL + 13 mL}}\\\\{\rm{ = 88 mL}}\\\\{\rm{ = 0}}{\rm{.088 L}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of excess N}}{{\rm{H}}_{\rm{3}}}{\rm{ solution = }}\frac{{{\rm{Number of moles}}}}{{{\rm{Total volume in litre}}}}\\\\ = \frac{{{\rm{0}}{\rm{.0085 moles}}}}{{{\rm{0}}{\rm{.088 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.096 M}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of formed N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{ solution = }}\frac{{{\rm{Number of moles}}}}{{{\rm{Total volume in litre}}}}\\\\ = \frac{{{\rm{0}}{\rm{.0065 moles}}}}{{{\rm{0}}{\rm{.088 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.0738 M}}\\\end{array}$

As ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ is a weak base it partially dissociates so at this point the solution is a buffer solution of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ and ${\rm{N}}{{\rm{H}}_{\rm{4}}}^ +$ .

Hence, the pOH of the solution can be calculated by using Henderson-Hasselbalch equation for base, which is written as follows:

${\rm{pOH = p}}{{\rm{K}}_{\rm{b}}}{\rm{ + log}}\frac{{{\rm{[N}}{{\rm{H}}_{\rm{4}}}^{\rm{ + }}{\rm{]}}}}{{{\rm{[N}}{{\rm{H}}_{\rm{3}}}{\rm{]}}}}$

The base dissociation constant of ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ , ${{\rm{K}}_{\rm{b}}}{\rm{ = 1}}{\rm{.8 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 5}}}}$

${\rm{p}}{{\rm{K}}_{\rm{b}}}{\rm{ = - log}}{{\rm{K}}_{\rm{b}}}$

Substitute the value of ${{\rm{K}}_{\rm{b}}}$ in the above equation, then ${\rm{p}}{{\rm{K}}_{\rm{b}}}$ for ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ = 4.75.

Substitute the values of ${\rm{p}}{{\rm{K}}_{\rm{b}}}$ , [ ${\rm{N}}{{\rm{H}}_{\rm{4}}}^ +$ ] and [ ${\rm{N}}{{\rm{H}}_{\rm{3}}}$ ] in the Henderson-Hasselbalch equation for base, then

$\begin{array}{c}\\{\rm{pOH = 4}}{\rm{.75 + log}}\frac{{0.0738}}{{0.096}}\\\\ = {\rm{ 4}}{\rm{.75 - 0}}{\rm{.1142}}\\\\{\rm{ = 4}}{\rm{.635}}\\\end{array}$

The pH of the buffer solution is calculated as follows:

$\begin{array}{c}\\{\rm{pH = }}14 - {\rm{pOH}}\\\\{\rm{ = 14 - 4}}{\rm{.635}}\\\\{\rm{pH = 9}}{\rm{.365}}\\\end{array}$

Therefore, the pH of the solution is 9.365.

3)

Given,

The concentration of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ solution is 0.35 M.

The volume of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ solution = 52.0 mL = 0.052 L.

The concentration of NaOH solution is 0.4 M.

The volume of NaOH solution added = 17 mL = 0.017 L.

$\begin{array}{c}\\{\rm{The number of moles of C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH solution = concentration }} \times {\rm{ volume in litre}}\\\\{\rm{ = 0}}{\rm{.35 moles /L }} \times {\rm{ 0}}{\rm{.052 L}}\\\\{\rm{ = 0}}{\rm{.0182 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The number of moles of NaOH solution = concentration }} \times {\rm{ volume in litre}}\\\\{\rm{ = 0}}{\rm{.4 moles /L }} \times {\rm{ 0}}{\rm{.017 L}}\\\\{\rm{ = 0}}{\rm{.0068 moles}}\\\end{array}$

The titration of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ with NaOH is a weak acid-strong base titration, in which NaOH is completely dissociated and ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ is partially dissociated.

The neutralization reaction is written as follows:

${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH(aq) + NaOH(aq) }} \to {\rm{ C}}{{\rm{H}}_{\rm{3}}}{\rm{COONa}}(aq){\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O (l)}}$

The net ionic equation is as follows:

${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH(aq) + O}}{{\rm{H}}^{\rm{ - }}}{\rm{(aq) }} \to {\rm{ C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ - }}}(aq){\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O (l)}}$

Hence, 1 mole of ${\rm{O}}{{\rm{H}}^ - }$ ions neutralizes 1 mole of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ solution to give 1 mole of conjugate base of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ .

Therefore, 0.0068 moles of NaOH solution neutralizes 0.0068 moles of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ solution to give 0.0065 moles of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ - }}}$ .

Thus, there will be excess of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ present in the solution.

$\begin{array}{c}\\{\rm{The number moles of excess C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH in the solution = 0}}{\rm{.0182 moles - 0}}{\rm{.0068 moles}}\\\\ = {\rm{ 0}}{\rm{.0114 moles}}\\\end{array}$

$\begin{array}{c}\\{\rm{The total volume of solution = 52 mL + 17 mL}}\\\\{\rm{ = 69 mL}}\\\\{\rm{ = 0}}{\rm{.069 L}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of excess C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH solution = }}\frac{{{\rm{Number of moles}}}}{{{\rm{Total volume in litre}}}}\\\\ = \frac{{{\rm{0}}{\rm{.0114 moles}}}}{{{\rm{0}}{\rm{.069 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.165 M}}\\\end{array}$

$\begin{array}{c}\\{\rm{The concentration of formed C}}{{\rm{H}}_{\rm{3}}}{\rm{COO - solution = }}\frac{{{\rm{Number of moles}}}}{{{\rm{Total volume in litre}}}}\\\\ = \frac{{{\rm{0}}{\rm{.0068 moles}}}}{{{\rm{0}}{\rm{.069 L}}}}{\rm{ }}\\\\{\rm{ = 0}}{\rm{.0985 M}}\\\end{array}$

As ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ is a weak acid it partially dissociates so at this point the solution is a buffer solution of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ and ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ - }}}$ .

Hence, the pH of the solution can be calculated by using Henderson-Hasselbalch equation for acid, which is written as follows:

${\rm{pH = p}}{{\rm{K}}_a}{\rm{ + log}}\frac{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ - }}}{\rm{]}}}}{{{\rm{[C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH]}}}}$

The acid dissociation constant of ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ , ${{\rm{K}}_{\rm{a}}}{\rm{ = 1}}{\rm{.8 }} \times {\rm{ 1}}{{\rm{0}}^{{\rm{ - 5}}}}$

${\rm{p}}{{\rm{K}}_{\rm{a}}}{\rm{ = - log}}{{\rm{K}}_{\rm{a}}}$

Substitute the value of ${{\rm{K}}_{\rm{b}}}$ in the above equation, then ${\rm{p}}{{\rm{K}}_{\rm{a}}}$ for ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ = 4.75.

Substitute the values of ${\rm{p}}{{\rm{K}}_a}$ , [ ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^{\rm{ - }}}$ ] and [ ${\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}$ ] in the Henderson-Hasselbalch equation for base, then

$\begin{array}{c}\\{\rm{pH = 4}}{\rm{.75 + log}}\frac{{0.0985}}{{0.165}}\\\\ = {\rm{ 4}}{\rm{.75 - 0}}{\rm{.224}}\\\\{\rm{ = 4}}{\rm{.526}}\\\end{array}$

Therefore, the pH of the solution is 4.526.

Ans: Part 1

Part 1

Answer

The pH of the given solution is 1.27.