Question

Solution of a monobasic acid has a pH = 5. If one mL of it is diluted to 1 litre, what will be the pH of the resulting solution?
a.) 3.45
b.) 6.96
c.) 8.58
d.) 10.25

Answer 1

Hint: We are diluting the solution. So,
${{\text{N}}_1}{{\text{V}}_1} = {{\text{N}}_2}{{\text{V}}_2}$
Also, $[{{\rm{H}}^ + }] < {10^{ - 7}}$, we have to consider the contribution of water.

Complete answer: Let the initial volume be ${{\rm{V}}_1}$. So, after dilution the volume becomes ${{\rm{V}}_2}$%
So, we have ${{\rm{V}}_1} = 1{\rm{ mL}}$
and ${{\rm{V}}_2} = {\rm{ 1 L = 1000 L}}$.
We know that, ${{\rm{N}}_1}{{\rm{V}}_1} = {{\rm{N}}_2}{{\rm{V}}_2}$%

So, if we first find ${{\rm{N}}_2}$, we can determine the pH of the resulting solution.
Initially, ${\rm{pH = 5 }}$ ${\rm{i}}{\rm{.e}}{\rm{. - log [}}{{\rm{H}}^ + }]{\rm{ = 5}}$
Therefore, ${\rm{[}}{{\rm{H}}^ + }]{\rm{ = 1}}{{\rm{0}}^{ - 5}}$.
Since the given acid is monobasic, ${{\rm{N}}_1} = {\rm{ 1}}{{\rm{0}}^{ - 5}}{\rm{M}}$
Substituting these values in ${{\rm{N}}_1}{{\rm{V}}_1} = {{\rm{N}}_2}{{\rm{V}}_2}$
we get,
${\rm{1}}{{\rm{0}}^{ - 5}}{\rm{ }} \times {\rm{ 1 = }}{{\rm{N}}_2}{\rm{ }} \times {\rm{ 1000}}$

\[{\rm{[}}{{\rm{H}}^ + }] = {10^{ - 8}}$$ \--------- (1) Since, $$[{{\rm{H}}^ + }] < {10^{ - 7}}$$, we have to consider the contribution of water. Water dissociates as $${\rm{2}}{{\rm{H}}_2}{\rm{O }} \to {\rm{ }}{{\rm{H}}_3}{{\rm{O}}^ + }{\rm{ + O}}{{\rm{H}}^ - }$$ Since, we have an acid, we cannot take $$[{{\rm{H}}_3}{{\rm{O}}^ + }]$$ and $$[{\rm{O}}{{\rm{H}}^ - }]$$ as $${10^{ - 7}}$$. So, let $$[{{\rm{H}}_3}{{\rm{O}}^ + }]$$ and $$[{\rm{O}}{{\rm{H}}^ - }]$$ be ‘x’. But we know that, $$[{{\rm{H}}_3}{{\rm{O}}^ + }].[{\rm{O}}{{\rm{H}}^ - }] = {10^{ - 14}}$$$$({\rm{x + 1}}{{\rm{0}}^{ - 8}}).({\rm{x}}) = {10^{ - 14}}$$ The equation we get is $${{\rm{x}}^2}{\rm{ + 1}}{{\rm{0}}^{ - 8}}{\rm{x - 1}}{{\rm{0}}^{ - 14}} = 0$$ We solve this equation using the formula $${\rm{x = }}\dfrac{{ - {\rm{b + }}\sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}$$ From this we get $${\rm{x = 9}}{\rm{.5 }} \times {\rm{ 1}}{{\rm{0}}^{ - 8}}{\rm{M}}$$ $${\rm{i}}{\rm{.e}}{\rm{. [}}{{\rm{H}}_3}{{\rm{O}}^ + }] = {\rm{ 9}}{\rm{.5 }} \times {\rm{ 1}}{{\rm{0}}^{ - 8}}{\rm{M}}$$ \------(2) Total$$ = {\rm{[}}{{\rm{H}}_3}{{\rm{O}}^ + }]{\rm{ + [}}{{\rm{H}}^ + }]$$ $$ = 9.5 \times {10^{ - 8}} + {10^{ - 8}}$$---from (1) and (2) $$ = (9.5 + 1) \times {10^{ - 8}}$$ $$ = 10.5 \times {10^{ - 8}}$$ Now, we find pH of resulting solution as $${\rm{pH = - log[10}}{\rm{.5}} \times {\rm{1}}{{\rm{0}}^{ - 8}}]$$ Solving log, we get $${\rm{pH = 6}}{\rm{.98}}$$ Hence the correct option is (b) 6.96. Additional information: An acid is said to be monobasic when it has only one hydrogen ion to donate i.e. only one hydrogen ion can be replaced during an acid-base reaction. Note: Do not forget that if $$[{{\rm{H}}^ + }] < {10^{ - 7}}$$, we have to consider the contribution of water. Also, since the acid is monobasic, the normality will be equal to hydrogen ion concentration.