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{N}}_2}{\rm{ = }}\dfrac{{{{10}^{ - 5}}}}{{1000}} = \dfrac{{{{10}^{ - 5}}}}{{{{10}^{ - 3}}}} = {10^{ - 8}}\
\[{\rm{[}}{{\rm{H}}^ + }] = {10^{ - 8}}\]
--------- (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.