Question

A solution of \[HCl\] has a\[pH = 5\]. If \[1mL\] of it is diluted to \[1\;litre\], what will be the \[pH\] of the resulting solution?

Answer 1

Hint: As per given when \[HCl\] is diluted to the solution then decrease its concentration and keeps the amount of\[HCl\] constant, but increases the total amount of solution, here in solution when the acid solution is diluted, then we can obtain the Concentration of the dilute solution by using the formula - \[{M_2} \times {V_2} = {M_1} \times {V_1}{\text{ }}\] and after that we will calculate the \[pH\] of resulting solution.

Complete step by step answer: All acids have a \[pH\] in the acid range, then the value is below \[7\]. When \[HCl\] is added to the solution, then the solution is acidic with \[pH\] is less than \[7\].
Now we have calculated the molarity of diluted solution, by using the formula
 \[{M_2} = \dfrac{{{M_1}{\text{ }} \times {\text{ }}{V_1}{\text{ }}}}{{{V_2}}}{\text{ }}\]
where
 \[{M_1} = \]Concentration in molarity (moles/Liters) of the concentrated solution
\[\;{V_1} = \] Volume of the concentrated solution,
\[{M_2} = \] Concentration in molarity of the dilute solution (after more solvent has been added).
 \[{V_2} = \] Volume of the dilute solution.
 First we find \[{M_2}\], then we determine the \[pH\] of resulting solution
given,
A solution of \[HCl\] has \[pH = 5\]
As we know , pH is the negative logarithm of the hydrogen ion concentration of a solution.
 It is expressed as: \[pH{\text{ }} = - {\text{ }}log{\text{ }}\left[ {{H^ + }} \right]\]
where \[\left[ {{H^ + }} \right]\;\] is the concentration of the hydrogen ion in the solution.
\[pH\] = \[ - {\text{ }}log{\text{ }}\left[ {{H^ + }} \right].\]
\[ - log{\text{ }}\left[ {{H^ + }} \right] = {10^{ - 5}}\]
\[\left[ {{H^ + }} \right] = {10^{ - 5}}\]
therefore the given acid, \[{M_1} = \]\[\;{10^{ - 5}}\]
Given, \[\;{V_1} = \] \[1ml\]
\[{V_2} = \] \[1L{\text{ }} = {\text{ }}1000L\]
This will give the hydronium ion concentration obtained by the ionization of \[HCl\]
\[{M_2} = \dfrac{{{M_1}{\text{ }} \times {\text{ }}{V_1}{\text{ }}}}{{{V_2}}}{\text{ }}\]
Now putting the values
\[ = \;\dfrac{{{{10}^{ - 5}}{\text{ }} \times {\text{ }}1}}{{1000}}\] \[ = {10^{ - 8}}M\]
since the value obtained from hydronium ion concentration from \[HCl\] is very small so we have to consider the contribution of water.
Since \[\left[ {{H^ + }} \right]\; < {10^{ - 7{\text{ }}}},\]
so let, \[\left[ {{H_3}{0^ + }} \right]\;\]and\[{\left[ {OH} \right]^ + }\] = \[x\]
we know that \[\left[ {{H_3}{0^ + }} \right]{\text{. }}\left[ {{\text{O}}{{\text{H}}^{\text{ + }}}} \right] = {10^{ - 14}}\left( {{\text{ }}x{\text{ }} + {\text{ }}{{10}^{ - 8}}} \right).\left( x \right)\]
The equation we get \[pH = \; - {\text{ }}log\left[ {10.5{\text{ }} \times {{10}^{ - 8}}} \right]\]
Solving the equation using the formulae \[\;\dfrac{{-\;\;b + \sqrt {{b^2} - 4ac} }}{{2a}}\]
Now we get, = \[9.5{\text{ }} \times {\text{ }}{10^{ - 8}}M\]
therefore \[\left[ {{H_3}{O^ + }} \right] = \;9.5{\text{ }} \times {10^{ - 8}}{\text{ }}M\]
Total \[ = \;\left[ {{H_3}{O^ + }} \right]\; + \;\;\left[ {{H^ + }} \right] = \;\left( {9.5{\text{ }} \times {{10}^{ - 8{\text{ }}}}M\; + {\text{ }}{{10}^{ - 8}}} \right)\]
= \[\left( {9.5{\text{ }} + {\text{ }}1} \right) \times {10^{ - 8}}\]
$\Rightarrow$ = \[(10.5\; \times {10^{ - 8}})\]
Now we find \[pH\]of resulting solution
\[pH = \; - {\text{ }}log\left[ {10.5{\text{ }} \times {{10}^{ - 8}}} \right]\]
By solving log we get,
\[pH = 6.9586\]

Note: In diluting an acidic solution the solution becomes less acidic, its pH increases and approaches \[7\] and dilution of basic solution becomes less basic, its pH decreases and approaches \[7\]. The solution becomes neutral in nature.