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

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.