Question

When \[\text{1ml}\] of $\text{0}\text{.1N}$ $\text{HCl}$ is added to $1$ liter of solution of \[\text{ }\!\!~\!\!\text{ pH}\] value of $4$, the \[\text{ }\!\!~\!\!\text{ pH}\] will be near:
(A)- $5$
(B)- $4.477$
(C)- $3$
(D)- $3.7$

Answer 1

Hint: \[\text{ }\!\!~\!\!\text{ pH}\] is used to define the acidity & basicity of any solution, if \[\text{ }\!\!~\!\!\text{ pH}\] is in between $0-7$ then solution will be acidic and if \[\text{ }\!\!~\!\!\text{ pH}\] is in between $7-14$ then solution will be basic. \[\text{ }\!\!~\!\!\text{ pH}\] is defined as the negative log of concentration of hydrogen ions i.e.
\[\text{pH=-log}\left[ {{\text{H}}^{\text{+}}} \right]\]

Complete step by step solution:
Given that, \[\text{ }\!\!~\!\!\text{ pH}\] of $1$ liter of solution = $4$ ……….. (i)
As we know that, \[\text{pH=-log}\left[ {{\text{H}}^{\text{+}}} \right]\]……………….(ii)
From (i) & (ii), we get
\[\text{-log}\left[ {{\text{H}}^{\text{+}}} \right]=4\]
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=1}{{\text{0}}^{\text{-4}}}$
Hence, concentration of hydrogen ions in the given solution is ${{10}^{-4}}$.
We, have to find out the \[\text{ }\!\!~\!\!\text{ pH}\] of resulting solution after the mixing of \[\text{1ml}\] of $\text{0}\text{.1N}$ $\text{HCl}$ to the given solution.
So, \[\text{ }\!\!~\!\!\text{ pH}\] or concentration of resulting solution will be calculated as follow:
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=}\dfrac{{{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{+}{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}}{{{\text{V}}_{\text{1}}}\text{+}{{\text{V}}_{\text{2}}}}$ ………………… (iii)
Here, \[{{\text{N}}_{\text{1}}}\text{=1}{{\text{0}}^{\text{-4}}}\]
\[{{\text{V}}_{\text{1}}}\text{=1L}\]
\[{{\text{N}}_{2}}\text{=}0.1\]
\[{{\text{V}}_{\text{2}}}\text{=1mL=0}\text{.001L}\]
On putting these values on the above (iii) equation we get,
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=}\dfrac{\text{1}{{\text{0}}^{-4}}\times \text{1+0}\text{.1}\times 0.001}{\text{1+0}\text{.001}}$
$\left[ {{\text{H}}^{\text{+}}} \right]\text{=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{M}$
So, concentration of hydrogen ions in the resulting solution is $\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{M}$, for calculating \[\text{ }\!\!~\!\!\text{ pH}\] putting this value in the (ii) equation we get,
\[\text{pH=-log}\left[ 2\times {{10}^{-4}} \right]\]
\[\text{pH=3}\text{.7}\]

Hence, option (D) is correct.

Additional information: For converting mili liter to liter, always divide the given quantity of liter by $1000$ because $\text{1L=1000mL}$ or
$\text{1mL=}\dfrac{\text{1}}{\text{1000}}\text{L}$.

Note: In this question, some of you may get confused why we put ${{10}^{-4}}$ in place of ${{\text{N}}_{\text{1}}}$, so the reason behind this is that normality, molarity, molality, etc. is used to define the concentration of any species in solution that’s why we use the value of the concentration of hydrogen in place of normality.