Question

\[\text{ 10 mL }\] of \[\text{ 0}\text{.1 N HCl }\] is added to \[\text{ 990 mL }\] solution of \[\text{NaCl}\]. The $\text{ pH }$ of the resulting solution is:
(A) Zero
(B) 3
(C) 7
(D) 10

Answer 1

Hint: The normality (N) of a solution is equal to the number of the equivalent of solute per litre of solution. It is given as:
$\text{Normality = }\dfrac{\text{No}\text{.Of equivalent of solute}}{\text{Volume in d}{{\text{m}}^{\text{3}}}}\text{ = }\dfrac{\text{Eq}\text{.}}{\text{V}}$
The \[\text{ pH}\] of the solution is equal to the$\text{ pH}=-\log \left[ {{\text{H}}^{+}} \right]\text{ }$. Hence, the concentration of $\text{ HCl }$ only contributes towards the $\text{ pH }$ of the solution.

Complete step by step solution:
We are given that,
\[\begin{align}
& \text{ Volume of HCl (}{{\text{V}}_{\text{HCl}}})\text{= 10 mL or 0}\text{.01 L} \\
& \text{ Normality of HCl (}{{\text{N}}_{\text{HCl}}})\text{ = 0}\text{.1 N} \\
& \text{ Volume of NaCl (}{{\text{V}}_{\text{NaCl}}})\text{= 10 mL or 0}\text{.01 L} \\
\end{align}\]
We have to find $\text{ pH }$ of the solution.
The normality of a solution is defined as the number of the equivalent of gram dissolved per unit volume in litre. The normality is represented by the term N.It can be represented as:
$\text{Normality = }\dfrac{\text{No}\text{.Of equivalent of solute}}{\text{Volume in d}{{\text{m}}^{\text{3}}}}\text{ = }\dfrac{\text{Eq}\text{.}}{\text{V}}$
Where, the eq.means the number of gram –equivalent of solute and V is the volume of solvent in which solute is dissolved.
The number of grams –equivalent of the hydrochloric acid is equal to the product of normality and the volume of the $\text{ HCl }$ solution. The above equation can be rearranged as follows:
$\text{Number of gram-equivalents of HCl = Normality }\!\!\times\!\!\text{ Volume}$
Let's use the equation to determine the no.of gram-equivalent of $\text{ HCl }$as follows,
$\text{Number of gram-equivalents of HCl = Normality }\!\!\times\!\!\text{ Volume}$
So, putting the values in the formula we get,
$\text{Number of gram-equivalents of HCl = N }\!\!\times\!\!\text{ V = 0}\text{.1 }\times \text{ 0}\text{.01 =0}\text{.001 }$
Now,
The total volume of the solution is equal to the volume of solution.i.e.volume of $\text{ NaCl }$ and $\text{ HCl }$.
$\begin{align}
& \text{Total volume of solution = }{{\text{V}}_{\text{NaCl}}}\text{ + }{{\text{V}}_{\text{HCl}}}\text{ } \\
& \therefore \text{Volume of solution = 990 + 10 }=\text{1000 mL or 1L} \\
\end{align}$
As NaCl is salt it will not contribute towards the \[\text{ pH}\] of the solution. Therefore, the pH of the solution will depend on the concentration of $\text{HCl}$ only.
Thus, the normality of $\text{HCl}$ is written as,
$\text{ }{{\text{N}}_{\text{HCl}}}\text{=}\dfrac{\text{Eq}\text{.}}{\text{V}}\text{=}\dfrac{\text{0}\text{.001}}{\text{1}}\text{ = 0}\text{.001 N}$
Now we have to find the value of\[\text{ pH}\]. The formula of \[\text{ pH}\] is given by:
\[\text{pH }=~\text{ }-\log [{{\text{H}}^{\text{+}}}]\text{ }=\text{ }-\text{log }\left( \text{0}\text{.001} \right)\text{ = 3}\]
Therefore, we can say that \[\text{ pH}\] of the resulting solution is 3.

Hence, (B) is the correct option.

Note: This \[\text{ pH}\] is associated with the species which can donate or accept the proton. In solution, the $\text{HCl}$ dissociates as follows:
$\text{ HCl(aq)}\to \text{ }{{\text{H}}^{\text{+}}}\text{(aq) + C}{{\text{l}}^{-}}\text{(aq) }$
Therefore, the \[\text{ pH}\] can be determined. However, the $\text{ NaCl }$ do not produce protons in the solution.
$\text{ NaCl(aq)}\to \text{ N}{{\text{a}}^{\text{+}}}\text{(aq) + C}{{\text{l}}^{-}}\text{(aq) }$
Thus, the $\text{ NaCl }$do not contribute towards\[\text{ pH}\] of the solution.