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

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,

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