Question

$10\mathrm{mL}$of $0.1N$$\mathrm{HCl}$ is added to $990mL$ solution of $NaCl$. The $\mathrm{pH}$ of the resulting solution is

• A. zero
• B. $3$
• C. $7$
• D. $10$

Answer 1

The correct option is B

$3$

Step 1: given data

Volume of $\mathrm{HCl}=10\mathrm{mL}$

The total volume of the solution is, $990+10=1000mLor1L$

The number of moles of HCl:

$\begin{array}{rcl}\mathrm{HCl}& =& \mathrm{Volume}\times \mathrm{Concentration}\\ & =& 10\times 0.1\\ & =& \text{1 millimoles}\end{array}$

Step 2: calculate the value of $\mathrm{pH}$

So, the concentration of $\left[H^{+}\right]$ ion$=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}}{\mathrm{Total}\mathrm{volume}\mathrm{of}\mathrm{solution}}$

$\begin{array}{rcl}\therefore \left[H^{+}\right]& =& \frac{1\times {10}^{-3}}{1}M\\ & =& {10}^{-3}M\\ & & \end{array}$

$$\begin{gathered} \therefore \mathrm{pH}=-\log \left[{10}^{-3}\right]\left[\because \log \left(m^n\right)=n\log \left(m\right)\right] \\ \Rightarrow \mathrm{pH}=3 \end{gathered}$$

Final Answer: hence, the correct option is (B), the $\mathrm{pH}$ of the resulting solution is $3$.