Question

Calculate the pH of a solution prepared by mixing $2.0\text{mL}$ of a strong acid $\left(HCI\right)$ solution of pH $3.0$ and $3.0\text{mL}$ of a strong base $\left(NaOH\right)$ of pH $10.0$. $(Take:lo{g}_{10}34=1.5)$

• A. $2.5$
• B. $3.5$
• C. $4.5$
• D. $6.5$

Answer 1

The correct option is B $3.5$

$\text{Number of milliequivalents of HCl}=2\times {10}^{-3}$
We know that $pH+pOH=14\Rightarrow 10+pOH=14pOH=10\left[O{H}^{-}\right]={10}^{-4}\text{M}\text{3 mL of}{10}^{-4}\text{M}NaOH=3\times {10}^{-4}$

$\text{Number of milliequivalents of NaOH}=3\times {10}^{-4}$
The resulting solution is acidic.
$\left[H^{+}\right]=\frac{(2\times {10}^{-3}-3\times {10}^{-4}\text{m. eq.})}{(2+3)\text{mL}}=3.4\times {10}^{-4}\text{M}orpH=-\log \left[H^{+}\right]$
$pH=-\log (3.4\times {10}^{-4})=3.5$