Question

Calculate the pH of a solution prepared by mixing 2.0 mL of a strong acid (HCI) solution of pH 3.0 and 3.0 mL of a strong base (NaOH) of pH 10.0

• 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 $pH+pOH=14\Rightarrow 10+pOH=14pOH=10\left[O{H}^{-}\right]={10}^{-4}M\text{3 mL of}{10}^{-4}MNaOH=3\times {10}^{-4}$

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