Question

$100mL$ of a strong acid solution of $pH=3$ is mixed with $900mL$ of another strong acid solution of $pH=4$. What will be the $pH$ of the resulting solution?
Take ${\log }_{10}\left(1.9\right)\approx 0.28$

• A. 3.28
• B. 4.28
• C. 1.72
• D. 3.72

Answer 1

The correct option is D 3.72

$pH=-{\log }_{10}[H{]}^{+}$
$pH=3[H{]}^{+}={10}^{-3}M$
$pH=4[H{]}^{+}={10}^{-4}M$

$\left[H^{+}\right]$ after mixing:
$\frac{{10}^{-3}\times 100+{10}^{-4}\times 900}{100+900}=1.9\times {10}^{-4}pH=-\left({\log }_{10}\right(1.9\times {10}^{-4})$
$pH=(4-\log 1.99)$
$pH=4-0.28=3.72$