Question

50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. Calculate the pH of the resulting solution.
Consider $log\left(2\right)=0.30$

• A. 1.3
• B. 4.2
• C. 12.7
• D. 11.7

Answer 1

The correct option is C 12.7

Millimoles of $HCl=0.1\times 50=5$ mmol
Millimoles of $NaOH=0.2\times 50=10$ mmol
1 mmol of $HCl$ is neutralized by 1 mmol of $NaOH$
$\therefore$ Resulting solution contains 5 mmol of NaOH in (50 + 50) = 100 mL of solution
$\therefore [OH{]}^{-}=\frac{5\times {10}^{-3}}{100\times {10}^{-3}}=5\times {10}^{-2}M$
$[OH{]}^{-}=5\times {10}^{-2}M$

$\left[H^{+}\right]=\frac{K_w}{O{H}^{-}}=\frac{{10}^{-14}}{5\times {10}^{-2}}=2\times {10}^{-13}$
$pH=-log\left[H^{+}\right]=-log(2\times {10}^{-13})$
$=(13-log2)=13-0.30=12.7pH=12.7$