Question

What will be the resultant pH when 150 mL of an aqueous solution of $HCl$ $(pH=2.0)$ is mixed with 350 mL of an aqueous solution of $NaOH(pH=12.0)$?
Take $log\left(2\right)=0.3$

• A. 8.6
• B. 12.6
• C. 11.6
• D. 10.6

Answer 1

The correct option is C 11.6

$pHofHCl=2$
$\therefore [H{]}^{+}={10}^{-2}M$
$pHofNaOH=12,$ $pOH=2$ $\therefore [OH{]}^{-}={10}^{-2}M$

$\begin{array}{cccccc}& HCl& +& NaOH⟶NaCl& +& H_{2}O\\ \text{meq. initial}& 150\times {10}^{-2}& & 350\times {10}^{-2}& 0& 0\\ \text{meq. final}& 0& & 2& 1.5& 1.5\end{array}$

$[OH{]}^{-}$ from $NaOH=\frac{2}{500}=4\times {10}^{-3}M$
$pOH=-log[OH{]}^{-}=-log(4\times {10}^{-3})$
$pOH=-\left(\right(2\times 0.3)-3)=2.4$
$pH=14-pOH=14-2.4=11.6$