Question

The $pH$ of $HCl$ is 1. $1.10ml$ of this solution is reacted with $40ml$ of $NaOH$ solution whose $pH=12$. The $pH$ of resulting solution is:

$\left(\log \right(1.2)=0.108)$

• A. $2.892$
• B. $1.892$
• C. $32.892$
• D. $11.842$

Answer 1

The correct option is D $11.842$

$\left[H^{+}\right]={10}^{-1}M$

moles of $HCl$ present$=1.10\times {10}^{-1}\times {10}^{-3}=1.1\times {10}^{-4}$

$\left[O{H}^{-}\right]={10}^{-(14-12)}={10}^{-2}M$

moles of $NaOH$ present$=40\times {10}^{-2}\times {10}^{-3}=4\times {10}^{-4}$

$\therefore$ moles of $O{H}^{-}$ left$=(4-1.1)\times {10}^{-4}=2.9\times {10}^{-4}$

Total Volume$=40+1.1=41.1ml$

$\therefore \left[O{H}^{-}\right]=\frac{2.9\times {10}^{-4}}{41.1\times {10}^{-3}}=0.00705$

$\therefore pOH=-\log 0.00705=2.1518$

$\therefore pH=11.8482$