Question

If $200mL$ of a $0.031\,molar\,solution\,of\,{H_2}S{O_4}$ are added to $84rall.$ of a $0.150M\,KOH$ solution, what is the $pH$ of the resulting solution?

Answer 1

Hint: In order to this question, to know the $pH$ of the final concluded solution, we will first find the initial number of moles of ${H^ + }$ and then initial number of moles of $O{H^ - }$ , and then we can find the $pH$ of the resulting solution.

Complete answer:
$mmol\,of\,{H^ + }(initial) = 200 \times 0.031 \times 2 = 124$
$mmol\,of\,O{H^ - }(initial) = 84 \times 0.15 = 12.6$
$mmol\,of\,O{H^ - }(left)after\,neutralisation = 0.2{[O{H^ - }]_{final}} = \frac{{0.2}}{{284}} = 7 \times {10^{ - 4}}M$
$pOH = 3.15\,$
and $pH = 10.85$

Note:
The $pH$ level of a solution shows whether it is acidic, alkaline or neutral. Neutral means it is neither acidic nor alkaline. On a scale of 0 to 14, a $pH$ level of 7 is neutral, a $pH$ level lower than 7 means a solution is acidic, and a $pH$ level greater than 7 means a solution is alkaline. Pure or distilled water has a $pH$ level of 7.