What is the $pH$ of the final solution...?

Question

A $25mL$ volume of $HBr(aq)$ at $0.050molL^-1$ concentration is mixed with a $10mL$ volume of $KOH(aq)$ at $0.020molL^-1$ concentration. What is the $pH$ of the final solution?

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Answer 1 · 2017-03-23T07:08:23

A stoichiometric equation is required:

$HBr(aq) + KOH(aq) rarr NaBr(aq) + H_2O(l)$

We get (finally) $pH=1.52$

And thus we need to find the amount of substance of both $"hydrogen bromide"$ , $"hydrobromic acid"$ , and $"potassium hydroxide"$ .

We use the relationship, $"Concentration"="Moles of solute"/"Volume of solution"$ , OR

$"Concentration"xx"Volume"="Moles of solute"$

$"Moles of HBr"=25xx10^-3cancelLxx0.050*mol*cancel(L^-1)=1.25xx10^-3*mol.$

$"Moles of KOH"=10xx10^-3cancelLxx0.020*mol*cancel(L^-1)=0.200xx10^-3*mol.$

Note that I converted the $mL$ volume to $L$ by using the relationship: $1*mL-=1xx10^-3*L$

Clearly, the hydrobromic acid is in excess. And given 1:1 stoichiometry, there are $((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)$ $HBr$ remaining.

So $[HBr]=((1.25*mol-0.200*mol)xx10^-3)/(35xx10^-3L)$

$=(1.05*molxx10^-3)/(35xx10^-3L)=0.030*mol*L^-1$ with respect to $HBr$ .

$pH=-log_10[H_3O^+]=-log_10(0.030)=1.52$