Question

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

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

Answer 1

A stoichiometric equation is required:

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

We get (finally) `pH=1.52`

Explanation:

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`