Question

What is `pH` in the following scenario?

`187.5*mL` of `0.60*mol*L^-1` `KOH` is reacted with `250.0*mL` of `0.30*mol*L^-1` `HNO_3`.

Answer 1

`pH~=13`

Explanation:

We assess the extent of the following reaction......

`KOH(aq) + HNO_3(aq) rarr H_2O(l) + KNO_3(aq)`....

There is thus 1:1 stoichiometry, but unequal molar quantities of `KOH` and `HNO_3`.

`"Moles of KOH"=187.5xx10^-3*Lxx0.60*mol*L^-1=0.1125*mol.`

`"Moles of nitric acid"=250.0xx10^-3*Lxx0.30*mol*L^-1=0.0750*mol`.

And thus with respect to `KOH` there is a concentration of

`(0.1125*mol-0.0750*mol)/(437.5xx10^-3*L)-=0.0857*mol*L^-1`.

Note that we assume (very reasonably) that the volumes are additive.......and really we must assume this.

And since `pOH=-log_10[HO^-]`, and since `pOH=-log_10(0.0857)=1.07`, and further since `pH+pOH=14`, `pH=12.93`.

Capisce?