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.

1 Answer
Aug 4, 2017

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?