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

Answer:

#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?