Question

In one glass - 100,0 mL 0,100M KOH solution, in other - 100,mL 0,100M HNO2 solution. What is pH of solution, which was get when both solutions were mixed up (consider that V of forthcoming solution is sum of original solutions - 200,0 ml)

Answer 1

The KOH and HNO2 exactly neutralize each other and the pH at the equivalence point is the pH of the hydrolyzed salt. It has a concn of 0.1M x (100 mL/200mL) = 0.05M

............NO2^- + HOH ==> HNO2 + OH^-
initial.....0.05.............0.......0
change.......-x..............x........x
equil......0.05-x............x........x

Kb for NO2^- = (Kw/Ka for HNO2) = (HNO2)(OH^-)/(NO2^-)

Substitute into the Kb expression above and solve for x = (OH^-), then convert to pH.

Answer 2

Why concentration of salt is 0,05?