Biochemistry
posted by Anonymous .
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)

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......0.05x............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. 
Why concentration of salt is 0,05?
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