What is the pH of a 1.00 molar solution of NaNO2(aq)? The Ka for HNO2 is 7.0*10^-4
(Please include how you solved the problem. Thanks!)
NaNO2 -----> Na+ + NO2-
NO2- ion undergoes hydrolysis.
NO2- + H2O <-----> HNO2 + OH-
1.0 0 0 initial
-x +x +x change
1.0-x x x equilibrium
Ka*Kb = 1.0*10^-14
Kb = (1.0*10^-14)/(7.0*10^-4) ==> 1.43*10^-11
Kb = [NHO2][OH-]/[NO2-]
1.43*10^-11 = x^2/(1)
x = 3.78*10^-6
pOH = -log(3.78*10^-6)
= 5.42
pH = 14 - 5.42
= 8.58
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