Question

What is the pH of a 0.15 M solution of potassium nitrite (KNO2)? (The Ka value for nitrous acid (HNO2)

is 4.0x10^-4).

Answer 1

You don0t need the concentration because it will be in equilibrium. That is, you may calculate it with the Equilibrium constant as follow:

KNO2(s) ----> K+(aq) + NO2-(aq) (in solution)

Forget K+ ion for now,

NO2-(aq) + H2O(l) <---> HNO2(aq) + OH-(aq)

NOTE that this salt (KNO2) will "free" OH- ions in solution, making it basic! Expect Basic pH

This equilibrium can be modeled as

Kb = [HNO2][OH-]/[NO2-]

But we don't have Kb, we got Ka = 4*10^-14

Recall from Equilibrium of water at 25ºC

Kw = Ka*Kb

Kw = 10^-14 (constant at 25ºC)

Ka = 4*10^-4 (you gave it)

Kb = Kw/Ka = (10^-14)/(4*10^-4) = 2.5*10^-11

Now, from the past equilbirum

Kb = [HNO2][OH-]/[NO2-]

Assume 1 mol of HNO2 per 1 mol of OH- (due to stoichiometry) lets call this "x"

[HNO2] = [OH-] = x

[NO2-] =  0.15 M (since all KNO2 goes into solution)

Substite all known values

Kb = [HNO2][OH-]/[NO2-]

2.5*10^-11 = x*x/(0.15)

solve for x

x = sqrt((2.5*10^-11)(0.15)) = 1.94*10^-6

Since [OH-] = x = 1.94*10^-6

pOH = -log(OH) = -log(1.94*10^-6) = 5.71

But we need pH

pH = 14- pOH = 14-5.71 = 8.29

which is a bsic pH, as expected

pH = 8.29