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Calculate the pH of 0.11MNaNO2; Ka for HNO2 is 4.6×10^−4. express the answer to two decimal...

Calculate the pH of 0.11MNaNO2; Ka for HNO2 is 4.6×10^−4.

express the answer to two decimal places

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Homework Answers

Answer #1

NaNO2 when mixed in water will form:

Na+ and NO2- ions

then

NO2- + H2O <--> HNO2 + OH-

this is a basic salt, it will free OH- ions

then

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

Kb = Kw/Ka = (10^-14)/(4.6*10^-4) =2.1739*10^-11

note that in equilibrium

[HNO2] = x = [OH-]

[NO2-] = 0.11 -x

then

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

2.1739*10^-11 = (x*x)/(0.11-x)

solve for x

x = [OH-] = 1.546*10^-6

pOH = -log(OH) = -log(1.546*10^-6) = 5.81079

pH = 14- pOH = 14-5.81079 = 8.18921

pH = 8.19

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