Question

How many grams of potassium nitrite need to be added to 250.0mL of a 0.1234 M HNO2 solution to produce a final buffer at pH= 3.65? Ka HNO2= 4.5 x 10 ^-4

Answer 1

From Henderson’s equation:

pH = pKa + log [salt]/[acid]

3.65 = - log 4.5 x 10^-4 + log [salt]/[acid]

3.65 = 3.35 + log [salt]/[acid]

0.3 = log [salt]/[acid]

2 = [salt]/[acid]

2[acid] = [salt]

Number of mmols of HNO₂ before addition of NaNO₂ = N x V = 0.1234 x 250.0 = 30.85

Number of mmols of HNO₂ after addition of NaNO₂ = 250.0 [acid]

Number of mmols of NaNO₂ = 2 x 250.0 [acid] = 500 [acid] = 500 x 0.1234 = 61.7 mmol

Number of moles of NaNO₂ = 0.0617

Mass of NaNO₂ = number of moles x molar mass = 0.0617 x 69 = 4.26 g