Question

A student is given 300.0 ml of a 1.10 M HNO₂ solution. The Ka of the acid is 4.5 X 10^-4. What mass of KNO₂ in grams should the student dissolve in the HNO₂ solution to turn it into a buffer with pH 3.57?

Answer 1

V = 300 ml

M = 1.10

Ka = 4.5*10^-4

KNO2 mass

pH = 3.57

This will form a buffer

use henderson hasselbach equations

pH = pka + log(Conjugate / acid)

3.57 = -log(4.5*10^-4) + log( conjugate / 1.10 )

3.57 = 3.35 + log( conj / 1.10 )

10^0.22 = [Conjugate]/1.1

1.659 = [Conjugate] / 1.1

[Conjugate] = 1.659*1.1 = 1.825

Since we have 300 ml...

M*V = moles

0.3*1.825 = 0.547 moles of KNO2

MW of KNO2 = 85.1g/gmol

mass = mol*MW = 0.547*85.1= 46.54 grams of KNO2 are needed