Question

What would be the pH of a buffer that contains 0.987 grams of Nitrous acid (HNO₂) and 1.242 grams of sodium Nitrite (NaNO₂), in 250ml of solution?(Ka of Nitrous acid = 4.5X10^-4)

Answer 1

Molar mass of HNO2,

MM = 1*MM(H) + 1*MM(N) + 2*MM(O)

= 1*1.008 + 1*14.01 + 2*16.0

= 47.018 g/mol

mass(HNO2)= 0.987 g

use:

number of mol of HNO2,

n = mass of HNO2/molar mass of HNO2

=(0.987 g)/(47.02 g/mol)

= 2.099*10^-2 mol

volume , V = 2.5*10^2 mL

= 0.25 L

use:

Molarity,

M = number of mol / volume in L

= 2.099*10^-2/0.25

= 8.397*10^-2 M

Molar mass of NaNO2,

MM = 1*MM(Na) + 1*MM(N) + 2*MM(O)

= 1*22.99 + 1*14.01 + 2*16.0

= 69 g/mol

mass(NaNO2)= 1.242 g

use:

number of mol of NaNO2,

n = mass of NaNO2/molar mass of NaNO2

=(1.242 g)/(69 g/mol)

= 1.8*10^-2 mol

volume , V = 2.5*10^2 mL

= 0.25 L

use:

Molarity,

M = number of mol / volume in L

= 1.8*10^-2/0.25

= 7.2*10^-2 M

HNO2 and NaNO2 forms buffer

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.347

use:

pH = pKa + log {[conjugate base]/[acid]}

pH = pKa + log {[NaNO2]/[HNO2]}

= 3.347+ log {7.2*10^-2/8.397*10^-2}

= 3.28

Answer: 3.28