Question

Calculate the pH of a solution made by dissolving 8.129 g of KNO2 into a 300.00 mL volumetric flask and adding water to the calibration mark. Ka of HNO2 = 7.2*10^-4

**Please show your work so I can fully understand how to solve for the pH given this info, thank you!

Answer 1

Molar mass of KNO2,

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

= 1*39.1 + 1*14.01 + 2*16.0

= 85.11 g/mol

mass(KNO2)= 8.129 g

number of mol of KNO2,

n = mass of KNO2/molar mass of KNO2

=(8.129 g)/(85.11 g/mol)

= 9.551*10^-2 mol

volume , V = 300.00 mL

= 0.3 L

Molarity,

M = number of mol / volume in L

= 9.551*10^-2/0.3

= 0.3184 M

This is concentration of KNO2

use:

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/7.2*10^-4

Kb = 1.389*10^-11

NO2- dissociates as

NO2- + H2O -----> KNO2 + OH-

0.3184 0 0

0.3184-x x x

Kb = [KNO2][OH-]/[NO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.389*10^-11)*0.3184) = 2.103*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.103*10^-6 M

use:

pOH = -log [OH-]

= -log (2.103*10^-6)

= 5.68

use:

PH = 14 - pOH

= 14 - 5.6772

= 8.32

Answer: 8.32