Question

A solution of nitrous acid, HNO2 is 6.3x10^-3 M, which is a weak acid with Ka=7.2x10^-4.

Calculate:

[HNO2]

[NO2]

pH

What is the concentration of a nitrous acid solution with a pH of 2.21

Answer 1

HNO2 = 6.3x10^-3M= 0.0063M             Ka= 7.2x10^-4

            HNO2 + H2O ----------------- H3O+ + NO2-

Initial       0.0063                                     0            0

change        -x                                           +x           +x

equilibrium 0.0063-x                                 +x            +x

    Ka= [H3O+][NO2-]/[HNO3]

7.2x10^-4 = x*x/(0.0063-x)

for solving the equation

x=0.0018

[HNO2]= 0.0063 - 0.0018 = 0.0045M

[NO2-]=0.0018M

[H3O+] = 0.0018M

[H3O+] = [H+] = 0.0018M

-log[H+] = -log(0.0018)

PH= 2.74

B)

PH= 2.21

-log[H+] = 2.21

[H+] = 10^-2.21

[H+]= 0.0062M

Ka= 7.2x10^-4

for weak acids

[H+] = square root of KaxC

[H+]^2= KaxC

C= [H+}^2/Ka = ( 0.0062)^2/7.2x10^-4

C= 5.34x10^-2M

Concentration of HNO2 solution = 5.34x10^-2M