Question

Calculate the pH of a 0.375 M aqueous solution of nitrous acid (HNO2, Ka = 4.5×10-4)...

Calculate the pH of a 0.375 M aqueous solution of nitrous acid (HNO2, Ka = 4.5×10-4) and the equilibrium concentrations of the weak acid and its conjugate base.

pH =   
[HNO2 ]equilibrium =    M
[NO2- ]equilibrium =    M

Homework Answers

Answer #1

HNO2 H+ + NO2-

Initial concentration: 0.375 M 0 0

equilibrium concentration: (0.375 -x)M x M xM (Let us assume x M is the equilibrium concentration of the conjugate base)

Ka = [H+]. [NO2-] / [ HNO2]

= x2 / (0.375 -x)

4.5 x 10 -4 =  x2 / (0.375 -x)

=  x2 / 0.375 (as x is very small)

x2 =    4.5 x 10 -4 x 0.375 = 1.6875 x 10 -4

   x = 1.299 x 10 -2

= 0.01299 M = 0.013 M

pH = - log [ H+] = - log [ 0.013] = 1.886 (answer)

Concentration of weak acid (HNO2) = (0.375 - x) = (0.375 - 0.013) M = 0.362 M (answer)

Concentration of conjugate base of weak acid (NO2-) = x = 0.013 M (answer)

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