Concentration of nitrous acid?

Question

At what solute concentration would nitrous acid have pH of 4.00? The Ka for the acid is 4 x 10^-4

Answer 1

HNO2, nitrous acid

HNO2 ~~> NO2- + H+

Ka = 4*10^(-4)

Ka = ([NO2-]*[H+])/[HNO2]

knowing pH = 4.00, you get [H+] = 10^(-4.00) and you know from the chemical equation above [H+] should be equal to [NO2-] = 10^(-4.00).

At equilibrium, [HNO2] = 0.000025 M = 2.5*10^(-5) M

But as an initial solute concentration, it is 1.25*10^(-4) M or 0.000125. The reason they are different is because HNO2 had to lose some concentration in order to reach equilibrium forming NO2- and H+. Therefore the value for [HNO2] in the equilibrium formula is (1.25*10^(-4) - [H+ or NO2-]) or (1.25*10^(-4) - 10^(-4.00)).

The solute concentration is 1.25*10^(-4) M

The concentration at equilibrium is 2.5*10^(-5)