Calculate the pH of a 0.250M solution of Sulfurous acid, H2SO3.

Answer 1

You will need the Ka for sulfurous acid to calculate the pH as it is a weak acid. Were you given a figure in the question?

H2SO3<> HSO3- + H+

Ka = 1.54×10−2 mol L^-1
(from a data book)

at start
H2SO3 HSO3- H+

0.250M 0M 0M

at equilibrium
0.250-x x x

Ka=[HSO3-][H+]/[H2SO3]

1.54×10−2 mol L^-1 = x^2/(0.250-x)

you can either solve the quadratic or assume that x is small with respect to 0.250 hence

1.54×10−2 mol L^-1 = x^2/(0.250 mol L^-1)

x^2=0.00385 mol^2 L^-2

hence find x

and pH=-log(x)

Answer 2

1.207