What is the pH of a 1.0 M H2SO3 solution?
(K1 = 1.2 x 10-2
, and K2 = 6.6 x 10-8
The majority of the H+ ions come from the first ionization (k1) and we can ignore k2.
............H2SO3 ==> H^+ + HSO3^-
I.............1 M ..............0..........0
C.............-x.................x..........x
E..............1-x...............x..........x
k2 = 0.012 = (H^+)(HSO3^-)/(H2SO3).
Plus the E line into the k2 expression and solve for (H^+), then convert to pH where pH = -log(H^+)
Post your work if you get stuck. Note that k2 is about 10^6 times weaker than k1; therefore, for every one HSO3^- that ionizes to affect the pH you have 1,000,000 H2SO3 molecules ionizing to affect the pH. That's why we can ignore the contribution of k2. Note, also, that the equation obtained when you substitute the E line is a quadratic and it will be necessary to solve the quadratic equation; i.e., you may not ignore the x in 1-x.