Question

1) What is the ph of a 0.005 M , solution of h2so4? ka2=1.2x10^-2 of HSo4-

Answer 1

Since the Ka1 value of H2SO4 is very high, all of the H2SO4 are dissociated into H+ and HSO₄^-(aq).

Now for the dissociation of HSO₄^-(aq),

------------ HSO₄^-(aq) ------ > SO₄^2-(aq) + H⁺(aq) ; Ka2 = 0.012

Init.conc: 0.005 M, ---------- 0 M, --------- 0.005 M

change: - x M, -------------- + x M, ------- + x M

eqm.conc:(0.005 - x) M, -- x M, -------- (0.005 + x) M

Ka2 = 0.012 = [SO₄^2-(aq)] *[H⁺(aq)] / [HSO₄^-(aq)] =  x * (0.005 + x) / (0.005 - x)

=> x² + 0.017x - 6.0*10^-5 = 0

=> x = 0.003 M

Hence [H⁺(aq)] = (0.005 + x) M = (0.005 + 0.003) M = 0.008 M

=> pH = - log[H⁺(aq)] = - log (0.008 M) = 2.10 (answer)