Calculate the hydronium ion concentration in 50.0 mL of 0.10M NaH2AsO4.
and we are given K1 = 6.0E-3, K2 = 1.1E-7 and K3 = 3.0E-12
I assumed these K values where given in respect to the Ka of acid H3AsO4.
(H^+) = sqrt[(k2(H2AsO4^-2)+Kw)/1+(H2AsO4^-2/k1)]
I think this can be simplified by calling (HAsO4^-2) = (salt) = C and this reduces to
(H^+) = sqrt[(k2C + kw)/1 + (C/k1)]
Since C/k1 is quite a bit larger than 1, this reduces to
(H^+) = sqrt (k1k2).
This latter one is the one I would use. For references on how this was derived, look in your text on how NaHCO3 is done.