What is the pH of a 0.1 M aqueous solution of NaH2PO4 Solution NaH2PO4 --------------> Na^+ (aq) + H2Po4^- (aq) Ka = 6.8*10^-8 Kb = Kw/Ka = 1*10^-14/6.8*10^-8 = 1.47*10^-7 H2PO4^- (aq) + H2O(l) ----------------> H3Po4(aq) + OH^- (aq) I 0.1 0 0 C -x +x +x E 0.1-x +x +x Kb = [H3PO4][OH^-]/[H2Po4^-] 1.47*10^-7 = x*x/0.1-x 1.47*10^-7 *(0.1-x) = x^2 x = 0.00012 [OH^-] = x = 0.00012M POH = -log[OH^-] = -log0.00012 = 3.92 PH = 14-POH = 14-3.92 = 10.08 .