Question

Phosphoric acid is triprotic with the step-wise dissociation shown below. A 1.34 M phosphoric acid solution is prepared. Calculate the equilibrium concentrations of the species indicated. Report all answers in scientific notation using 2 sig figs. Rather than use the quadratic equation, you can make assumptions like we do in class (ie if K<<1, then 0.1-x -0.1) H3PO HO H.PO H3O Kai = 7.5c10-3 Preview Preview Preview H.PO H20 HPo? + H50+ K - 6.2:10 -* Preview Preview HPO H2O + PO H,O K 4.8x10-13 Preview Preview What is the pH of this solution? Report pH as a number and not in scientific notation.

Answer 1

----H3PO4(aq) + H2O(l) ----------> H3O^+ (aq) + H2PO4^- (aq)

I ---- 1.34M ---------------------------- 0 -------------   0

C---- - x------------------------------- +x --------------- +x

E---- 1.34-x ------------------------- +x --------------   +x

       Ka1   = [H3O^+][H2PO4^-]/[H3PO4]

      7.5*10^-3 = x*x/(1.34-x)

   7.5*10^-3(1.34-x) = x^2

   x = 0.097

[H3Po4] = 1.34-x = 1.34-0.097 = 1.243M

[H3O^+]   = x    = 0.097M

[H2Po4^-] = x   = 0.097M

H2Po4^- (aq) + H2O(l) -----------> HPo4^2- (aq) + H3O^+ (aq)

   Ka2   = [HPO4^2-][H3O^+]/[H2Po4^-]

   6.2*10^-8 = [HPO4^2-]*0.097/0.097

[HPO4^2-]   = 6.2*10^-8 M

HPO4^2- (aq) + H2O(l) ---------> H3O^+ (aq) + PO4^3-(aq)

   Ka3   = [H3O^+][PO4^3-]/[HPO4^2-]

    4.8*10^-13 = 0.0097[PO4^3-]/6.2*10^-8

    [PO4^3-]   = 4.8*10^-13*6.2*10^-8/0.0097 = 3.06*10^-18 M

PH = -log[H3O^+]

       = -log0.0097

        = 2.0132 >>>>>answer

         = 2.0132*10^0