Question

Calculate the pH at 25°C of a 0.65 M aqueous solution of phosphoric acid (H3PO4). (Ka1, Ka2, and Ka3 for phosphoric acid are 7.5 × 10−3, 6.25 × 10−8, and 4.8 × 10−13, respectively.)

Please show how you do the quadratic equation. I am struggling with this in particular. Thank you!

Answer 1

------- H3PO4(aq) ----------> H^+ (aq)    +    H2PO4^- (aq)

I ----- 0.65 ---------------------- 0 ------------------- 0

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

E--- 0.65-x -------------------- +x ------------------ +x

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

          7.5*10^-3   = x*x/(0.65-x)

         7.5*10^-3(0.65-x) = x^2

          x^2 + 7.5*10^-3x -0.004875 = 0

        x^2 +0.0075x-0.004875 = 0

    x = -b- b^2-4ac/2a   , x = -b+ b^2-4ac/2a  

x = -0.07367                   , x = 0.066

[OH^-]   = x   = 0.066M

POH   = -log[OH^-]

            = -log0.066

            = 1.18

PH      = 14-POH

           = 14-1.18

            = 12.82>>>>>answer