Question

A 12.0 mass % solution of H3PO4 in water has a density of 1.104 g/mL. Calculate the pH in the solution.

Ka1= 7.1 × 10−3

Ka2 = 6.3 × 10−8

Ka2 = 4.2 × 10−13

Answer 1

molarity = mass % x 10 x density / molar mass

             = 12.0 x 10 x 1.104 / 98

            = 1.35 M

H3PO4 ------------------------> H2PO4-   + H+

1.35                                         0                 0

1.35-x                                     x                    x

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

7.1 x 10^-3 = x^2 / 1.35-x

x^2 + 7.1 x 10^-3 x - 9.585 x 10^-3 = 0

x = 0.0944

[H+] = 0.0944 M

pH = -log [H+]

pH = -log (0.0944)

pH = 1.02