Question

We will abbreviate malonic acid CH2(CO2H)2, a diprotic acid, as H2A (pK1 = 2.847 and pK2 = 5.696). Find the pH in (a) 0.200 M H2A; and (b) 0.200 M NaHA.

I also need to make sure the answers are in the correct number of significant digits, thank you!

Answer 1

a)

[H2A] = 0.200 M

H2A -----------------> HA-   +    H+

0.2                          0            0

0.2- x                      x             x

Ka1 = [HA-][H+] / [H2A]

1.42 x 10^-3 = x^2 / 0.2 - x

x = 0.0162

[H+] = 0.0162 M

pH = -log[H+] = -log (0.0162 )

pH = 1.79

b)

0.200 M NaHA

here

pH = pKa1 + pKa2 / 2

     = 2.847 + 5.696 / 2

     = 4.27

pH = 4.27