Question

# For the diprotic weak acid H2A, Ka1 = 3.2 × 10-6 and Ka2 = 5.4 ×...

For the diprotic weak acid H2A, Ka1 = 3.2 × 10-6 and Ka2 = 5.4 × 10-9. What is the pH of a 0.0550 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

pH=

[H2A]=

[A2-]=

For the diprotic weak acid H2A,

Ka1 = 3.2 * 10^-6 and Ka2 = 5.4 *10^-9

M = 0.055 of H2A

In equilibrium:

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

3.2 * 10^-6 = x^2 / (0.055-x)

[H2A] = 0.055

solve for x:

x = 4.17*10^-4

[H2A] = 0.055 -  4.17*10^-4 = 0.054583

[H2A] =0.054583 M

Now substitute this data in the second equlibrium

Ka2 = [H+][A-]/[HA]

5.4 *10^-9. = ( 4.17*10^-4+x)(x) / ( 4.17*10^-4-x )

x is relatively small so we can ignore it... if you want an exact solution you may calculate for x and then substitute here:

5.4 *10^-9*(( 4.17*10^-4-x )) =  (  4.17*10^-4+x)(x)

( 5.4 *10^-9)( 4.17*10^-4) - ( 5.4 *10^-9)x = ( 4.17*10^-4)x + x^2

x^2 + ( 4.17*10^-4 + 5.4*10^-9)x -(2.2518*10^-12)= 0

x = 5.39*10^-9

[A-2] = 5.39*10^-9

[H+] = 3.984*10^-4+5.39*10^-9

[H+] = 3.984*10^-4

pH = - log ( 3.984*10^-4)

ph = 3.39968