HA

H

^{+} + A

^{-}At equilibrium CA0-x x x

H2O

H

^{+} + OH

^{-}At equilibrium something y y

[tex]

K_w=[H^+][OH^-] \\

K_w=(x+y)y (Eq. a) \\

K_a=\frac{[H^+][A^-]}{[HA]} \\

Ka=\frac{(x+y).x}{C_{A0}-x} (Eq. b) \\

K_w=10^{-14} \\

K_a=1.8 \times 10^{-5} \\

C_{A0}=0.57 \times 10^{-9} \\

[/tex]

Solve (Eq. a) and (Eq. b) simultaneously in x and y to get

[tex]

x=5.6684 \times 10^{-10} \\

y=9.9717 \times 10^{-8} \\

[H^+] = x+y = 1.0028 \times 10^{-7} \\

[/tex]

By Borek's shortcut

[tex]

[H^+] = \sqrt{C_{A0} \times K_a} = 1.013 \times 10{-7} \\

[/tex]

QED?

Two wrongs make a right? The Kw effect compensates for neglecting the HA dissociation?

An analytical closed form solution might be more elegant though.....Still working on it.