| Calculating [H+] and [A2-] at Equilibrium in a Solution of a Weak Diprotic Acid H2A |
| Suppose that for the diprotic acid, H2A, Ka1
= 1.0 x 10-5 and Ka2 = 1.0 x 10-9.
|
| What are the values of [H+] and [A2-]
at equilibrium
in 0.100 mol/L H2A?
|
|
H2A <--------> H+
+
HA-
Initial 0.1 M ----- ----- Changes -x +x +x |
| Corrections
caused by second equilibrium 0 +y -y The second conjugate of H+ at the expense of HA- Final 0.1
-x
x+y x-y Since y is very
very small x+yx and x-yx.
|
| Ka1 = (x)(x) = 1.0 x 10-5
0.1 - x |
| 1.0 x 10-6 - 1.0 x 10-5x = x2
x2 + 1.0 x 10-5x -1.0 x 10-6
= 0
(use the quadratic equation on this)
|
| x = 9.95 x 10-4 mol/L
Concentration of [A2-] at equilibrium Ka2 = [H+][A2-]
but
[H+] = [HA-]
Ka2 = [A2-] = A2- [Ka2] = Ka2 = 1.0 x 10-9
mol/L
|