H3PO4(aq) + H2O(l) → ← H3O+(aq) + H2PO4–(aq)
Ka1 = [H3O+]e[H2PO4–]e [H3PO4]e = 7.1×10–3
Initial0.037 0 0
Change–x +x +x
Equilibrium0.037 – x x x
Can we approximate? 0.037/0.0071 = 5.2 NO!
7.1×10–3 = (x)(x) (0.037 – x)
2.63×10–4 – 7.1×10–3x = x2
x2 + 7.1×10–3x – 2.63×10–4 = 0
We will need to use the quadratic equation to give:
x = 1.3×10–2 or x = –2.0×10–2
Only the positive value makes sense chemically, so
[H3O+] = [H2PO4–] = 1.3×10–2 M
H2PO4–(aq) + H2O(l) → ← H3O+(aq) + HPO42–(aq)
Ka2 = [H3O+]e[HPO42–]e [H2PO4–]e = 6.3×10–8
Initial0.013 0.013 0
Change–x +x +x
Equilibrium0.013 – x 0.013 + x x
Can we approximate? 0.013/6.3×10–8 = 2.1×105 YES!
6.3×10–8 = (0.013)(x) (0.013)
x = 6.3×10–8
[H3O+] = 0.013 + x = 0.013 + 0.000000063 = 0.013 M
The approximation was justified.
[H2PO4–] = 0.013 – x = 0.013 – 0.000000063 = 0.013 M
[HPO42–]e = 6.3×10–8 ( = Ka2) M
Since Ka3 (4.2×10–13) is much smaller than Ka2, there will be no significant contribution to the hydronium ion concentration from hydrogen phosphate ion.
The pH of the solution will be due only to Step 1:
pH = –log(1.3×10–2) = 1.89
This value is at the low end of our expected range, and since phosphoric acid is a reasonably strong weak acid, the answer is sensible.