Set up an ICE table.

Assume x is neglible and 0.45 – x ≅ 0.45 M

\(\displaystyle K_a\;=\;\frac{x^2}{0.45}\;=\;2.4\times 10^{-3}\)
 
Solve for x.

\(\displaystyle x\;=\;\sqrt{0.45\times\;(2.4\times 10^{-3})}\;=\;0.033\;M\)
 
Check assumption.

\(\displaystyle \frac{0.033\;M}{0.45\;M}\times\;100\;=\;7.3\%\)
 
7.3% > 5% — we need to solve for x using the quadratic formula.

\(\displaystyle \frac{x^2}{0.45\;-\;x}\;=\;2.4\times 10^{-3}\)
 
Multiply, collect all like terms, and set the equation to zero.

x² + (2.4 x 10^-3)x – 0.00108 = 0

a = 1, b = 2.4 x 10^-3, c = -0.00108

Use the quadratic formula.

\(\displaystyle \frac{-b\;\pm\;\sqrt{b^2\;-\;4ac}}{2a}\)
 
\(\displaystyle \frac{-2.4\times 10^{-3}\;\pm\;\sqrt{(2.4\times 10^{-3})^2\;-\;4\times 1\times\;-0.00108}}{2\times\;1}\)
 

\(\displaystyle \frac{-2.4\times 10^{-3}\;\pm\;0.0658}{2}\)
 
x = 0.0317 M, x = -0.0341

x = 0.0317 M

pH = -log(0.0317) = 1.50

The second dissociation will be neglible because K_a2 = 4.8 x 10^-9. This is a very small value. We can solve for x, the hydronium concentration, just to make sure.

HSeO₄^– (aq) + H₂O (l) ⇄ SeO₄^2- (aq) + H₃O⁺ (aq)

\(\displaystyle K_{a2}\;=\;\frac{[SeO_4^{2-}][H_3O^+]}{[HSeO_4^-]}\;=\;4.8\times 10^{-9}\)
 

\(\displaystyle x\;=\;\sqrt{0.0317\times\;4.8\times 10^{-9}}\;=\;1.2\times 10^{-5}\;M\)
 
Check the assumption.

Set up an ICE table for the first dissocation

Assume x is negligible and 0.25 – x ≅ 0.25

\(\displaystyle \frac{x^2}{0.25}\;=\;1.0\times 10^{-7}\)
 
x = 1.6 x 10^-4

Check assumption using 5% rule.

\(\displaystyle \frac{1.6\times 10^{-4}}{0.25}\times\;100\;=\;0.064\%\)
 
0.064% < 5% -- our assumption is valid.
[H₂S] = 0.25 – 1.6 x 10^-4 = 0.25 M
[H₃O⁺] = 1.6 x 10^-4 M
[HS^–] = 1.6 x 10^-4 M

The second dissociation is:

HS^– (aq) + H₂O (l) ⇄ S^2- (aq) + H₃O⁺ (aq)      K_a2 = 1.0 x 10^-19

Set up an ICE table. Use the concentrations from the first dissociation.

Assume x is negligible and 1.6 x 10^-4 M – x ≅ 0.25 1.6 x 10^-4 M

\(\displaystyle \frac{(1.6\times 10^{-4})x}{1.6\times 10^{-4}}\;=\;1.0\times 10^{-19}\)
 
x = 1.0 x 10^-19 M

Check assumption using 5% rule

\(\displaystyle \frac{1.0\times 10^{-19}}{1.6\times 10^{-4}}\times\;100\;=\;6.25\times 10^{-14}\;\%\)
 
6.25 x 10^-14 % < 5% -- our assumption is valid, and we are able to neglect x. The concentrations of all species are:
[H₂S] = 0.25 – 1.6 x 10^-4 = 0.25 M
[H₃O⁺] = 1.6 x 10^-4 M
[HS^–] = 1.6 x 10^-4 M
[S^2-] = 1.0 x 10^-19 M

The pH of the solution is:

The first dissociation essentially goes to completion. The concentrations of H₃O⁺ and HSeO₄^– are 2.5 M.

The second dissociation has pK_a2 = 1.70. K_a = 10^-1.70 = 0.01995 M.

HSeO₄^– (aq) + H₂O (l) ⇄ SeO₄^2- (aq) + H₃O⁺ (aq)

Set up an ICE table.

Assume 2.5 M – x and 2.5 M + x ≅ 2.5 M

\(\displaystyle \frac{x(2.5)}{2.5}\;=\;0.01995\)
 
x = 0.01995 M

Check the assumption

K_a1 = 10^-2.89 = 1.3 x 10^-3
K_a2 = 10^-4.40 = 3.98 x 10^-5

The chemical equation for the first dissociation is:

C₄H₆O₆ (aq) + H₂O (l) ⇄ C₄H₅O₆^– (aq) + H₃O⁺ (aq)

Set up an ICE table

Assume x is negligible and 0.65 M – x ≅ 0.65 M

\(\displaystyle \frac{x^2}{0.65}\;=\;1.3\times 10^{-3}\)
 
x = 0.029 M

Check assumption.

\(\displaystyle \frac{0.029\;M}{0.65\;M}\times\;100\;=\;4.5\%\)
 
4.5% < 5% -- our assumption is valid.
pH = -log(0.029) = 1.54

We can look at the second dissociation to see if it makes a difference in the pH.

C₄H₅O₆^– (aq) + H₂O (l) ⇄ C₄H₄O₆^2- (aq) + H₃O⁺ (aq)

The concentration of hydronium ion is 0.029 M + x, [C₄H₄O₆^2-] = x, and [C₄H₅O₆^–] = 0.029 – x.

We will assume 0.029 – x and 0.029 + x is ≅ 0.029 M

\(\displaystyle \frac{x(0.029)}{0.029}\;=\;3.98\times 10^{-5}\)
 
x = 3.98 x 10^-5 M

Check assumption.

\(\displaystyle \frac{3.98\times 10^{-5}\;M}{0.029\;M}\times\;100\;=\;0.14\%\)
 
0.14% < 5% and 0.029 - 3.98 x 10^-5 M = 0.029 M