Answer
$[H_3O^+] = 0.026 \space M$
$[HS{O_4}^-] = 0.014 \space M$
$[S{O_4}^{2-}] = 6.0 \times 10^{-3} \space M$
Work Step by Step
1. Draw the ICE table for the first and second reactions:
$$\begin{vmatrix}
Compound& [ H_2SO_4 ]& [ HS{O_4}^- ]& [ H_2O ]\\
Initial& 0.020 & 0 & 0 \\
Change& -0.020& +0.020& +0.020\\
Equilibrium& 0& 0.020& 0.020 \\
\end{vmatrix}$$
Since $H_2SO_4$ is a strong acid, it will react completely, so x = 0.020
$$\begin{vmatrix}
Compound& [ HS{O_4}^- ]& [ S{O_4}^{2-} ]& [ H_3O^+ ]\\
Initial& 0.020 & 0 & 0.020 \\
Change& -x& +x& +x\\
Equilibrium& 0.020 -x& 0 +x& 0.020 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ S{O_4}^{2-} ][ H_3O^+ ]}{[ HS{O_4}^- ]}$$
$$K_a = \frac{(x)( 0.020 + x)}{0.020 - x}$$
3. Assuming $0.020 \gt\gt x:$
$$K_a = \frac{(x)( 0.020 )}{0.020}$$
$$x = \frac{K_a \times [ HS{O_4}^- ]_{initial}}{ 0.020 } = \frac{ 1.1 \times 10^{-2} \times 0.020 }{ 0.020 } $$
$x = 0.011 $
4. Test if the assumption was correct:
$$\frac{ 0.011 }{ 0.020 } \times 100\% = 55.0 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{( 0.020 + x)(x)}{[ HS{O_4}^- ]_{initial} - x}$$
$$K_a [ HS{O_4}^- ] - K_a x = 0.020 x + x^2$$
$$x^2 + (K_a + 0.020 ) x - K_a [ HS{O_4}^- ] = 0$$
$$x_1 = \frac{- ( 1.1 \times 10^{-2} + 0.020 )+ \sqrt{( 1.1 \times 10^{-2} + 0.020 )^2 - 4 (1) (- 1.1 \times 10^{-2} ) ( 0.020 )} }{2 (1)}$$
$$x_1 = 6.0 \times 10^{-3} $$
$$x_2 = \frac{- ( 1.1 \times 10^{-2} + 0.020 )- \sqrt{( 1.1 \times 10^{-2} + 0.020 )^2 - 4 (1) (- 1.1 \times 10^{-2} ) ( 0.020 )} }{2 (1)}$$
$$x_2 = -0.037 $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 6.0 \times 10^{-3} $$
6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 6.0 \times 10^{-3} + 0.020 = 0.026 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.026 ) = 1.59 $$
$[H_3O^+] = 0.026 \space M$
$[HS{O_4}^-] = 0.020 - x = 0.014 \space M$
$[S{O_4}^{2-}] = x = 6.0 \times 10^{-3} \space M$
