#### Answer

$[H_3O^+] = 0.21 \space M$
$[HS{O_4}^-] = 0.19 \space M$
$[S{O_4}^{2-}] =0.011 \space M$

#### Work Step by Step

1. Draw the ICE table for the first and second reactions:
$$\begin{vmatrix}
Compound& [ H_2SO_4 ]& [ HS{O_4}^- ]& [ H_2O ]\\
Initial& 0.20 & 0 & 0 \\
Change& -0.20& +0.20& +0.20\\
Equilibrium& 0& 0.20& 0.20 \\
\end{vmatrix}$$
Since $H_2SO_4$ is a strong acid, it will react completely, so x = 0.20
$$\begin{vmatrix}
Compound& [ HS{O_4}^- ]& [ S{O_4}^{2-} ]& [ H_3O^+ ]\\
Initial& 0.20 & 0 & 0.20 \\
Change& -x& +x& +x\\
Equilibrium& 0.20 -x& 0 +x& 0.20 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ S{O_4}^{2-} ][ H_3O^+ ]}{[ HS{O_4}^- ]}$$
$$K_a = \frac{(x)( 0.20 + x)}{0.20 - x}$$
3. Assuming $ 0.20 \gt\gt x:$
$$K_a = \frac{(x)( 0.20 )}{[ HS{O_4}^- ]_{initial}}$$
$$x = \frac{K_a \times [ HS{O_4}^- ]_{initial}}{ 0.20 } = \frac{ 1.1 \times 10^{-2} \times 0.20 }{ 0.20 } $$
$x = 0.011 $
4. Test if the assumption was correct:
$$\frac{ 0.011 }{ 0.20 } \times 100\% = 5.5 \%$$
The percent is approximately 5%, therefore, the difference between this and the actual number is not significant for a 2 significant figures sum.
5. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 0.011 + 0.20 = 0.21 $$
$[H_3O^+] = 0.21 \space M$
$[HS{O_4}^-] = 0.20 - 0.011 = 0.19 \space M$
$[S{O_4}^{2-}] = x = 0.011 \space M$