Answer
(d) $[X^{2-}] = 1.2 \times 10^{-11} \space M$
Work Step by Step
1. $K_{a1} \gt \gt K_{a2}$. Thus, we can neglect the effect of the second ionization in the $[H_3O^+]$.
2. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ H_2X ]& [ HX^- ]& [ H_3O^+ ]\\
Initial& 0.150 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.150 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
3. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HX^- ][ H_3O^+ ]}{[ H_2X ]}$$
$$K_a = \frac{(x)(x)}{[ H_2X ]_{initial} - x}$$
4. Assuming $ 0.150 \gt\gt x:$
$$K_a = \frac{x^2}{[ H_2X ]_{initial}}$$
$$x = \sqrt{K_a \times [ H_2X ]_{initial}} = \sqrt{ 4.5 \times 10^{-6} \times 0.150 }$$
$x = 8.2 \times 10^{-4} $
5. Test if the assumption was correct:
$$\frac{ 8.2 \times 10^{-4} }{ 0.150 } \times 100\% = 0.55 \%$$
6. The percent is less than 5%. Thus, it is correct to say that $x = 8.2 \times 10^{-4} $
7. $$[H_3O^+] = x = 8.2 \times 10^{-4} $$
8. Now, let's consider the second ionization to find the concentration of $X^{2-}$.
9. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HX^- ]& [ X^{2-} ]& [ H_3O^+ ]\\
Initial& 8.2 \times 10^{-4} & 0 & 8.2 \times 10^{-4} \\
Change& -x& +x& +x\\
Equilibrium& 8.2 \times 10^{-4} -x& 0 +x& 8.2 \times 10^{-4} +x\\
\end{vmatrix}$$
10. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ X^{2-} ][ H_3O^+ ]}{[ HX^- ]}$$
$$K_a = \frac{(x)( 8.2 \times 10^{-4} + x)}{[ HX^- ]_{initial} - x}$$
11. Assuming $ 8.2 \times 10^{-4} \gt\gt x:$
$$K_a = \frac{(x)( 8.2 \times 10^{-4} )}{[ HX^- ]_{initial}}$$
$$x = \frac{K_a \times [ HX^- ]_{initial}}{ 8.2 \times 10^{-4} } = \frac{ 1.2 \times 10^{-11} \times 8.2 \times 10^{-4} }{ 8.2 \times 10^{-4} } $$
$x = 1.2 \times 10^{-11} $
12. Test if the assumption was correct:
$$\frac{ 1.2 \times 10^{-11} }{ 8.2 \times 10^{-4} } \times 100\% = 1.5 \times 10^{-6} \%$$
13. The percent is less than 5%. Thus, it is correct to say that $x = 1.2 \times 10^{-11} $
14. In conclusion: $[X^{2-}] = 1.2 \times 10^{-11} \space M$
