## Chemistry: Molecular Approach (4th Edition)

(d) $[X^{2-}] = 1.2 \times 10^{-11} \space M$
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: $[X^{2-}] = 1.2 \times 10^{-11} \space M$