Answer:
The first dissociation occurred at 0.00311 M and the second at 0.0000484 M.
Explanation:
From the given information:
The ICE table can be computed as follows:
H2A → HA⁻ + H⁺
Initial 0.10 0 0
Change -x x x
Equilibrium 0.10 - x x x
[tex]Ka_1 = \dfrac{[HA^-][H^+]}{[H_2A]}[/tex]
[tex]1.0\times 10^{-4}= \dfrac{[x][x]}{[0.10-x]}[/tex]
[tex]1.0\times 10^{-4}= \dfrac{(x)^2}{(0.10-x)}[/tex]
By solving for x;
x² = (1.0 × 10⁻⁴ × 0.1)
x = [tex]\sqrt{1\times 10^{-5}}[/tex]
x = [H⁺] =[HA⁻] = 0.00311 M
The acid then further its dissociation again, So;
The ICE table can be computed as follows:
HA⁻ → A⁻ + H⁺
Initial 0.00311 0 0.00311
Change -x x x
Equilibrium 0.00311 - x x 0.00311 + x
[tex]Ka_2 = \dfrac{[A^-][H^+]}{[HA]}[/tex]
[tex]5.0 \times 10^{-5} = \dfrac{(0.00311+x)x}{(0.00311-x)}[/tex]
By solving for x;
x = [H⁺] = 0.0000484 M
Therefore, the first dissociation occurred at 0.00311 M and the second at 0.0000484 M.
