Answer
$K_{a1} = 0.052 $
$K_{a2} = 5.3 \times 10^{-5}$
Work Step by Step
1. Draw the ICE table for the first reaction:
$$\begin{vmatrix}
Compound& [ H_2C_2O_4 ]& [ HC_2O_4^- ]& [ H_3O^+ ]\\
Initial& 1.05 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 1.05 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
- Draw the ICE table for the second reaction:
$$\begin{vmatrix}
Compound& [ HC_2O_4^- ]& [ H_3O^{+} ]& [ C_2O_4^{2-} ]\\
Initial& x & x & 0 \\
Change& -y& +y& +y\\
Equilibrium& x -y& x +y& 0 +y\\
\end{vmatrix}$$
x = change in the first reaction, y = change in the second reaction.
$y = [C_2O{_4}^{2-}] = 5.3 \times 10^{-5} \space M$
$[H_3O^+] = 10^{-pH} = 10^{-0.67} = 0.21 \space M$
$x + y = [H_3O^+] = 0.21$
$x + 5.3 \times 10^{-5} = 0.21$
$x = 0.21 - 5.3 \times 10^{-5} = 0.21$
2. Write the expression for $K_a1$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HC_2O_4^- ][ H^+ ]}{[ H_2C_2O_4 ]}$$
$$K_a = \frac{(x)(x)}{[ H_2C_2O_4 ]_{initial} - x}$$
We know that x = 0.21
4. Substitute the value of x and calculate the $K_a1$:
$$K_{a1} = \frac{( 0.21 )^2}{ 1.05 - 0.21 }$$
$K_{a1} = 0.052 $
5. Write the Ka2 expression, and substitute:
$$K_{a2} = \frac{[C_2O_4^{2-}][H_3O^+]}{[HC_2O_4^-]} = \frac{(y)(x+y)}{x-y}$$
$$K_{a2} = \frac{(5.3 \times 10^{-5})(0.21 + 5.3 \times 10^{-5})}{0.21 - 5.3 \times 10^{-5}} = 5.3 \times 10^{-5}$$
