## General Chemistry: Principles and Modern Applications (10th Edition)

$K_{a1} = 0.052$ $K_{a2} = 5.3 \times 10^{-5}$
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: $$K_{a2} = \frac{[C_2O_4^{2-}][H_3O^+]}{[HC_2O_4^-]} = \frac{(y)(x+y)}{x-y}$$ $$K_{a2} = \frac{(5.3 \times 10^{-5})(0.21 + 5.3 \times 10^{-5})}{0.21 - 5.3 \times 10^{-5}} = 5.3 \times 10^{-5}$$