# Chapter 16 - Exercises - Page 771: 26

$$[X^{2-}] = K_{a2}$$ This relationship exists if the ionization caused by the second proton is much smaller than the same from the first one, which happens if $K_{a2}$ is much smaller than $K_{a1}$.

#### Work Step by Step

When calculating the equilibrium concentrations of the solution after the first ionization we get this ICE table: \begin{vmatrix} & [HX^-] & [H_3O^+] & [X^{2-}] \\ Initial & x_1 & x_1 & 0 \\ Change & -x_2 & + x_2 & +x_2 \\ Equil & x_1 - x_2 & x_1 + x_2 & x_2 \end{vmatrix} $x_1$ represents the ionization of the first proton, and $x_2$ represents the ionization of the second one. -Writing the $K_a$ expression: $$K_{a2} = \frac{(x_1 + x_2)(x_2)}{(x_1-x_2)}$$ If $x_1 \gt \gt x_2$: $$K_{a2} = \frac{(x_1)(x_2)}{x_1} = x_2 = [X^{2-}]$$

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