Answer
The ratio of $[HP{O_4}^{2-}]$ to $[H_2P{O_4}^-]$ is equal to: 1.6.
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 7.40}$
$[H_3O^+] = 3.981 \times 10^{- 8}$
2. Write the $K_a$ equation, and find the ratio:
$K_a = \frac{[H_3O^+][HP{O_4}^{2-}]}{[H_2P{O_4}^-]}$
$6.3 \times 10^{-8} = \frac{3.981 \times 10^{-8}*[HP{O_4}^{2-}]}{[H_2P{O_4}^-]}$
$\frac{6.3 \times 10^{-8}}{3.981 \times 10^{-8}} = \frac{[HP{O_4}^{2-}]}{[H_2P{O_4}^-]}$
$1.6 = \frac{[HP{O_4}^{2-}]}{[H_2P{O_4}^-]}$
