Chemistry and Chemical Reactivity (9th Edition)

The $[HP{O_4}^{2-}]$ to $[H_2P{O_4}^-]$ ratio must be equal to $1.6$.
We find: $pH = pK_a + log(\frac{[Base]}{[Acid]})$ $7.4 = 7.20 + log(\frac{[Base]}{[Acid]})$ $7.4 - 7.20 = log(\frac{[Base]}{[Acid]})$ $0.2 = log(\frac{[Base]}{[Acid]})$ $10^{0.2} = \frac{[Base]}{[Acid]}$ $1.6 = \frac{[Base]}{[Acid]}$