## Chemistry and Chemical Reactivity (9th Edition)

The ${[H_2{PO_4}^{-}]}$ to ${[HP{O_4}^{2-}]}$ ratio must be equal to $1.6$;
1. Calculate $[H_3O^+]$: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 7}$ $[H_3O^+] = 1 \times 10^{- 7}$ 2. Write the $K_a$ equation, and find the ratio: $K_a = \frac{[H_3O^+][H{PO_4}^{2-}]}{[H_2P{O_4}^-]}$ $6.2 \times 10^{-8} = \frac{1 \times 10^{-7}*[H{PO_4}^{2-}]}{[H_2P{O_4}^-]}$ $\frac{6.2 \times 10^{-8}}{1 \times 10^{-7}} = \frac{[H{PO_4}^{2-}]}{[H_2P{O_4}^-]}$ $\frac{1 \times 10^{-7}}{6.2 \times 10^{-8}} = \frac{[H_2{PO_4}^{-}]}{[HP{O_4}^{2-}]}$ $1.6 = \frac{[H_2{PO_4}^{-}]}{[HP{O_4}^{2-}]}$