## Chemistry and Chemical Reactivity (9th Edition)

$[H_2PO{_4}^-] = 7.7 \times 10^{-3}M$ and $[HP{O_4}^{2-}] = 1.2 \times 10^{-2}M$
As we calculated in the last exercise, at pH = 7.4: $1.6 = \frac{[Base]}{[Acid]} = \frac{[HP{O_4}^{2-}]}{[H_2PO{_4}^-]}$ So: $1.6[H_2PO{_4}^-] = [HP{O_4}^{2-}]$ If: $[H_2PO{_4}^-] + [HP{O_4}^{2-}] = 2.0 \times 10^{-2}M$: $[H_2PO{_4}^-] +1.6[H_2PO{_4}^-] = 2.0 \times 10^{-2}M$ $2.6[H_2PO{_4}^-] = 2.0 \times 10^{-2}M$ $[H_2PO{_4}^-] = \frac{2.0 \times 10^{-2}}{2.6}$ $[H_2PO{_4}^-] = 7.7 \times 10^{-3}M$ And: $1.6 * (7.7 \times 10^{-3}) = [HP{O_4}^{2-}] = 1.2 \times 10^{-2}M$