Answer
$[H_2PO{_4}^-] = 7.7 \times 10^{-3}M$
and
$ [HP{O_4}^{2-}] = 1.2 \times 10^{-2}M$
Work Step by Step
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:
$1.6 * (7.7 \times 10^{-3}) = [HP{O_4}^{2-}] = 1.2 \times 10^{-2}M$
