Answer
$[D^+] \approx 2.9833 \times 10^{-8}M$
$[OD^-] \approx 2.9833 \times 10^{-8}M$
Work Step by Step
1. Because it is a neutral solution:
$[D^+]=[OH^-]$
2. Multiplying the concentrations of [D+] and [OD-], we obtain Kw.
$[D^+]×[OD^−]=Kw$
$[D^+]×[OD^−]=8.9×10^{−16}$
3. For math purpose, let's call [D+] by [OD-], because they have the same value.
$[D^+]×[D^+]=8.9×10^{−16}$
$[D^+]^2=8.9×10^{−16}$
$[D^+]=2.9833 \times 10^{−8} M$
4. Because $[OD^-] = [D^+]$:
$[OD^−]=2.9833\times 10^{−8} M$
