Đáp án đúng: C
16,9
$\displaystyle \begin{array}{l}{{H}_{2}}PO_{4}^{-}\,+\,O{{H}^{-}}\,\xrightarrow{{}}\,HPO_{4}^{2-}\,+\,\,{{H}_{2}}O\\\,\,\,\,0,1\,\,\to \,\,\,0,1\,\,\,\,\xrightarrow{{}}\,\,\,\,\,0,1\end{array}$
$\begin{array}{l}HPO_{4}^{2-}\,+\,O{{H}^{-}}\,\xrightarrow{{}}\,PO_{4}^{3-}\,+\,{{H}_{2}}O\\\,\,0,05\,\leftarrow \,0,05\,\,\,\,\,\xrightarrow{{}}\,0,05\end{array}$
Sau phản ứng dung dịch chứa 0,1 mol K+; 0,15 mol Na+; 0,05 mol$HPO_{4}^{2-}$; 0,05 mol$PO_{4}^{3-}$
$\Rightarrow \,m\,=\,0,1.39\,+\,0,15.23\,+\,0,05.(96\,+\,95)\,=\,16,9\,gam$