${n_{{H_3}P{O_4}}} = 0,1 \times 2 = 0,2\,mol$
Cho H3PO4 td KOH, các phản ứng xảy ra:
$\begin{gathered}
1)\,{H_3}P{O_4} + O{H^ - } \to {H_2}P{O_4}^ - + {H_2}O \hfill \\
2){H_3}P{O_4} + 2O{H^ - } \to HP{O_4}^{2 - } + 2{H_2}O \hfill \\
3){H_3}P{O_4} + 3O{H^ - } \to P{O_4}^{3 - } + 3{H_2}O \hfill \\
\end{gathered} $
TH1: Xảy ra pt (1) và (2)
$\left\{ \begin{gathered} K{H_2}P{O_4}\,(X):\,2a\,mol \hfill \\ {K_2}HP{O_4}(Y):\,3a\,mol \hfill \\ \end{gathered} \right.$
$ \to \left\{ \begin{gathered} {n_{{H_3}P{O_4}}} = 2a + 3a = 0.2 \to a = 0,04\,mol \hfill \\ {n_{O{H^ - }}} = {n_{{H_2}P{O_4}^ - }} + 2{n_{HP{O_4}^{2 - }}} = 8a = 0,32\,mol\, \hfill \\ \end{gathered} \right.$
$ \to {V_{KOH}} = 0.32/4 = 0,08\,l$
$\left\{ \begin{gathered} {C_{{M_{{K_2}HP{O_4}}}}} = \frac{{3a}}{{0.08 + 0.1}} = 0,67M \hfill \\ {C_{{M_{K{H_2}P{O_4}}}}} = \frac{{2a}}{{0.08 + 0.1}} = 0,44M \hfill \\ \end{gathered} \right.$
TH2: Xảy ra pt (2), (3)
$\left\{ \begin{gathered} {K_3}P{O_4}\,(Y):\,3a\,mol \hfill \\ {K_2}HP{O_4}(X):\,2a\,mol \hfill \\ \end{gathered} \right.$
$ \to \left\{ \begin{gathered} {n_{{H_3}P{O_4}}} = 2a + 3a = 0.2 \to a = 0,04\,mol \hfill \\ {n_{O{H^ - }}} = 3{n_{P{O_4}^{3 - }}} + 2{n_{HP{O_4}^{2 - }}} = 13a = 0,52\,mol\, \hfill \\ \end{gathered} \right.$
$\begin{gathered} \to {V_{KOH}} = 0.52/4 = 0,13\,l \hfill \\ \left\{ \begin{gathered} {C_{{M_{{K_3}P{O_4}}}}} = \frac{{3a}}{{0.13 + 0.1}} = 0,52M \hfill \\ {C_{M{K_2}HP{O_4}}} = \frac{{2a}}{{0.13 + 0.1}} = 0,35M \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $
