Đáp án:
10,5% và 15,975%
Giải thích các bước giải:
\(\begin{array}{l}
{m_{{H_3}P{O_4}}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{100 \times 39,2}}{{100}} = 39,2(g)\\
{n_{{H_3}P{O_4}}} = \dfrac{m}{M} = \frac{{39,2}}{{98}} = 0,4\,mol\\
{m_{NaOH}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{100 \times 25}}{{100}} = 25g\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{25}}{{40}} = 0,625\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{{H_3}P{O_4}}}}} = \dfrac{{0,625}}{{0,4}} = 1,5625\\
1 < T < 2 \Rightarrow \text{ Tạo 2 muối $NaH_2PO_4$ và $Na_2HPO_4$}\\
{H_3}P{O_4} + NaOH \to Na{H_2}P{O_4} + {H_2}O\\
{H_3}P{O_4} + 2NaOH \to N{a_2}HP{O_4} + 2{H_2}O\\
{n_{Na{H_2}P{O_4}}} = a(mol);{n_{N{a_2}HP{O_4}}} = b(mol)\\
a + b = 0,4\\
a + 2b = 0,625\\
\Rightarrow a = 0,175;b = 0,225\\
{m_{{\rm{dd}}spu}} = 100 + 100 = 200g\\
{C_\% }Na{H_2}P{O_4} = \dfrac{{0,175 \times 120}}{{200}} \times 100\% = 10,5\% \\
{C_\% }N{a_2}HP{O_4} = \dfrac{{0,225 \times 142}}{{200}} \times 100\% = 15,975\%
\end{array}\)
