Đáp án + Giải thích các bước giải:

$a\big)$

$P_2O_5+3H_2O\to 2H_3PO_4$

$3KOH+H_3PO_4\to K_3PO_4+3H_2O$

$2KOH+H_3PO_4\to KH_2PO_4+2H_2O$

$b\big)$

Đổi $100ml=0,1l$

$n_{P_2O_5}=\frac{7,1}{142}=0,05(mol)$

$n_{KOH}=\frac{80.10\%}{56}=\frac{1}{7}(mol)$

Theo PT: $n_{H_3PO_4}=2n_{P_2O_5}=0,1(mol)$

$\to C_{M\,H_3PO_4}=\frac{0,1}{0,1}=1M$

$c\big)$

$n_{H_3PO_4(PU2)}=\frac{0,1}{2}=0,05(mol)$

Xét tỉ lệ: $\frac{n_{KOH}}{n_{H_3PO_4}}=\frac{\frac{1}{7}}{0,05}=2,86$

$\to$ Tạo muối $K_3PO_4:x(mol);KH_2PO_4:y(mol)$

$V_{dd\,KOH}=\frac{80}{1,25}=64(ml)=0,064(l)$

Theo PT: $\begin{cases} 3x+2y=\frac{1}{7}\\ x+y=0,05 \end{cases}$

$\to\begin{cases} x=\frac{3}{70}(mol)\\ y=\frac{1}{140}(mol) \end{cases}$

$V_{dd\,spu}=0,05+0,064=0,114(l)$

$\to\begin{cases} C_{M\,K_3PO_4}=\dfrac{\frac{3}{70}}{0,114}\approx 0,375M\\ C_{M\,KH_2PO_4}=\dfrac{\frac{1}{140}}{0,114}\approx 0,063M \end{cases}$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{P_2}{O_5} + 3{H_2}O \to 2{H_3}P{O_4}\\
b)\\
{n_{{P_2}{O_5}}} = \dfrac{{7,1}}{{142}} = 0,05\,mol\\
{n_{{H_3}P{O_4}}} = 2{n_{{P_2}{O_5}}} = 0,1\,mol\\
{C_M}{H_3}P{O_4} = \dfrac{{0,1}}{{0,1}} = 1M\\
c)\\
{n_{KOH}} = \dfrac{{80 \times 10\% }}{{56}} = \dfrac{1}{7}\,mol\\
T = \dfrac{{{n_{KOH}}}}{{{n_{{H_3}P{O_4}}}}} = \dfrac{{\dfrac{1}{7}}}{{0,05}} = 2,86\\
2 < T < 3 \Rightarrow \text{ Tạo 2 muối : $K_3PO_4,K_2HPO_4$ } \\
3KOH + {H_3}P{O_4} \to {K_3}P{O_4} + 3{H_2}O\\
2KOH + {H_3}P{O_4} \to {K_2}HP{O_4} + 2{H_2}O\\
hh:{K_3}P{O_4}(a\,mol),{K_2}HP{O_4}(b\,mol)\\
\left\{ \begin{array}{l}
3a + 2b = \frac{1}{7}\\
a + b = 0,05
\end{array} \right.\\
 \Rightarrow a = \dfrac{3}{{70}};b = \dfrac{1}{{140}}\\
{V_{{\rm{dd}}KOH}} = \dfrac{{80}}{{1,25}} = 64\,ml\\
{V_{{\rm{dd}}}} = 0,05 + 0,064 = 0,114l\\
{C_M}{K_3}P{O_4} = \dfrac{{\dfrac{3}{{70}}}}{{0,114}} = 0,38M\\
{C_M}{K_2}HP{O_4} = \dfrac{{\dfrac{1}{{140}}}}{{0,114}} = 0,063M
\end{array}\)