Bảo toàn P:
$n_{P_2O_5}=0,5n_{P}=\dfrac{0,5.3,1}{31}=0,05 mol$
$\Rightarrow$ Mỗi phần có $0,025 mol P_2O_5$
- P1:
$m_{dd}=m_{P_2O_5}+500=503,55g$
$m_{H_3PO_4(dd)}=500.6\%=30g$
$n_{H_3PO_4}$(thêm)$=2n_{P_2O_5}=0,05 mol$
$\Rightarrow C\%B=\dfrac{100.(30+0,05.98)}{503,55}=6,93\%$
- P2:
$n_{H_3PO_4}=2n_{P_2O_5}=0,05 mol$
$n_{NaOH}=0,4.0,3=0,12 mol=n_{OH^-}$
$\dfrac{n_{OH^-}}{n_{H_3PO_4}}=2,4$
$\Rightarrow$ Tạo muối: $Na_2HPO_4$ (x mol), $Na_3PO_4$ (y mol)
Bảo toàn Na: $2x+3y=0,12$ (1)
Bảo toàn P: $x+y=0,1$ (2)
(1)(2)$\Rightarrow$ nghiệm âm