\(m_{ddKOH}=1,045.47,85=50\left(g\right)\)
\(\Rightarrow m_{KOH}=\dfrac{5,6.50}{100}=2,8\left(g\right)\)
\(n_{KOH}=0,05\left(mol\right)\)
\(m_{ddH_3PO_4}=1,08.18,25=19,71\left(g\right)\)
\(\Rightarrow m_{H_3PO_4}=\dfrac{14,7.19,71}{100}=2,89737\left(g\right)\)
\(n_{H_3PO_4}=0,03\left(mol\right)\)
\(\dfrac{n_{KOH}}{n_{H_3PO_4}}=\dfrac{0,05}{0,03}=1,67\)
=> Tạo muối NaH2PO4 và muối Na2HPO4
\(NaOH\left(0,03\right)+H_3PO_4\left(0,03\right)--->NaH_2PO_4\left(0,03\right)+H_2O\)\(\left(1\right)\)
\(NaOH\left(0,02\right)+NaH_2PO_4\left(0,02\right)--->Na_2HPO_4\left(0,02\right)+H_2O\)\(\left(2\right)\)
Sau phản ứng, muối thu được là: \(\left\{{}\begin{matrix}NaH_2PO_4:0,03-0,02=0,01\left(mol\right)\\Na_2HPO_4:0,02\left(mol\right)\end{matrix}\right.\)
\(m dd sau =50+19,71=69,71(g)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaH_2PO_4}=1,72\%\\C\%_{Na_2HPO_4}=4,07\%\end{matrix}\right.\)