Gọi \(\left\{{}\begin{matrix}C_{M\left(dd.H_2SO_4\right)}=aM\\C_{M\left(dd.NaOH\right)}=bM\end{matrix}\right.\)
- TN1:
\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,12a\left(mol\right)\\n_{NaOH}=0,04b\left(mol\right)\end{matrix}\right.\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,04b--->0,02b
=> \(C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,12a-0,02b}{0,12+0,04}=0,1M\)
=> 0,12a - 0,02b = 0,016 (1)
- TN2:
\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,06a\left(mol\right)\\n_{NaOH}=0,08b\left(mol\right)\end{matrix}\right.\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,12a<----0,06a
=> \(C_{M\left(NaOH.dư\right)}=\dfrac{0,08b-0,12a}{0,06+0,08}=0,16M\)
=> 0,08b - 0,12a = 0,0224 (2)
(1)(2) => a = 0,24; b = 0,64
Chỗ "Mặt khác..." là trộn với bao nhiêu ml dung dịch NaOH thế bn ?
