Đáp án:
Bài 1:
$H_2SO_4 \to 2H^+ + SO_4^{2-}$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$
$MgCl_2 \to Mg^{2+} + 2Cl^-$
Bài 2:
a) $[H^+]=0,01M$, $[Cl^-]=0,01M$
b)
$[H^+]=0,1M$, $[SO_4^{2-}]=0,05M
Giải thích các bước giải:
Bài 1:
$H_2SO_4 \to 2H^+ + SO_4^{2-}$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$
$MgCl_2 \to Mg^{2+} + 2Cl^-$
Bài 2:
a) $n_{HCl}=\frac{0,0365}{36,5}= 10^{-3}$
$Cm_{HCl}=\frac{10^{-3}}{0,1}= 0,01M$
$HCl \to H^+ + Cl^-$
$0,01$→ $0,01$ $0,01$ (mol/l)
b)
$Cm_{H_2SO_4}=\frac{0,0025}{0,05}=0,05M$
$H_2SO_4 \to 2H^+ + SO_4^{2-}$
$0,05$ → $0,1$ $0,05$ (mol/l)
