Bỏ qua sự điện li của nước.
a,
$n_{HCl}=\frac{11,2.2}{0,082.273}=1 mol$
$\Rightarrow m_{dd}=1.36,5+238,5=275g$
$\Rightarrow V_{dd}=275:1,1=250ml=0,25l$
$HCl\to H^+ + Cl^-$
$n_{H^+}=n_{Cl^-}=1 mol$
$\Rightarrow [H^+]=[Cl^-]=\frac{1}{0,25}=4M$
b,
$n_{CaCl_2}=n_{CaCl_2.6H_2O}=\frac{25}{111+18.6}=0,114 mol$
$m_{H_2O}=300g$
$m_{dd}=25+300=325g$
$\Rightarrow V_{dd}=325:1,08=300,9ml= 0,3009l$
$CaCl_2\to Ca^{2+}+ 2Cl^-$
$\Rightarrow n_{Ca^{2+}}=0,114 mol; n_{Cl^-}=0,114.2=0,228 mol$
$[Ca^{2+}]=\frac{0,114}{0,3009}= 0,38M$
$[Cl^-]=\frac{0,228}{0,3009}=0,76M$