a)Đổi: $V_{HCl}$=200 ml=0,2 l
$n_{HCl}$=0,2.0,2=0,04 (mol)
HCl+NaOH→NaCl+$H_{2}$O
1----1--------1------1--------(mol)
0,04-0,04-----0,04---0,04----(mol)
$V_{NaOH}$=$\frac{0,04}{0,1}$=0,4 (l)
$V_{dd NaCl}$=0,2+0,4=0,6 (l)
$C_{M_{dd NaCl}}$=$\frac{0,04}{0,6}$=$\frac{1}{15}$ (M)
b)2HCl+$Ca(OH)_{2}$→$CaCl_{2}$+$H_{2}$O
--2------1---------------1------------1---------(mol)
--0,04---0,02------------0,02---------0,02-----(mol)
$m_{dd Ca(OH)_{2}}$=$\frac{0,02.74}{5%}$=29,6 (g)
$m_{dd CaCl_{2}}$=200.1+29,6=229,6 (g)
$C%_{dd CaCl_{2}}$=$\frac{0,02.111}{229,6}$.100%≈0,97%