$C_{M_{HCl}}=x$
$n_{KOH}=0,1 mol$
$n_{HCl}=0,1x$
- TH1: chỉ tạo muối
$\Rightarrow m_{KCl}=74,5.0,1=7,45g$ (loại)
- TH2: dư kiềm
$KOH+HCl\to KCl+H_2O$
$\Rightarrow n_{KCl}=0,1x$
$n_{KOH}=0,1-0,1x$
$\Rightarrow 56(0,1-0,1x)+74,5.0,1x=6,525$
$\Leftrightarrow x=0,5$