Gọi $C_{M_{HCl}}=x$
$n_{K^+}=n_{OH^-}=0,1 (mol)$
$n_{H^+}=n_{Cl^-}=0,1x (mol)$
$H^++OH^-\to H_2O$
- TH1: không dư kiềm
$\Rightarrow 0,1=0,1x\Leftrightarrow x=1$
$m_{KCl}=0,1.74,5=7,45g$ (Loại)
$\Leftrightarrow x=0,74$
- TH2: dư kiềm
$m_{K^+}+m_{Cl^-}=0,1.39+0,1x.35,5=3,9+3,55x(g)$
$\Rightarrow m_{OH^-\text{dư}}=6,525-3,9-3,55x=2,625-3,55x(g)$
$n_{OH^-\text{pứ}}=n_{H^+}=0,1x$
$\Rightarrow n_{OH^-\text{dư}}=0,1-0,1x$
$\Rightarrow 17(0,1-0,1x)=2,625-3,55x$
$\Leftrightarrow x=0,5$