`a)n_{HCl}=0,25.1,5=0,375(mol)`

`KOH+HCl→KCl+H_{2}O`

`0,375`   `0,375`   `0,375`          / `mol`

`⇒V_{ddKOH}=\frac{0,375}{2}=0,1875`(lít)

`b)V_{ddspứ}=0,1875+0,25=0,4375` (lít)

`⇒C_{M_{ddKCl}}=\frac{0,375}{0,4375}=\frac{6}{7}(M)` 

`c)`

`NaOH+HCl→NaCl+H_{2}O` 

`0,375`      `0,375`                       / `mol`

`⇒m_{NaOH}=0,375.40=15(g)`

`⇒C_{%_{ddNaOH}}=15:10%=150(g)`