Câu 25:
$n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,25+0,25.0,5.2=0,5(mol)$
$n_{OH^-}=n_{NaOH}+2n_{Ca(OH)_2}=0,2V+0,3.2V=0,8V(mol)$
$n_{H^+}=n_{OH^-}$
$\to 0,8V=0,5$
$\to V=0,625(l)$
Câu 26:
$n_{H^+}=n_{HCl}+2n_{H_2SO_4}+n_{HNO_3}=0,1+0,2.0,5.2+0,1.1,5=0,45(mol)$
$n_{OH^-}=n_{NaOH}+n_{KOH}=C_M+0,05(mol)$
$n_{OH^-}=n_{H^+}\to C_M+0,05=0,45$
$\to C_M=0,4(M)$
Câu 27:
$n_{H^+}=0,25.2+0,25.0,5.2=0,75(mol)$
$n_{OH^-}=0,25.0,5+0,25.x.2=0,5x+0,125(mol)$
Sau phản ứng dư $H^+$
$\to n_{H^+\text{dư}}=0,75-(0,5x+0,125)=0,625-0,5x(mol)$
$V_{dd\text{spu}}=0,25+0,25=0,5l$
$\to 0,625-0,5x=0,5.0,5$
$\to x=0,75(M)$
