Bài 5 :
a) $n_{NaOH} = 0,1.0,1 = 0,01(mol) ; n_{Ba(OH)_2} = 0,1.0,1 = 0,01(mol)$
$NaOH + HCl \to NaCl + H_2O$
$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = n_{NaOH} + 2n_{Ba(OH)_2} = 0,03(mol)$
$V_{dd\ HCl} = \dfrac{0,03}{0,3} = 0,1(lít)$
b)
$m_{dd\ NaOH} = 50.1,109 = 55,45(gam) ; m_{dd\ HCl} = 50.1,047 = 52,35(gam)$
$n_{NaOH} = \dfrac{55,45.8\%}{40} = 0,1109(mol)$
$n_{HCl} = \dfrac{52,35.10\%}{36,5} = 0,14(mol)$
$NaOH + HCl \to NaCl + H_2O$
Theo PTHH : $n_{HCl\ pư} = n_{NaCl} = n_{NaOH} = 0,1109(mol)$
$n_{HCl\ dư} = 0,14 - 0,1109 = 0,0291(mol)$
$V_{dd} = 0,05 + 0,05 = 0,1(lít)$
$C_{M_{NaCl}} = \dfrac{0,1109}{0,1} = 1,109M ; C_{M_{HCl\ dư}} = \dfrac{0,0291}{0,1} = 0,291M$
$[Na^+] = 1,109M ; [Cl^-] = 1,109 + 0,291 = 1,4M ; [H^+] = 0,291M$
