Đáp án:
a)
[Na^+]=0,025M$
[Cl^-]=0,025M$
b)
[K^+]=0,36M$
[Ba^{2+}]=0,08M$
[OH^-]= 0,52M$
Giải thích các bước giải:
a)
$n_{NaCl}=\frac{2,925}{58,5}=0,05mol$
$Cm_{NaCl}=\frac{0,05}{2}=0,025M$
$NaCl \to Na^+ + Cl^-$
$0,025$ → $0,025$ $0,025$ (mol/l)
[Na^+]=0,025M$
[Cl^-]=0,025M$
b)
$n_{KOH}=0,3.1,2=0,36mol$
$n_{Ba(OH)_2}=0,2.0,4=0,08mol$
$KOH \to K^+ + OH^-$
$0,36$→ $0,36$ $0,36$ (mol)
$Ba(OH)_2 \to BA^{2+} + 2OH^-$
$0,08$ → $0,08$ $0,16$ (mol/l)
[K^+]=0,36M$
[Ba^{2+}]=0,08M$
[OH^-]= 0,36 + 0,16 =0,52M$