Đáp án:

a,

$$n_{Ba}=\dfrac{13,7}{137}=0,1$$ mol 

$$Ba+2H_2O\to Ba(OH)_2+H_2$$ 

$$\Rightarrow n_{Ba(OH)_2}=n_{Ba}=0,1$$ mol 

$$\Rightarrow C_{M_{Ba(OH)_2}}=\dfrac{0,1}{2}=0,05M$$ 

b,

$$Ba(OH)_2+2HCl\to BaCl_2+2H_2O$$ 

Tại điểm tương đương, ta có:

$$2V_{Ba(OH)_2}.C_{M_{Ba(OH)_2}}=V_{HCl}.C_{M_{HCl}}$$ 

$$\Rightarrow 2.50.0,05=60.a$$ 

$$\Rightarrow a=0,083M$$