TN1:

n_{Na_2O}={15,5}/{62}=0,25(mol)

Na_2O+H_2O->2NaOH

Theo PT: n_{NaOH}=2n_{Na_2O}=0,5(mol)

400ml=0,4l

->C_{M\ NaOH}={0,5}/{0,4}=1,25M

TN2:

200ml=0,2l

->n_{NaOH}=0,2.1,25=0,25(mol)

2NaOH+H_2SO_4->Na_2SO_4+2H_2O

Theo PT: n_{Na_2SO_4}=n_{H_2SO_4}=1/2n_{NaOH}=0,125(mol)

->m_{dd\ H_2SO_4}={0,125.98}/{19,6\%}=62,5(g)

->V_{dd\ H_2SO_4}={62,5}/{1,2}\approx 52,08(ml)=0,05208(l)

V_{dd\ B}=0,05208+0,2=0,25208(l)

->C_{M\ B}=C_{M\ Na_2SO_4}={0,125}/{0,25208}\approx 0,496M

n_{Na_2O}= {15,5}/{62}=0,25(mol)

$$\rm \ PTHH: Na_2O+ H_2O →2NaOH$$

Theo pt : n_{NaOH} = 2.n_{Na_2O}=0,5(mol)$$

$$\rm \ PTHH: 2NaOH + H_2SO_4→ Na_2SO_4 + 2H_2O$$

 (mol) 0,5 →0,25→0,25

+

C_{M_{NaOH}}= {0,5}/{0,4}=1,25M

+

m_{ dd H_2SO_4} = {0,25.98}/{19,6%}=125(g)

→ V_{dd H_2SO_4} = 125.1,2=150(ml) = 0,15(l)

+

C_{M_{Na_2SO_4}}= {0,25}/{0,2}= 1,25M