Question

Trung hòa 200ml dd H2SO4 1M bằng dd NaOH 20%

a. Tính khối lượng dd NaOH cần dùng?

b. Nếu trung hòa dd 200 ml H2SO4 1M bằng dd KOH 5.6% ( D= 1.045g/ml) thì cần bao nhiêu ml dung dịch KOH?

Answer 1

nH2SO4 = 0.2*1=0.2 mol

2NaOH + H2SO4 --> Na2SO4 + H2O

0.4________0.2

mNaOH = 0.4*40=16g

2KOH + H2SO4 --> K2SO4 + H2O

0.4______0.2

mKOH= 0.4*56=22.4g

mddKOH = 22.4*100/5.6=400g

VddKOH = 400/1.045=382.77ml

Answer 2

\(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (1)

a) Theo PT1: \(n_{NaOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,4\times40=16\left(g\right)\)

b) H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O (2)

Theo PT2: \(n_{KOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{KOH}=0,4\times56=22,4\left(g\right)\)

\(\Rightarrow m_{ddKOH}=\frac{22,4}{5,6\%}=400\left(g\right)\)

\(\Rightarrow V_{ddKOH}=\frac{400}{1,045}=382,78\left(ml\right)\)

Answer 3