could someone please answer this (1 Viewer)

[ta1g]

0.02 L of 0.05mol/L potassium hydroxide was titrated with sulphuric acid four times. The first titration was a rough one, so the mean volume of acid used was found from the other three. It was 0.0205 L. Find the concentration of the sulphuric acid?

 

[serge]

0.02 L of 0.05mol/L potassium hydroxide was titrated with sulphuric acid four times. The first titration was a rough one, so the mean volume of acid used was found from the other three. It was 0.0205 L. Find the concentration of the sulphuric acid?

2KOH + H2S04 --> 2H20 +K2SO4

so for every mole that reacts you need

2n(KOH)= 1n(H2SO4)

n=cv

2 x 0.02 x 0.05 = 0.205 x c(H2SO4)

c= 9.8 x10^-3

tell me if i messed up people...

 

[Dreamerish*~]

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

KOH: 0.02 x 0.05 = 0.001 moles
H₂SO₄: 0.0205 x M moles

Molar ratio = 2:1

. : 0.001 moles of KOH will react with 0.0005 moles of H₂SO₄.

0.0005 moles of H₂SO₄ in 0.0205 L.

. : M = 0.0005/0.0205 = 0.024 (2 d.p.)

I sure hope I'm right.

 

[Dreamerish*~]

Alternatively, this is a formula which I should (but rarely) use. Probably because it becomes confusing in these cases:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Since the molar ratio is 2:1,

C₁V₁/C₂V₂ = 2 (₁ being KOH and ₂ being H₂SO₄)

C₁V₁ = 2C₂V₂

0.05 x 0.02 = 2(C₂ x 0.0205)

Working it out, C₂ = 0.024 (2 d.p.)

Gosh, I hate this method. If not for the answer being the same as my previous one, I'm bloody confused.

 

[gosh]

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

KOH: 0.02 x 0.05 = 0.001 moles
H₂SO₄: 0.0205 x M moles

Molar ratio = 2:1

. : 0.001 moles of KOH will react with 0.0005 moles of H₂SO₄.

0.0005 moles of H₂SO₄ in 0.0205 L.

. : M = 0.0005/0.0205 = 0.024 (2 d.p.)

I sure hope I'm right.

yeah that looks right