Titration problem (1 Viewer)

[Irskin]

Question 17 (b) in the 2005 STANSW Trial Paper:

A student used the titration process to determine the concentration of a potassium hydroxide solution. She found that 25.00 mL of a 0.1 mol/L sulfuric acid solution reacted with exactly 10.00 mL of the potassium hydroxide solution. Calculate the concentration of the potassium hydroxide solution.

The answers say it is 0.125 mol/L but i am confident it is 0.5 mol/L

What do you guys come up with?

 

[mitsui]

i got 0.5 too

 

[pLuvia]

Isn't it,
H₂SO₄+2KOH --> K₂SO₄+2H₂O
Using the formula c₁V₁=c₂V₂

Since the moles of potassium hydroxide to sulfuric acid is 2:1
then 2c₁V₁=c₂V₂
c₁=0.1(0.025)/2(0.01)
=0.125mol/L

How did you guys do it?

 

[roosterman]

This one is in my notes.

E.g. A student used the titration process to determine the concentration of a potassium hydroxide solution. She found that 25.00 mL of a 0.1 mol L-1 sulfuric acid solution reacted with exactly 10.0 mL of the potassium hydroxide solution.
This one is in my notes.

Calculate the concentration of the potassium hydroxide solution.

H2SO4 + 2KOH * K2SO4 + 2H2O
1 mol : 2 mol

25.00 mL of a 0.1 mol L-1 sulfuric acid
C = n/v
0.1 = n/0.025
.: n = 0.1 x 0.025
= 0.0025 mol of H2SO4

Since 0.0025 mol of H2SO4 reacted,
n(KOH) = 0.0025 x 2 = 0.005 mol.

C = n/v
C(KOH) = 0.005 / 0.01
= 0.5 M

 

[renton]

thats interesting, both ways seem 2 make sense. someone help to try and figure this one out

 

[renton]

Isn't it,
H₂SO₄+2KOH --> K₂SO₄+2H₂O
Using the formula c₁V₁=c₂V₂

Since the moles of potassium hydroxide to sulfuric acid is 2:1
then 2c₁V₁=c₂V₂
c₁=0.1(0.025)/2(0.01)
=0.125mol/L

How did you guys do it?

ok i figured out where u made a little mistake.

H2SO4 + 2KOH - - - - > 2H2O + K2SO4

therefore,

2 x n(H2SO4) = n(KOH) i think you had the 2 x on the wrong side

2 x c(H2SO4) x v(H2SO4) = c(KOH) x v(KOH)
2 x 0.1 x 0.025 = c(KOH) x 0.01

so,

c(KOH) = 2 x 0.1 x 0.025 / 0.01 = 0.5 mol L-1

 

[mitsui]

wait i am still confused =@

H2SO4 + 2KOH
is 1:2

so it means in that equilibrium, one mole of H2SO4 reacted with 2 moles of KOH, so n(KOH) should be 2x n(H2SO4)??

 

[renton]

uh oh, im confused. didnt i say that 2 x n(H2SO4) = n(KOH)

 

[mitsui]

sry, yes u did. xD

i was talking about the book's (& pluvia, whose answer is 99% times right) answer. =@

 

[pLuvia]

sry, yes u did. xD

i was talking about the book's (& pluvia, whose answer is 99% times right) answer. =@

lol, not 99%

Ah I see where I went wrong Thanks renton.

 

[Riviet]

1 mole sulfuric acid reacts with 2 moles of KOH

that is,

1 mole sulfuric acid : 2 moles KOH

We calculate that 0.025 x 0.1 (0.0025) moles of sulfuric acid reacted, but the mole ratio means that we need to multiply the moles of sulfuric acid by 2 to obtain the number of moles of KOH.

.'. n_KOH=2 x 0.0025=0.005

.'. c=n/v=0.005/0.001
=0.5 mol/L

I hope that clears things up.