Solved: Titration Calc: Find KOH vol to Neutralize H2SO4

[ajassat]

Homework Statement

2.0g of pure undiluted H_sSO₄ is cautiously added to water and the solution diluted to 500 cm³. What volume of 0.1M potassium hydroxide, KOH, would exactly neutralise 25cm³ of the acid solution?

Homework Equations

concentration = number of moles/volume
mass in grams = number of moles*relative formula mass

The Attempt at a Solution

H₂SO₄
2.0g/500cm³
Divide by 20 to get values for 25cm³
0.1g/25cm³

Conversion of grams to M
mass in grams = number of moles*relative formula mass
0.1g = number of moles*98
0.1g/98 = number of moles = 1.02moles

We now have

1.02M/25cm³

Let us work out the concentration...

25/100 * 1.02 = 0.0255molar

Now let us work out the volume

Concentration = number of moles/volume
Concentration*volume = number of moles
volume = number of moles/concentration
volume = 1.02/0.0255
volume = 40cm³

Is this correct?

 

[Bohrok]

After you calculate the moles of H₂SO₄, I think you have problems. 1.02M/25cm³ makes no sense with molarity over volume, but it should be 1.02 moles/25 cm³. You also don't use the molarity of the KOH.
Write a balanced equation first.

You should start this way, which has the moles you calculated.

[tex]25 cm^3 solution \left(\frac{1.02 mol H_2SO_4} {500 cm^3 solution}\right)[/tex]

You have moles of the acid, then multiply it by conversion factors using the balanced equation and the molarity of the base, and you will get the volume of the base.

 

[ajassat]

After you calculate the moles of H₂SO₄, I think you have problems. 1.02M/25cm³ makes no sense with molarity over volume, but it should be 1.02 moles/25 cm³. You also don't use the molarity of the KOH.
Write a balanced equation first.

You should start this way, which has the moles you calculated.

[tex]25 cm^3 solution \left(\frac{1.02 mol H_2SO_4} {500 cm^3 solution}\right)[/tex]

You have moles of the acid, then multiply it by conversion factors using the balanced equation and the molarity of the base, and you will get the volume of the base.

Can you show me the complete process?

 

[Bohrok]

Write the balanced equation first.

 

[ajassat]

write the balanced equation first.

2koh + h₂so₄ --> k₂so₄ + 2h₂o

 

[ajassat]

It is a 2:1 ratio.

 

[ajassat]

If H₂SO₄ is 1.02 then KOH is 2.04moles

 

[ajassat]

volume = concentration/number of moles
volume = 0.1M/2.04 (KOH) * 1000
volume = 49?

 

[Bohrok]

So you know you need twice as much KOH as H₂SO₄ given from
[tex]
25 cm^3 solution \left(\frac{1.02 mol H_2SO_4} {500 cm^3 solution}\right)
[/tex]

which gives you moles of KOH. The molarity of the base is 0.1M KOH or [tex]\frac {0.1 mol KOH} {1 L KOH}[/tex]. How can you use this fraction with your moles of KOH from above to give you the volume of KOH solution?

Edit: Looks like you don't need these steps. Read the next post.

 

[Bohrok]

If H₂SO₄ is 1.02 then KOH is 2.04moles

1.02 is the moles of H₂SO₄ in the 500 cm³ of acid solution, but you need to know how many moles are in 25 cm³ of solution since that is how much is being neutralized by the base. You need to calculate this from what I wrote above. That will give you the actual number of moles of acid you're working with in 25 cm³ that's being neutralized. Then the 2:1 ratio you said is correct and you should get the correct answer.

 

[ajassat]

0.051 moles in 25cm^3. Therefore 0.102 moles for KOH.

Revised volume calculation

V = 0.1/0.102

 

[Bohrok]

Yes, that looks right.

 

[ajassat]

I need to multiply the answer by 1000 right?

 

[Bohrok]

That would give you mL of solution, but the question didn't say in what units to give the volume. You should be okay with leaving it in L unless you're expected to use a certain unit when the question doesn't specify.

 

[Borek]

No!

First of all - 2 g of H₂SO₄ is not 1.02 mole. Thus everything calculated later is wrong.

Won't hurt to start here:

http://www.titrations.info/titration-calculation

 

[Bohrok]

Conversion of grams to M
mass in grams = number of moles*relative formula mass
0.1g = number of moles*98
0.1g/98 = number of moles = 1.02moles

You do have a mistake here. 0.1/98 is .00102, so that is the actual number of moles of H₂SO₄. This is also wrong

volume = concentration/number of moles

Concentration = moles/volume, so volume = moles/concentration

I hadn't checked your work thoroughly before, but I think these are all the mistakes you had.