Ph calculations

I’m unsure on this question

Calculate the pH of the solutions produced in the following way:

20 cm3 of 1.0 M H2SO4 with water added to make the volume up to 100 cm3

Original post by Blue_skies124

I’m unsure on this question

Calculate the pH of the solutions produced in the following way:

20 cm3 of 1.0 M H2SO4 with water added to make the volume up to 100 cm3

The definition of pH requires that you work out the hydrogen ion concentration.
How many moles of hydrogen ions are there?
What is their final concentration?

Reply 2

Blue_skies124

OP

17

I’ve got no idea
I’m really stuck on this question I just don’t know where to start

Original post by Blue_skies124

I’ve got no idea
I’m really stuck on this question I just don’t know where to start

You start with the equation for sulfuric acid dissolving in water.
Then you use the formula moles = molarity x volume to work out the moles of sulfuric acid and hence the moles of hydrogen ions.

Reply 4

Blue_skies124

OP

17

But wouldn’t you multiply 20cm3 with the old volume over the new volume?

Original post by Blue_skies124

But wouldn’t you multiply 20cm3 with the old volume over the new volume?

You can't simply use ratios because sulfuric acid is diprotic (releases two hydrogen ions per molecule)

Original post by Blue_skies124

But wouldn’t you multiply 20cm3 with the old volume over the new volume?

You can use ratios if you factor in the diprotic nature of sulfuric acid. This will allow you to find the hydrogen ion concentration. Then you take the negative of the log to get the pH.

Reply 7

NewbieNord

3

1. moles H+ before dilution; 0.02 x 2 (it's diprotic) = 0.04 mol
2. concentration H+ after dilution; 0.04 / 0.1 = 0.4 moldm-3
3. ph = -log(0.4) = 0.40

Reply 8

Blue_skies124

OP

17

Where did you get the 0.02 mol from to multiply it by 2

Original post by Blue_skies124

Where did you get the 0.02 mol from to multiply it by 2

20cm³ = 0.02 dm³

mol = molarity x volume = 1.0 x 0.02 dm³ = 0.02 mol

Reply 10

Blue_skies124

OP

17

Actually that makes sense now. So the 0.02 mol comes from 0.02 dm3 x 1M but where did you get the 0.1 when dividing 0.04

Original post by Blue_skies124

Actually that makes sense now. So the 0.02 mol comes from 0.02 dm3 x 1M but where did you get the 0.1 when dividing 0.04

Molarity = mol/vol

The volume = 100cm³ = 0.1 dm³

Reply 12

Blue_skies124

OP

17

I finally managed to get to the answer but I think I did it a bit differently.
So I worked out the concentration of H+ ions in the original H2SO4 solution which was: 1x2 since it’s diprotic
Then I worked out the concentration of H+ in diluted solution: 2x 20/100 2x the old volume/new volume
Then I worked out the ph Which was 0.40
I don’t know if my working out is still correct though

Original post by Blue_skies124

I finally managed to get to the answer but I think I did it a bit differently.
So I worked out the concentration of H+ ions in the original H2SO4 solution which was: 1x2 since it’s diprotic
Then I worked out the concentration of H+ in diluted solution: 2x 20/100 2x the old volume/new volume
Then I worked out the ph Which was 0.40
I don’t know if my working out is still correct though

Your method is fine.

Ph calculations

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