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I don't understand this simple pH calculation

Calculate the pH of the solutions formed in the following way.

addition of 25 cm3 of water to 100 cm3 of 0.100 mol dm-3 H2SO4

I did:

H+ in h2so4 = 0.1 x 2 = 0.2

H+ in diluted h2so4 = 0.2 x 25/125 = 0.04

ph = -log(0.04) = 1.40

but the answer book says 0.80

help please x
Calculate the pH of the solutions formed in the following way.

addition of 25 cm3 of water to 100 cm3 of 0.100 mol dm-3 H2SO4

I did:

H+ in h2so4 = 0.1 x 2 = 0.2

H+ in diluted h2so4 = 0.2 x 25/125 = 0.04

ph = -log(0.04) = 1.40

but the answer book says 0.80

help please x
Reply 2
addition of 25 to 100 means the H2SO4 will be 80% of its initial conc, i.e. total volume = 125 cm3
EDIT : think what you aren't clear on here is that H2SO4 is given in MolDm-3, and not Mol

Mol of H2SO4 in first solution = 0.01

New concentration of H2SO4 = 0.01/125 x 1000 = 0.08.

H2SO4 is a strong, dibasic acid. Can you do the rest?
(edited 9 years ago)
Reply 4
Original post by Azurefeline
Mol of H2SO4 in first solution = 0.01

New concentration of H2SO4 = 0.01/125 x 1000 = 0.08.

H2SO4 is a strong, dibasic acid. Can you do the rest?


Its 0.1
Original post by Maker
Its 0.1

No, it isn't.
Reply 6
Original post by Azurefeline
No, it isn't.


You are right.

Its easier to do this though

0.1x100/125 = 0.08mol/l
(edited 9 years ago)
Original post by Maker
You are right.

Its easier to do this though

0.1x100/125 = 0.08mol/l

A difference in thought processes there, that's all. To me, my way is a lot easier, perhaps not for everybody.

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