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# Calculate the total volume of solution formed? H2SO4 problems A2 Chemistry

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1. A 25.0 cm3 sample of 0.0850 mol dm–3 sulphuric acid was placed in a beaker.
Distilled water was added until the pH of the solution was 1.25
Calculate the total volume of the solution formed. State the units.
................................ ................................ ................................ ...........
Can you please tell me the answer, with monoprotic acids its easy, but this is diprotic so a little bit more difficult, please comment below your total volumes in cm3, I have got a total volume I need to cross check it with others! Thanks
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Just quoting in Puddles the Monkey so she can move the thread if needed
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(Original post by Puddles the Monkey)
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3. (Original post by Sniperdon227)
A 25.0 cm3 sample of 0.0850 mol dm–3 sulphuric acid was placed in a beaker.
Distilled water was added until the pH of the solution was 1.25
Calculate the total volume of the solution formed. State the units.
................................ ................................ ................................ ...........
Can you please tell me the answer, with monoprotic acids its easy, but this is diprotic so a little bit more difficult, please comment below your total volumes in cm3, I have got a total volume I need to cross check it with others! Thanks
Hi! I moved this to the Chemistry forum for you - you're more likely to get an answer here
4. Hi,

A ph is given in Q
So you convert that to get [h]+ conc
Then work out moles of hcl by doing 0.0850 x 25/1000
Then do v = n/c which gives volume to dm3

I didn't have calculator on me but that should be the answer
5. (Original post by Sniperdon227)
A 25.0 cm3 sample of 0.0850 mol dm–3 sulphuric acid was placed in a beaker.
Distilled water was added until the pH of the solution was 1.25
Calculate the total volume of the solution formed. State the units.
................................ ................................ ................................ ...........
Can you please tell me the answer, with monoprotic acids its easy, but this is diprotic so a little bit more difficult, please comment below your total volumes in cm3, I have got a total volume I need to cross check it with others! Thanks
Sulphuric acid only completely dissociates for the first proton, the pka of the second dissociation is 2 i.e. not a strong acid
6. (Original post by ahsan_ijaz)
Hi,

A ph is given in Q
So you convert that to get [h]+ conc
Then work out moles of hcl by doing 0.0850 x 25/1000
Then do v = n/c which gives volume to dm3

I didn't have calculator on me but that should be the answer
Where'd you get HCl from?
7. (Original post by Sniperdon227)
Where'd you get HCl from?
Taking sulphuric acid as a strong acid for both protons I get 51 cm^3 of water added
8. (Original post by langlitz)
Taking sulphuric acid as a strong acid for both protons I get 51 cm^3 of water added
using [H+]new = [H+]old x (old volume/total volume)

so 10^-1.25= 0.0850 x (0.025/x) where x is what we want to find out

What i need to know is when doing '10^-1.25' do I get 2[H+] or 1[H+] and is 0.0850 2[H+] or 1[H+]
9. (Original post by Sniperdon227)
using [H+]new = [H+]old x (old volume/total volume)

so 10^-1.25= 0.0850 x (0.025/x) where x is what we want to find out

What i need to know is when doing '10^-1.25' do I get 2[H+] or 1[H+] and is 0.0850 2[H+] or 1[H+]
This is how I did it:
n=cv = 0.025*0.085= 2.125x10^-3 moles
Each molecules contributes 2 H+ ions so there are 4.25x10^-3 moles of H+ ions from the sulphuric acid.

mol dm^-3
v=n/c = 4.25x10^-3/0.056 = 0.076 dm^3 = 76 cm^3 is the total volume
So 76-25 = 51 cm^3 added
10. (Original post by langlitz)
This is how I did it:
n=cv = 0.025*0.085= 2.125x10^-3 moles
Each molecules contributes 2 H+ ions so there are 4.25x10^-3 moles of H+ ions from the sulphuric acid.

mol dm^-3
v=n/c = 4.25x10^-3/0.056 = 0.076 dm^3 = 76 cm^3 is the total volume
So 76-25 = 51 cm^3 added
Got ya so when calculating the moles of H2SO4 ie n=0.0850x25x10^-3 thats equal to 1[H+] so you have to multiply by Two then proceed cheers for that I get it now

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