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

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.
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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

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

You can also find the Exam Thread list for A-levels here and GCSE here. :dumbells:

Just quoting in Puddles the Monkey so she can move the thread if needed :h:

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Reply 2

Fox Corner

Original post by Sniperdon227

Hi! I moved this to the Chemistry forum for you - you're more likely to get an answer here :smile:

Reply 3

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

Reply 4

langlitz

Original post by Sniperdon227

Sulphuric acid only completely dissociates for the first proton, the pka of the second dissociation is 2 i.e. not a strong acid

Reply 5

Sniperdon227

Original post by ahsan_ijaz

Where'd you get HCl from?

Reply 6

langlitz

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

Reply 7

Sniperdon227

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+]

Reply 8

langlitz

Original post by Sniperdon227

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.

pH = -log_{10}[H^+] => [H^+]=10^{-pH}=10^{-1.25}=0.056

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

Reply 9

Sniperdon227

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.

pH = -log_{10}[H^+] => [H^+]=10^{-pH}=10^{-1.25}=0.056

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