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    Calculate the pH of the solution formed when 25.0cm^3 of 0.150 mol dm^-3 aqueous sulfuric acid are added to 30.0cm^3 of 0.200 mol dm^-3 aqueous potassium hydroxide at 25 Celcius. Assume the the sulfuric acid is fully dissociated.

    Any ideas? I don't really know what i'm supposed to be working out.

    I was thinking something like this:

    work out moles of each
    find out which is in excess (I worked out it was KOH)
    work out total volume
    work out concentration of excess c=(nx1000)/V
    Use kw to get [H+]
    use -log10([H+])

    The problem is it came to 0.388. Which is obviously impossible for a weak base.
    What am I doing wrong?
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    Did you count the fact that suphuric acid gives 2 protons per molecule instead of just one?
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    All seems okay tbh. Prob just an arithmetic error.
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    (Original post by Jaredss)
    Did you count the fact that suphuric acid gives 2 protons per molecule instead of just one?
    I don't think I did. What part of the calculation should I have changed?
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    (Original post by IFondledAGibbon)
    I don't think I did. What part of the calculation should I have changed?
    x2 conc of H+, gives acid in excess.
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    (Original post by Hippysnake)
    x2 conc of H+, gives acid in excess.
    so I should have done 0.150x2= 0.3

    I'll run it through see what I get...
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    (Original post by Hippysnake)
    x2 conc of H+, gives acid in excess.
    came out as pH=1.57

    makes sence...

    Another note: does anyone have the CHEM4 june 2010 mark scheme? That is where the question is from...
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      I worked it out as H2SO4 is in excess not KOH and then came out with 2.82.

      Edit: Sorry my calculation was incomplete. 1.57 is what I got.
     
     
     
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