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# A little help with the old pH, please? Watch

1. 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?
2. Did you count the fact that suphuric acid gives 2 protons per molecule instead of just one?
3. All seems okay tbh. Prob just an arithmetic error.
4. (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?
5. (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.
6. (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...
7. (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...
8. 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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Updated: December 19, 2010
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