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frodo00
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The answer is 1.60, I got ph to be 11.80

the question is calculate the ph change when water is added to 25cm3 of 0.250 of NaOH to prepare 1.00dm3 of solution (Kw=1.00*10-14 mol2dm-6)


So I did... 0.250 X 25/1000=6.25X10-3

Then I worked the hydrogen ions by dividing kw by 6.25X10-3 and then I put that answer into -log... to give 11.80 (rounded)


Where have i gone wrong?
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BobbJo
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use Kw and given [OH-]
initial [H+] is found
initial pH is found (13.4)
find [OH-] after dilution
final [H+] is found
final pH is found (11.8)

hence change is 1.60
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frodo00
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(Original post by BobbJo)use Kw and given [OH-]
initial [H+] is found
initial pH is found (13.4)
find [OH-] after dilution
final [H+] is found
final pH is found (11.8)
hence change is 1.60

Thank you! i always forget the initial ph..
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