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    Can someone please show me how to work through question 1? I should be able to do the rest, I just need some guidance on how to solve it.
    Thanks

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    * ive worked out moles for OH and H. minused them to give me the excess OH.
    where do I go on from here?
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    Great!

    You've calculated the moles of OH- ions, so you need to work out the concentration of OH- by using:
    Moles = concentration x volume (check units!)

    Does this look familiar?
    Kw = [H+][OH-]

    The value of Kw = 1x10^-14 (You don't have to remember this, it's given to you).

    Re-arrange above equation to get [H+].

    Then finally use pH = -log[H+]
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    (Original post by Bulletzone)
    Great!

    You've calculated the moles of OH- ions, so you need to work out the concentration by using Moles = concentration x volume (check units!)

    Kw = [H+][OH-]
    The value of Kw = 1x10^-14 (You don't have to remember this, it's given to you).

    Re-arrange above equation to get [H+].

    Then finally use pH = -log[H+]
    Ah this is helpful, Thank you!
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    (Original post by Lia22)
    Ah this is helpful, Thank you!
    Absolutely no problem

    Feel free to PM me if you're stuck on anything chemistry related, I'll do my best to help as it is possibly my favourite subject.
    It's the only science that makes sense to me.:woo: :woo: :woo:
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    Instead of using Kw to calculate the pH when an alkali is in excess, try using pOH.

    pOH = - log [OH-]

    pKw = - log Kw (at 25 oC, i.e. when Kw = 1.00 x 1014, pKw = 14.00)

    pH = pKw - pOH, i.e. = 14.00 - pOH
 
 
 
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