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    1. Calculate the pH of the following
    0.1 mol dm-3 KOH
    and
    0.05 mol dm-3 Ba(oH)2
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    (Original post by aladdin818)
    1. Calculate the pH of the following
    0.1 mol dm-3 KOH
    and
    0.05 mol dm-3 Ba(oH)2
    Hey I'll move this to the chemistry section as you will be more likely to get some help there

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    (Original post by aladdin818)
    1. Calculate the pH of the following
    0.1 mol dm-3 KOH
    and
    0.05 mol dm-3 Ba(oH)2
    Find pOH by doing the -log of [0.1]

    As K to OH is 1 to 1.

    Wait no yes, pH + pOH = 14
    So 14 - pOH = pH
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    Exactly what Dinasaurus said.

    Antoher method would be the following:

    Since Kw = [H+][OH-] and we know that Kw (the ionic product of water) is 1.00 x10^-14 mol^2 dm^-6 and that [OH-] is 0.1 mold dm^-3, then [H+] = Kw / [OH-] = 1.00 x10^-14 / 0.1 = 1.00 x10^-13 and so pH = -log[H+] = -log(1.00 x10^-13) = 13

    I prefer the pOH method, but just wanted to show you the other method
 
 
 
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