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    Calculate the pH of the solution formed when 3.5g of impure NaOH (98.7% purity) is dissolved in water and made up to 100cm^3 and then 25cm^3 of 0.35 mol dm^3 dibasic acid is added

    98.7% x 3.5g=3.4545g

    no of mol of NaOH=0.086mol

    conc of NaOH=0.863M

    dibasic=2 hydrogen ions
    0.35x2x 25/125= 0.14

    -log0.14=0.85

    what have i done wrong?
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    You need to first work out the amount (no. of mol) of OH- and H+ based on the concentrations and volumes - be careful with dm3 and cm3 (to convert cm3 to dm3, divide by 1000).

    These will react with each other to form water: H+(aq) + OH-(aq) --> H2O(l). This means that whichever one there is less of will be completely neutralised, for example if you had 1mol of H+ and 3mol of OH- you would end up with 0mol of H+ and 2mol OH-. You then need to calculate the concentration of the one left over.

    After this, you'll have to do different calculations depending on if you were left with H+ or OH-:
    If you're left with H+, you can just use pH = -log[H+] to get pH.
    If you're left with OH-, you can either use Kw = [OH-][H+] so [H+] = Kw/[OH-] to find [H+] and then use the pH formula, or you can calculate pOH = -log[OH-] and then subtract 14 to get pH. You'll probably have only learnt one of these methods, so use whichever you're familiar with.
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    (Original post by lizardlizard)
    You need to first work out the amount (no. of mol) of OH- and H+ based on the concentrations and volumes - be careful with dm3 and cm3 (to convert cm3 to dm3, divide by 1000).

    These will react with each other to form water: H+(aq) + OH-(aq) --> H2O(l). This means that whichever one there is less of will be completely neutralised, for example if you had 1mol of H+ and 3mol of OH- you would end up with 0mol of H+ and 2mol OH-. You then need to calculate the concentration of the one left over.

    After this, you'll have to do different calculations depending on if you were left with H+ or OH-:
    If you're left with H+, you can just use pH = -log[H+] to get pH.
    If you're left with OH-, you can either use Kw = [OH-][H+] so [H+] = Kw/[OH-] to find [H+] and then use the pH formula, or you can calculate pOH = -log[OH-] and then subtract 14 to get pH. You'll probably have only learnt one of these methods, so use whichever you're familiar with.
    I don't quite understand could you explain it any simpler?
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    (Original post by will'o'wisp)
    I don't quite understand could you explain it any simpler?
    Which parts are you having trouble with?
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    (Original post by lizardlizard)
    Which parts are you having trouble with?
    The first bit

    i managed to get an answer of 13.21 to 2dp is this ok?
 
 
 
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