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# more pH calculations watch

1. 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?
2. 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.
3. (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?
4. (Original post by will'o'wisp)
I don't quite understand could you explain it any simpler?
Which parts are you having trouble with?
5. (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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