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pH in a titration

Really stuck on this question, Ive been given the answer i just don't know how to get there:

Calculate the pH in a titration when 10cm^3 of 0.1M sodium hydroxide have been added to 10cm^3 of 0.25M ethanoic acid (Ka= 1.76x10^-5)

Final answer should be : 4.58
Reply 1
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Jtw95
OP
Original post by keromedic
10cm^3 of 0.1M sodium hydroxide have been added to 10cm^3 of 0.25M so you know hat 0.75M of NaOh is remaining. That's what 'provides' the PH. You know ka. what's the problem?

Im still baffled hahah
Original post by Jtw95
Im still baffled hahah

I read that wrong sorry. I was incorrect.
The ethanoic acid is in excess of 0.15 mol
Reply 3
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Jtw95
OP
Original post by keromedic
I read that wrong sorry. I was incorrect.
The ethanoic acid is in excess of 0.15 mol

So how would you work out the pH from there ( Sorry, im tired and its a saturday night lol)
CH3CH2OH⇌CH3CH2O−+H+CH_3CH_2OH \rightleftharpoons CH_3CH_2O^-+H^+
∴kc=[CH3CH2O−][H+]CH3CH2OH\therefore k_c=\dfrac{[CH_3CH_2O^-][H^+]}{CH_3CH_2OH}.
(edited 10 years ago)
Reply 5
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Jtw95
OP
Original post by keromedic
CH3CH2OH⇌CH3CH2O−+H+CH_3CH_2OH \rightleftharpoons CH_3CH_2O^-+H^+
∴kc=[CH3CH2O−][H+]CH3CH2OH\therefore k_c=\dfrac{[CH_3CH_2O^-][H^+]}{CH_3CH_2OH}.


Can you change the top line to (h+)^2 ?
Reply 6
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Borek
4
Original post by Jtw95
Can you change the top line to (h+)^2 ?


No! You can do it only in solutions in which concentrations are described by the dissociation stoichiometry (amount of acetate anion equals amount of H+ then), that's not the case here.

The simplest method of calculating pH in this case is to assume neutralization was stoichiometric (that is, amount of acetate equals exactly that of added strong base, and amount of acid is exactly that of the excess reagent), then to plug concentrations of acetic acid and acetate into Henderson-Hasselbalch equation.
Reply 7
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Jtw95
OP
Original post by Borek
No! You can do it only in solutions in which concentrations are described by the dissociation stoichiometry (amount of acetate anion equals amount of H+ then), that's not the case here.

The simplest method of calculating pH in this case is to assume neutralization was stoichiometric (that is, amount of acetate equals exactly that of added strong base, and amount of acid is exactly that of the excess reagent), then to plug concentrations of acetic acid and acetate into Henderson-Hasselbalch equation.


I was thinking this when I was plugging the numbers in. Thanks!

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