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I really need help with a titration problem?
What will the pH in the titration of 50 mL of 0.10 M acetic acid (CH3COOH) when 25 mL of 0.10 M NaOH has been added at 25 degrees celsius. (Note that this is the half equivalence point; Ka for acetic acid is 1.8 x 10^-5)
I keep looking for help online but I just don't get it. Can someone please, very neatly, CLEARLY and nicely explain to me how to solve this problem? I'd really appreciate it. Thanks in advance.
2 Answers
- ?Lv 78 years agoFavorite Answer
HC2H302 + NaOH = Na+ + C2H302- + H20
The balanced equation shows that the acetic acid and the NaOH react on a one to one mole bases.
Original moles acetic acid = .10 Molar times .05 liters = .005 moles
Original moles of NaOH added = .10 molar times .025 liters = .0025 moles
Surplus moles of acetic acid surviving. .005 moles minus .0025 moles aced neutralized = .0025 moles
Moles acetate ion created should also be .0025 moles.
pH = pKa plus log (A) over (HA)
pKa = -log Ka which is 4.74
New concentration acid = .0025 moles acid over new total volume of .075 Liters = .033 Molar
New concentration acetate ion .0025 moles ion over .075 Liters also = .0033 molar
pH = pKa of 4.74 plus log (.0033) over (.0033)
pH = 4.74 plus zero = 4.74
Source(s): retired chem teacher USA - 8 years ago
Just take the third derivative of the titration heptaionic 50ml if we include CH3COOH of its partial function constantly at 0.10 at 25. Lastly, we'll include the Deram Law by 1.8*10^-5 so your answer is that the pH answer is 3.384
Source(s): Chemistry Teacher
