Question

Pure ethanoic acid (25.0cm3, CH3COOH), pure ethanol (35.0 cm3, C2H5OH) and pure water (20.0cm3, H2O) were mixed in a sealed flask at room temperature. The flask was placed in a heated water bath and maintained at 323K until equilibrium was established.
CH3COOH(aq) + C2H5OH(aq) ↔ CH3 COOC2H5(aq) + H2O(l)
When equilibrium was attained the molar concentration of ethanoic acid was determined by titration as follows. A sample of the equilibrium mixture was taken (10.0cm3) and titrated with a standard solution of aqueous potassium hydroxide (KOH; 1.00mol dm-3). The volume of aqueous KOH required for complete reaction with ethanoic acid in the equilibrium sample of the reaction mixture was 25.5cm3.
(a) (i) Predict the value pH of the titration mixture at the equivalence point. Explain your answer.
(ii) What chemical indicator might you use for this titration?

Answer 1

This problem is one of misdirection. You can ignore everything at the beginning and start with "A sample of ......"
mols KOH used = 0.0255 x 1M = 0.0255
mols CH3COOH titrated = 0.0255 = mols CH3COOK formed at the equivalence point.
(CH3COOK) = 0.0255 mols/total volume
total volume = 10 cc titrated + 25.5 titration volume = 35.5 cc = 0.0355 so
(CH3COOK) = 0.0255mols/0.0355L = ? approximately 0.7 M
The pH at the equivalence point will be determined by the hydrolysis of the salt.
....CH3COO^- + HOH ==> CH3COOH + OH^-
I....0.7..................0.......0
C.....-x..................x.......x
E...0.7-x................x.........x

Kb for acetate = (Kw/Ka for acetic acid) = (CH3COOH)(OH^-)/(CH3COO^-)
Substitute and solve for x = OH^- then convert to pH. You can make a choice about the indicator to use

Answer 2

Is this: (CH3COOK) = 0.0255mols/0.0355L = ? approximately 0.7 M wrong?

moles/volume=concentration right? which would make this = 0.7183 moldm3 and then to get the pH...-log(0.7183)?