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?
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
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)?
(a) (i) The pH of the titration mixture at the equivalence point can be predicted to be slightly basic. This is because the reaction between ethanoic acid and aqueous potassium hydroxide (KOH) produces potassium ethanoate (CH3COOK), which is the conjugate base of ethanoic acid. Therefore, the presence of the conjugate base will lead to a slightly basic pH at the equivalence point.
(ii) A suitable indicator for this titration would be phenolphthalein. Phenolphthalein is commonly used as an indicator in acid-base titrations because it changes color around pH 8.2-10.0. Since the equivalence point in this titration will result in a solution with a slightly basic pH, phenolphthalein will provide a clear color change from colorless to pink as the reaction nears completion.
To predict the value of pH at the equivalence point in this titration, we need to understand the reaction that occurs between ethanoic acid (CH3COOH) and potassium hydroxide (KOH). The equation for this reaction is:
CH3COOH + KOH → CH3COOK + H2O
At the equivalence point, the moles of ethanoic acid will be equal to the moles of potassium hydroxide. Since the reaction is a 1:1 stoichiometric ratio, the number of moles of ethanoic acid and potassium hydroxide will be the same.
Now, let's consider the species present at the equivalence point. We have potassium acetate (CH3COOK) and water (H2O). Potassium acetate is the salt of a weak acid (acetic acid) and a strong base (potassium hydroxide). Therefore, it will undergo hydrolysis in water, resulting in the formation of acetic acid and hydroxide ions.
CH3COOK + H2O ⇌ CH3COOH + OH-
The presence of hydroxide ions (OH-) in the solution will increase the pH. Hence, at the equivalence point, the pH will be higher (basic) due to the presence of the hydrolysis products.
(ii) For this titration, a suitable indicator would be phenolphthalein. Phenolphthalein is commonly used for acid-base titrations involving a strong acid and a strong base. It changes color in the pH range of 8.2 to 10.0, transitioning from colorless to pink. Since the pH at the equivalence point is expected to be basic, phenolphthalein will provide a clear and noticeable color change, indicating the end point of the titration.