Subject – equilibria
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?
b)
(iii) Write a balanced equation for the reaction of aqueous potassium hydroxide (KOH(aq)) with aqueous ethanoic acid (CH3COOH(aq)).
(iv) Calculate the amount of unreacted aqueous ethanoic acid (CH3COOH(aq)) in the sample taken for titration with aqueous potassium hydroxide(KOH(aq)). Clearly show the steps in your calculation.
v) Hence calculate the amount of aqueous ethanoic acid CH3COOH(aq)) in the sealed flask at equilibrium. Clearly show the steps in your calculation and comment on any assumptions that you make.
The questions which follow will require the use of the following data. The questions relate to the experiment described at the beginning of questions
Substance ===============Density at room temperature (g cm-3)
Ethanoic acid========= 1.05
Ethanol ============ 0.790
Water ============= 1.00
Element======== Molar mass (g mol-1)
Oxygen=========== 16.0
Carbon============ 12.0
Hydrogen========== 1.00
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To answer the questions, let's break them down one by one.
a) (i) To predict the value of pH at the equivalence point, we need to consider the reaction that is taking place. In this case, it is the reaction between ethanoic acid (CH3COOH) and aqueous potassium hydroxide (KOH). This reaction produces a salt, potassium acetate (CH3COOK), and water (H2O).
The equivalence point is reached when the moles of acid react completely with an equal number of moles of base. In this case, since the reaction is 1:1, the moles of ethanoic acid reacting with the moles of potassium hydroxide will be equal.
Since potassium acetate is a salt of a weak acid (ethanoic acid) and a strong base (potassium hydroxide), it will hydrolyze in water to produce hydroxide ions (OH-) and ethanoic acid (CH3COOH). This hydrolysis reaction will lead to an increase in the concentration of hydroxide ions in the titration mixture, resulting in an alkaline pH at the equivalence point.
Therefore, the predicted value of pH at the equivalence point would be greater than 7, indicating an alkaline solution.
(ii) The chemical indicator that can be used for this titration is phenolphthalein. Phenolphthalein is commonly used in acid-base titrations and changes color in the pH range of around 8.2 to 10.0. At the equivalence point, when the solution becomes alkaline due to the hydrolysis of the potassium acetate, phenolphthalein will change from colorless to pink, indicating the completion of the reaction.
b) (iii) The balanced equation for the reaction of aqueous potassium hydroxide (KOH(aq)) with aqueous ethanoic acid (CH3COOH(aq)) is:
CH3COOH(aq) + KOH(aq) → CH3COOK(aq) + H2O(l)
(iv) To calculate the amount of unreacted aqueous ethanoic acid (CH3COOH(aq)) in the sample taken for titration with aqueous potassium hydroxide (KOH(aq)), we need to use the stoichiometry of the balanced equation and the volume and concentration of KOH used.
From the balanced equation, we can see that 1 mole of ethanoic acid reacts with 1 mole of KOH. The volume of KOH used is 25.5 cm3, and the concentration of KOH is 1.00 mol dm-3.
Using the relationship between volume, concentration, and moles (n = V * C), we can calculate the moles of KOH used.
moles of KOH = (25.5 cm3 * 1.00 mol dm-3) / 1000 cm3 = 0.0255 moles
Since the reaction is 1:1, the moles of ethanoic acid present in the sample would also be 0.0255 moles.
Next, we need to convert the moles of ethanoic acid to mass using the molar mass of ethanoic acid (CH3COOH), which is 60.0 g mol-1.
mass of ethanoic acid = 0.0255 moles * 60.0 g mol-1 = 1.53 grams
Therefore, the amount of unreacted aqueous ethanoic acid in the sample taken for titration with KOH is 1.53 grams.
v) To calculate the amount of aqueous ethanoic acid (CH3COOH(aq)) in the sealed flask at equilibrium, we need to consider the volume and concentration of the original ethanoic acid solution, as well as the dilution that occurred due to the addition of ethanol and water.
The initial volume of pure ethanoic acid is 25.0 cm3, and the final volume of the equilibrium mixture is 80.0 cm3 (sum of the volumes of ethanoic acid, ethanol, and water).
Using the formula of dilution (C1V1 = C2V2), we can calculate the concentration of ethanoic acid in the final equilibrium mixture.
Initial concentration of ethanoic acid:
C1 = moles / volume = unknown
Final concentration of ethanoic acid:
C2 = moles / volume = 0.0255 moles / (80.0 cm3 / 1000 cm3/dm3) = 0.319 mol dm-3
Now, we can use the dilution formula to find the initial concentration of ethanoic acid.
C1 * 25.0 cm3 = 0.319 mol dm-3 * 80.0 cm3 / 1000 cm3/dm3
C1 = (0.319 mol dm-3 * 80.0 cm3) / (1000 cm3/dm3 * 25.0 cm3) = 0.255 mol dm-3
Therefore, the initial concentration of aqueous ethanoic acid in the sealed flask at equilibrium is 0.255 mol dm-3.
Assumptions made in the calculation:
- The volume changes due to dilution are negligible compared to the initial volumes.
- The volumes are at room temperature and assume the same density as the substances provided in the data.
It's important to note that this answer is based on the information provided in the question, and any additional information or parameters should be considered if available.