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A mixture of 0.50 mol of ethanoic acid and 1.00 mol of ethanol was shaken for a long time to reach equilibrium. The whole mixture was titrated quickly with 1.00 mol dm3 sodium hydroxide and 80cm3 of alkali were required.

1. Write an equation for the reaction between ethanoic acid and ethanol.
2. Explain why the reaction mixture was titrated quickly.
3. By making use of the titration results and the equation in (1), Calculate:
a. How many moles of ethanoic acid remained at equilibrium
b. How many moles of ethanoic acid had reacted.
c. How many moles of ethanol were left in the equilibrium mixture.
d. Calculate a value for Kc for the reaction.

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1 answer
  1. CH3COOH + HOCH2CH3 =>CH3COOC2H5 + H2O

    3a. mol CH3COOH = M x L = about 0.80.
    b. initial = 0.500-0.08 ending = 0.420
    c. 1.00-0.08 = 0.92
    d. Substitute and calculate Kc.

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