Question

When $ 3{\text{ g}} $ of ethanoic acid and $ 2.3{\text{ g}} $ of ethanol were equilibrated at $ {\text{100}}{{\text{ }}^ \circ }C $ for an hour and then quickly cooled in an ice bath. $ {\text{50 c}}{{\text{m}}^3} $ of $ {\text{1 mole d}}{{\text{m}}^{ - 3}} $ aqueous sodium hydroxide were added. When the mixture was titrated with $ {\text{1 mole d}}{{\text{m}}^{ - 3}} $ hydrochloric acid, $ {\text{33}}{\text{.3 c}}{{\text{m}}^3} $ of acid were required.
 $ {C_2}{H_5}OH(l){\text{ }} + {\text{ }}C{H_3}COOH(l){\text{ }} \rightleftharpoons {\text{ }}C{H_3}COO{C_2}{H_5}(l){\text{ }} + {\text{ }}{H_2}O(l) $
Why was the mixture cooled in an ice bath and how do you find $ K_c $ with this information?

Answer 1

Hint: Here we will find the moles of ethanoic acid and ethanol with the help of given mass of ethanoic acid and ethanol respectively. Then we will find the moles of sodium hydroxide. After the titration with hydrochloric acid we will find the amount of sodium hydroxide left and then with this amount of sodium hydroxide we can find the equilibrium constant of the reaction.

Complete Step By Step Answer:
We are given with the reaction as:
 $ {C_2}{H_5}OH(l){\text{ }} + {\text{ }}C{H_3}COOH(l){\text{ }} \rightleftharpoons {\text{ }}C{H_3}COO{C_2}{H_5}(l){\text{ }} + {\text{ }}{H_2}O(l) $
Since we know that number of moles is the ratio of given mass and molar mass of the compound, therefore we can find the number of moles of ethanoic acid and ethanol initially as:
Given mass of ethanoic acid $ = {\text{ }}3{\text{ g}} $
Molar mass of ethanoic acid $ = {\text{ }}\left( {12 + \left( {1 \times 3} \right) + 12 + 16 \times 2 + 1} \right){\text{ = 60 g}} $
 $ \Rightarrow $ Moles of ethanoic acid $ = {\text{ }}\dfrac{{{\text{3 g}}}}{{60{\text{ g}}}}{\text{ = 0}}{\text{.05 mole}} $
Similarly for ethanol:
Given mass of ethanol $ {\text{ = }}2.3{\text{ g}} $
Molar mass of ethanol $ = {\text{ }}\left( {12 \times 2 + \left( {1 \times 5} \right) + 16 + 1} \right){\text{ = 46 g}} $
 $ \Rightarrow $ Moles of ethanol $ = {\text{ }}\dfrac{{{\text{2}}{\text{.3 g}}}}{{46{\text{ g}}}}{\text{ = 0}}{\text{.05 mole}} $
Now when we add sodium hydroxide then some of the moles of sodium will be excess in reaction while some react with hydrochloric acid. Now we will find moles of sodium hydroxide.
The volume of sodium hydroxide given is $ {\text{50 c}}{{\text{m}}^3} $ which is equal to $ 0.050{\text{ d}}{{\text{m}}^3} $ and the number of moles will be:
 $ \Rightarrow $ Number of moles $ = {\text{ 0}}{\text{.050 d}}{{\text{m}}^3}{\text{ }} \times {\text{ 1 mole d}}{{\text{m}}^{ - 3}}{\text{ = 0}}{\text{.050 mole}} $
Now we will find the moles of sodium hydroxide which will be consumed by titration with hydrochloric acid. The reaction will be written as:
 $ NaOH + HCl \to NaCl + {H_2}O $
Therefore we can say that equal moles of hydrochloric acid and sodium hydroxide will be consumed while titration. Thus moles of sodium hydroxide will be equal to moles of hydrochloric acid as given in question. Volume of hydrochloric acid is $ {\text{33}}{\text{.3 c}}{{\text{m}}^3} $ which is equal to $ {\text{0}}{\text{.0333 d}}{{\text{m}}^3} $ .
 $ \Rightarrow $ Moles of sodium hydroxide $ = {\text{ 0}}{\text{.0333 d}}{{\text{m}}^3}{\text{ }} \times {\text{ 1 mole d}}{{\text{m}}^{ - 3}}{\text{ = 0}}{\text{.0333 mole}} $
After titration moles of sodium hydroxide left $ = {\text{ 0}}{\text{.0500 - 0}}{\text{.0333 = 0}}{\text{.0167 mole}} $
 $ \Rightarrow $ Thus we can write the reaction as:
 $ {C_2}{H_5}OH(l){\text{ }} + {\text{ }}C{H_3}COOH(l){\text{ }} \rightleftharpoons {\text{ }}C{H_3}COO{C_2}{H_5}(l){\text{ }} + {\text{ }}{H_2}O(l) $

S.NO	$ \left[ {{C_2}{H_5}OH} \right] $	$ \left[ {C{H_3}COOH} \right] $	$ \left[ {C{H_3}COO{C_2}{H_5}} \right] $	$ \left[ {{H_2}O} \right] $
$ t = 0 $	$ 0.0500 $	$ 0.0500 $	$ 0 $	$ 0 $
$ t = {t_{eq.}} $	$ 0.0500 - 0.0333 = 0.0167 $	$ 0.0500 - 0.0333 = 0.0167 $	$ 0.0333 $	$ 0.0333 $

Therefore we can write the value of equilibrium constant $ \left( {{{\text{K}}_c}} \right) $ as:
 $ {{\text{K}}_c}{\text{ = }}\dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right]{\text{ }}\left[ {{H_2}O} \right]}}{{\left[ {{C_2}{H_5}OH} \right]{\text{ }}\left[ {C{H_3}COOH} \right]}} $
 $ \Rightarrow {{\text{K}}_c}{\text{ = }}\dfrac{{{\text{0}}{\text{.0333 }} \times {\text{ 0}}{\text{.0333}}}}{{{\text{0}}{\text{.0167 }} \times {\text{ 0}}{\text{.0167}}}} $
 $ \Rightarrow {{\text{K}}_c}{\text{ = 4}} $
The value of the equilibrium constant is approximately four and we have to add cold water because the reaction is reversible. Therefore to prevent the reaction getting reversed we add cold water to it.

Note:
The number of moles which are calculated above are round-off values. We have taken values up to two decimal places. There might be some difference in decimal value but it would not affect our answer. It must be noted that for finding moles of sodium hydroxide we have to convert the units of the volume. The moles of hydrochloric acid and sodium consumed while titration is equal.