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sesenic [268]
3 years ago
11

1. In an equilibrium experiment, acetic acid (which is a weak acid) is mixed with sodium acetate (a soluble salt), with methyl o

range as an indicator. Explain this phenomenon by using the common ion effect. Include equations in your explanation.
Chemistry
1 answer:
koban [17]3 years ago
4 0

Explanation:

It is known that acetic acid is a weak acid. It's equilibrium of dissociation will be represented as follows.

          CH_{3}COOH(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COO^{-}(aq) + H_{3}O^{+}(aq)

On the other hand, sodium acetate (CH_{3}COONa) is a salt of weak acid, that is, CH_{3}COOH and strong base, that is, NaOH. Therefore, aqueous solution of sodium acetate will be basic in nature.

Since, acetic acid is a weak acid but still it is an acid. So, when methyl orange is added in a solution of acetic acid then it given a reddish-orange color because of its acidity.

When sodium acetate is mixed into this solution then it will dissociate as follows.

            CH_{3}COO^{-}Na^{+}(aq) \rightleftharpoons CH_{3}COO^{-}(aq) + Na^{+}(aq)

As both solutions are liberating acetate ion upon dissociation. Hence, it is the common ion.

So, when more acetate ions will increase from dissociation of sodium acetate the according to Le Chatelier's principle the equilibrium will shift on left side.

As a result, there will be decrease in the concentration of hydronium ions. As a result, there will be increase in the pH of the system.

Hence, color of methyl orange will change from reddish orange to yellow. This shift in equilibrium is due to the common ion which is CH_{3}COO^{-} ion.

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A) What volume of butane (C 4 H 10 ) can be produced at STP, from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas
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From the stoichiometry of the reaction, carbon is in excess and 5.856 g s left over.

<h3>What is the volume of butane produced?</h3>

The reaction can be written as; 4C(s) + 5H2(g) -----> C4H10(g)

Number of moles of C =  13.45 g/1 2g/mol = 1.12 moles

If 1 mole of hydrogen occupies 22.4 L

x moles of hydrogen occupies  17.65 L

x = 0.79 moles

Now;

4 moles of carbon reacts with 5 moles of hydrogen

1.12 moles of carbon reacts with  1.12 moles * 5 moles/4 moles

= 1.4 moles of hydrogen

Hence hydrogen is the limiting reactant here and carbon is in excess.

If 4 moles of carbon reacts with 5 moles of hydrogen

x moles of carbon reacts with 0.79 moles of hydrogen

x = 0.632 moles

Number of moles of carbon unreacted =  1.12 moles -  0.632 moles

= 0.488 moles

Mass of carbon unreacted = 0.488 moles * 12 g/mol

= 5.856 g

Volume of butane produced is obtained from;

5 moles of hydrogen produces 1 mole of butane

0.79 moles of hydrogen produces 0.79 moles *  1 mole/ 5 moles

= 0.158 moles

1 mole of butane occupies 22.4 L

0.158 moles of butane occupies 0.158 moles * 22.4 L/ 1 mole

= 3.53 L

Learn more about stoichiometry:brainly.com/question/9743981

#SPJ1

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