Describe two ways you could make this equilibrium shift toward reactants that haven't already been discussed in this problem. Explain your reasoning.

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Chapter1: Chemical Foundations
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Check Your Understanding
Describe two ways you could make this equilibrium shift toward reactants that haven't already
been discussed in this problem. Explain your reasoning.
Answer
Possibilities include removing HCN from the system or adding an acid (other than HCN),
which would increase [H+].
Transcribed Image Text:Check Your Understanding Describe two ways you could make this equilibrium shift toward reactants that haven't already been discussed in this problem. Explain your reasoning. Answer Possibilities include removing HCN from the system or adding an acid (other than HCN), which would increase [H+].
Solution
The equilibrium is
CH3 COOH(aq) 2H*(aq)+CH3COO (aq)
(a) Sodium acetate is added and will dissociate into Na* and CH3 COO¯. The sodium ion is not
involved in this equilibrium, but by adding CH; COO¯,we increase the concentration of a
product. This in turn increases the value of Q, so Q > K. Equilibrium must shift toward
reactants, so additional CH3 COOH will form.
(b) Additional acetic acid is added. Because CH3 COOH is a reactant, the increase in
concentration of the acid decreases the value of Q. Q becomes less than K, so equilibrium
shifts toward products. Some of the added CH3 COOH will dissociate, and the concentration of
both Ht and CH3 COO will increase.
(c) Sodium hydroxide is added. Again the sodium ion does not play a significant role in this
perturbation, and at first glance, it might seem like the hydroxide ion doesn't either. But if you
use a little bit of the chemical intuition you've developed by this point in the course, you
should realize that OH- and H+ ions will react with one another to form water. This will
decrease the concentration of H+. Because the concentration of a product is reduced, the size
of the numerator decreases, making Q < K. Thus the equilibrium shifts to form more
products.
Transcribed Image Text:Solution The equilibrium is CH3 COOH(aq) 2H*(aq)+CH3COO (aq) (a) Sodium acetate is added and will dissociate into Na* and CH3 COO¯. The sodium ion is not involved in this equilibrium, but by adding CH; COO¯,we increase the concentration of a product. This in turn increases the value of Q, so Q > K. Equilibrium must shift toward reactants, so additional CH3 COOH will form. (b) Additional acetic acid is added. Because CH3 COOH is a reactant, the increase in concentration of the acid decreases the value of Q. Q becomes less than K, so equilibrium shifts toward products. Some of the added CH3 COOH will dissociate, and the concentration of both Ht and CH3 COO will increase. (c) Sodium hydroxide is added. Again the sodium ion does not play a significant role in this perturbation, and at first glance, it might seem like the hydroxide ion doesn't either. But if you use a little bit of the chemical intuition you've developed by this point in the course, you should realize that OH- and H+ ions will react with one another to form water. This will decrease the concentration of H+. Because the concentration of a product is reduced, the size of the numerator decreases, making Q < K. Thus the equilibrium shifts to form more products.
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