Reaction Quotient and Equilibrium Shift

Question

Which of the following conclusions is best supported by the evidence that for the weak acid equilibrium $$HA ightleftharpoons H^+ + A^-$$, the reaction quotient $$Q = \frac{[H^+][A^-]}{[HA]}$$ calculated from the given concentrations (yielding $$Q = 1.0 	imes 10^{-3}$$) is greater than the acid dissociation constant $$K_a = 1.8 	imes 10^{-5}$$?
 
| Species   | Concentration (M)              |
|-----------|--------------------------------|
| $$H^+$$   | $$1.0 	imes 10^{-3}$$         |
| $$A^-$$   | 0.10                           |
| $$HA$$    | 0.10                           |
| $$K_a$$   | $$1.8 	imes 10^{-5}$$         |

Accepted Answer

Since $$Q > K_a$$, the equilibrium will shift to the left, favoring the formation of undissociated acid $$HA$$.

Answer

A reaction quotient greater than the dissociation constant indicates that the system is at equilibrium with no net directional shift.

Answer

Because $$Q > K_a$$, the reaction must shift to the right in order to produce more $$H^+$$ and $$A^-$$.

Answer

Despite $$Q$$ exceeding $$K_a$$, the system's shift direction cannot be determined because common ion effects might override the equilibrium.

Rank	User	Correct Count	Attempt Count	Total Time	Score
#1	Pelisx	2	2	1m 12s	128
#2	202401067	1	1	0m 29s	71
#3	malyn27	1	1	0m 34s	66
#4	70217102drcdsc	1	1	0m 40s	60
#5	misbeih69	1	1	0m 40s	60
#6	jasonl0404sosa	1	1	0m 44s	56
#7	kkat.s.music	1	1	0m 58s	42
#8	Derinchiay2013	4	5	6m 32s	-2
#9	ggerzon	0	1	0m 04s	-14
#10	jicereza09	1	3	4m 50s	-210

Species	Concentration (M)
$H^+$	$1.0 \times 10^{-3}$
$A^-$	0.10
$HA$	0.10
$K_a$	$1.8 \times 10^{-5}$

Reaction Quotient and Equilibrium Shift

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