acid dilution and ka qn (1 Viewer)

Masaken

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i'm trying to explain the effect on Ka and hence equilibrium if i dilute a solution of HF(aq) to one third of its original concentration. the way i understand it, since it's 1/3 of the original concentration, the concentrations of each individual species in the Ka expression are all divided by 1/3 (and hence multiplied by 3), thus resulting in Q = Ka x 3.

is this correct? because the worked solutions says that the concentrations of each individual species in the Ka expression are divided by 3 (and hence multiplied by 1/3), which results in Q = Ka x 1/3, which doesn't make sense to me, unless i've done something wrong?
 

Bob99

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i'm trying to explain the effect on Ka and hence equilibrium if i dilute a solution of HF(aq) to one third of its original concentration. the way i understand it, since it's 1/3 of the original concentration, the concentrations of each individual species in the Ka expression are all divided by 1/3 (and hence multiplied by 3), thus resulting in Q = Ka x 3.

is this correct? because the worked solutions says that the concentrations of each individual species in the Ka expression are divided by 3 (and hence multiplied by 1/3), which results in Q = Ka x 1/3, which doesn't make sense to me, unless i've done something wrong?
when you are diluting, equilibrium will shift towards the side to replace the diluted chemical. In this case your equation would be HF ⇌ H + F
As equilibum shift to the left, it will dissociate less, hence Q will decrease.

FYI last time i did chem was in november and i have forgotten quite a bit of the theory. so i might be wrong
 
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wizzkids

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Whoa. hold on a minute. Let's get this straight, Ka does not change.
Ka is a constant for practically any dilution.
The only thing that can change Ka is a change of temperature.
OK, so what does a change of dilution do to the equilibrium? It changes the percent dissociation of HF molecules, but Ka does not change.
The equilibrium will shift so that a greater percentage of HF molecules are dissociated to H+ and F- ions.
Can you see what is going on now?
 

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