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Say we have the equilibrium $$\ce{A_{(aq)} + 2B_{(aq)} <=> C_{(aq)} + D_{(aq)}}$$ meaning the formula for the equilibrium constant is $$\ce{K_{c} = \frac{[C][D]}{[A][B]^2}}$$ My understanding is that adding more water would shift the equilibrium to the left (in the same way that decreasing the pressure would shift the equilibrium to the left if it was a gaseous equilibrium).

Why is this the case? What mathematically (using the equation) and physically (in terms of kinetics?) means that $\ce{[A]}$ and $\ce{[B]}$ must increase?

And why does the equilibrium constant remain the same?

Apologies if this is a duplicate, but I couldn't find a specific answer to this question.

Many thanks!

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    $\begingroup$ Mathematically, say, if you dilute everything with equal amount of water, all concentrations become 2 times less; guess what would happen to $\ce{\frac{[A][B]^2}{[C][D]}}$? $\endgroup$ – Ivan Neretin Jun 3 '16 at 11:50
  • $\begingroup$ Sorry, I wrote the $K_c$ equation wrong. But halving the values of $\ce{[A], [B], [C], and [D]}$ would increase the value of $K_c$ ($\ce{\frac{[C][D]}{[A][B]^2}}$), which I know isn't the case. $\endgroup$ – user234501 Jun 3 '16 at 12:01
  • $\begingroup$ That's right; now what else do you have to change in order fo $K_c$ to actually stay constant? $\endgroup$ – Ivan Neretin Jun 3 '16 at 12:23
  • $\begingroup$ The concentrations of A, B, C, and D have to change. But why does Kc have to remain the same? $\endgroup$ – user234501 Jun 3 '16 at 12:25
  • $\begingroup$ That's right, too, and you seem to know which way they have to change. As for why the equilibrium constant is the way it is, see chemistry.stackexchange.com/questions/6924/… or chemistry.stackexchange.com/questions/14836/… $\endgroup$ – Ivan Neretin Jun 3 '16 at 12:37

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