A + B ==> C + D at equilibrium
If more C is added to the equilibrium mixture, the reaction will shift to the left. You KNOW concn A will increase, you KNOW concn B will increase, and you KNOW concn D will decrease. What many students wonder about, is what happens to C. Since the reaction shifts to the left, the equilibrium mixture will become smaller in C BUT we added to C to make it shift in the first place. The FINAL result is that C actually increases. It increases, first, because we added more C to make it do the shift thing, then it decreases some BECAUSE of the shift thing, but the two don't balance each other out. To make up some numbers, and I will use moles because I don't want to go through the confusion of getting the volume into this, suppose the original equilibrium was established and the moles C, at equilibrium, is 0.4. Now we introduce, from an external source, 0.2 mole C. Nothing else changes. A will increase, B will increase, D will decrease, but C changes, too. It shifts to the left so some of that 0.4 will be lost, say 0.1 will be lost, but we added 0.2 so the total C now will be 0.4-0.1+0.2 = 0.5. So C actually is larger than when we started with the original equilibrium but it isn't 0.4 + 0.2 (it isn't 0.6 moles) because some of it was lost when the reaction shifted to the left. Look in your chemistry text. Some texts I've see make real calculations to show what the equilibrium concns are at equilibrium, then they add some material from an external source and recalculate the new concns when the system returns to equilibrium. Those examples make the point about this very issue.