How can the equilibrium shift, while Kc remains constant? Consider the following reversible reaction.
$$\ce{Cr2O7^2-(aq) + H2O(l) &lt;=&gt; 2 CrO4^2-(aq) + 2 H+(aq)}$$
What will happen to the position of equilibrium and the value of $K_c$ when more $\ce{H+}$ ions are added at constant temperature?
$$
\begin{array}{lcc}
\hline
  &amp; \text{Position of equilibrium} &amp; \text{Value of}~K_c \\
\hline
\text{A.} &amp; \text{shifts to the left} &amp; \text{decreases} \\
\text{B.} &amp; \text{shifts to the right} &amp; \text{increases} \\
\text{C.} &amp; \text{shifts to the right} &amp; \text{does not change} \\
\text{D.} &amp; \text{shifts to the left} &amp; \text{does not change} \\
\hline
\end{array}
$$

The answer given is D.
I understand that when more $\ce{H+}$ ions are added, the reaction shifts in reverse. So I've understood that this is known as having the equilibrium shift to the left, toward the reactants.
What confuses me is why $K_c$ doesn't change. If the equilibrium shifts to the left, I suppose that must mean more reactants are formed in relation to the products, so $K_c$ would decrease.
But the answer key says that $K_c$ remains unchanged. This is coherent with what I've been taught, but I don't understand how the equilibrium can change yet have $K_c$ remain constant.
I'm just afraid I've got some of the theory mixed up, so I wanted to get some clarity. Could someone explain this please?
$K_c$ is related to the ratio of reactants to products at equilibrium.

If the reaction is currently at equilibrium, and you add more products then the reaction is now out of equilibrium and the reverse reaction will happen until it is back in equilibrium.

I don't like the wording of "equilibrium shifts to the left" myself, I would say that reverse reactions occurs to restore equilibrium. But $K_c$ doesn't change based on reactants/products concentrations since its the ratio at equilibrium. $K_c$ will normally depend on temperature though.

Based on the options you have, $K_c$ not changing and the reverse reaction occurring, the closest answer would be D. (If you take equilibrium shifting to the left to mean the reaction goes in that direction.)