An acid-base equilibrium system is created by dissolving 0.10 mol HF in water to a volume of 1.0 L. What is the effect of adding 0.050mol F(aq) to this solution?

The equation for the equilibrium is

HF\,\rightleftharpoons\,H^+\,+\,F^-

You know the concentration of HF. You haven't given me any Ka for HF so you can't solve this quantitatively.

Remember that

K_a=\,\frac{[H^+][F^-]}{[HF]}

So, from that you can calculate the concentration of [F-] and figure out what happens to the equilibrium (it will drive it to the left).