A 1-liter solution contains 0.494 M hydrofluoric acid and 0.371 M potassium fluoride. Addition of 0.408 moles of hydrochloric acid will: (Assume that the volume does not change upon the addition of hydrochloric acid.) a. Raise the pH slightly
b. Lower the pH slightly
c. Raise the pH by several units
d. Lower the pH by several units
e. Not change the pH
f. Exceed the buffer capacity

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

And the concentration of HF after the reaction of KF with HCl is is:

Now, from the equilibrium of equation (1) we have:

 (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!