Answer:
b) Adding 0.075 moles of HCl
Step-by-step explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa (Weak base with its conjugate acid).
The buffer of the problem is the acetic acid / lithium acetate.
The addition of any moles of the acid and the conjugate base will not destroy the buffer, just would change the pH of the buffer. Thus, a and c will not destroy the buffer.
The addition of an acid (HCl) or a base (NaOH), produce the following reactions:
HCl + LiC₂H₃O₂ → HC₂H₃O₂ + LiCl
The acid reacts with the conjugate base to produce the weak acid.
And:
NaOH + HC₂H₃O₂ →NaC₂H₃O₂ + H₂O
The base reacts with the weak acid to produce conjugate base.
As the buffer is 1.0L, the moles of the species of the buffer are:
HC₂H₃O₂ = 0.300 moles
LiC₂H₃O₂ = 0.045 moles
The reaction of HCl with LiC₂H₃O₂ consume all LiC₂H₃O₂ -because there are an excess of moles of HCl that react with all LiC₂H₃O₂-
As you will have just HC₂H₃O₂ after the reaction, the addition of b destroy the buffer.
In the other way, 0.0500 moles of NaOH react with the HC₂H₃O₂ but not consuming all HC₂H₃O₂, thus d doesn't destroy the buffer.