A 1.0 L buffer solution is 0.300 M HC2H3O2 and 0.045 M LiC2H3O2. Which of the following actions will destroy the buffer? a) Adding 0.050 moles of HC2H3O2
b) Adding 0.075 moles of HCl
c) Adding 0.0500 moles of LiC2H3O2
d) Adding 0.050 moles of NaOH
e) None of the above will destroy the buffer.

b) Adding 0.075 moles of HCl

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.