Question

A 1.0 L buffer solution contains .30 moles acetic acid and .10 moles lithium acetate. which of the following actions will destroy the buffer?

A 1.0 L buffer solution contains 0.30 moles acetic acid (CH3CO2H) and 0.10 moles lithium acetate (CH,COLI). Which of the following actions will destroy the buffer? A. Addition of 0.10 moles of NaOH B. Addition of 0.10 moles of HCI

Answer 1

ANSWER : Addition of 0.10 mol HCl

Buffer solution contains weak acid and its conjugate base. Given buffer solution contain weak acid CH3COOH and conjugate base CH3COO - .

Consider reaction of NaOH with buffer solution.

CH3COOH + OH - $\rightarrow$ CH3COO - + H2O

Let's use ICE table.

moles	CH3COOH	OH -	CH3COO -
I	0.30	0.10	0.10
C	-0.10	-0.10	+0.10
E	0.20	0.00	0.20

After addition of NaOH , solution contain equal amounts of CH3COOH and CH3COO - . This solution acts as a good buffer.

Now, consider reaction of HCl with buffer solution.

CH3COO - + HCl $\rightarrow$ CH3COOH + Cl -

Let's use ICE table.

moles	CH3COO -	HCl	CH3COOH
I	0.10	0.10	0.30
C	-0.10	-0.10	+0.10
E	0.00	0.00	0.40

After addition of 0.10 mole HCl , solution contains only CH3COOH. There is no conjugate base remained in the solution. Hence, this solution is a acid solution not a buffer solution.

i e addition of 0.10 mol HCl destroys the buffer solution.