Respuesta :
Answer:
adding 0.06 moles of LiOH will destroy the buffer
Explanation:
Step 1: analyzing the buffer
We have in a 1L buffer solution 0.06M HC2H3O2, which is acetic acid, and is a weak acid.
We have a 0.250M LiC2H3O2, which is lithium acetate, and is a conjugate base.
Step 2 : Calculating moles
mole acetic acid = 0.06M x 1L = 0.06 mol
mole Lithium acetate = 0.250M x 1L = 0.250 mole
⇒ we can say that there is roughly 4-5 times more base than acid
⇒Since both are weak ( respectively acid and base), adding one of those will not be enough to destroy the buffer, but instead we should think of adding a strong acid or base.
Step3: Adding acid or base
If we add a strong acid like HCL (0.06 moles) this will react with 0.06 moles of the weak base.
⇒0.25 mole conjugate base - 0.06 mole acid = 0.19 mole base remaining
⇒0.06 mole weak acid + 0.06 mole acid = 0.12 mole acid formed
⇒ We still have a buffer here, so adding 0.06 moles of a strong acid like HCl wil not destroy the buffer either.
If we add 0.06 moles LiOH this will react with 0.06 moles of the weak acid. This means there will be remain 0 weak acid, but 100% conjugate base.
Since we saw from the start that the buffer has 4-5 x more base than acid, it seems logical that adding more base will destroy the buffer.
The correct answer is adding 0.06 moles of LiOH will destroy the buffer
