Question

Which of the following cannot be a buffer solution?

Group of answer choices

1.0-L solution, in which 0.10 mol acetic acid (CH₃COOH) and 0.050 mol potassium hydroxide (KOH) are dissolved.

0.50-L solution, in which 0.25 mol ammonia (NH₃) and 0.18 mol hydrochloric acid (HCl) are dissolved.

0.75-L solution, in which 1.0 mol hydrofluoric acid (HF) and 2.0 mol sodium hydroxide (NaOH) are dissolved.

1.0-L solution, in which 3.0 mol lithium hydroxide (LiOH) and 5.0 mol formic acid (HCOOH) are dissolved.

0.75-L solution, in which 0.50 mol ammonia (NH₃) and 0.50 mol ammonium chloride (NH₄⁺Cl^-).

Answer 1

uswer, - lol Solution in which 0.10 mo ocetic and and 0.0somo) koly are dissolved - It will form bu w w hоw whex 0+ Kон 3 үwley of acekc and so some cutic and will be left and after reaction of CH₂ coom +KOH - CH₂COok (conjegelt base of orehe and) is produced i so it will form butter . so any solution having concoutiation of Strong and I base is less than concentraten (me) of weak and I base butter will form. so only *0.75L-solution in which 1.0 mol Hf and q.O mol haoy are dissolved will not form buffer as no MF will be Naoy left after reaction with op mer of HF. (mores of Naom