Buffer Solution question (mult. choice).

  • Thread starter Thread starter RESmonkey
  • Start date Start date
  • Tags Tags
    Buffer Choice
AI Thread Summary
A buffer solution requires a weak acid and its conjugate base or a weak base and its conjugate acid. In the discussion, option D (0.2 mole of CH3COOH and 0.1 mole of NaOH) is identified as a buffer because acetic acid partially dissociates, allowing for equilibrium with its conjugate base, sodium acetate. Option E is dismissed as a buffer since HBr is a strong acid that fully dissociates, lacking the necessary equilibrium for buffering. The conversation also touches on the buffer behavior of NH3/NH4Cl, highlighting the importance of partial dissociation in weak acids and bases. Understanding these principles is crucial for identifying buffer solutions effectively.
RESmonkey
Messages
25
Reaction score
0
New here, hello :)

Which of following mixtures will be a buffer when dissolved in a liter of water?

a. .1 mole of Ba(OH)2 and .2 mole of HBr
b. .3 mole of KCl and .3 mole of HCl
c. .4 mole of NH3 and .4 mole of HCl
d. .2 mole of CH3COOH and .1 mole of NaOH
e. .2 mole of HBr and .1 mole of NaOH

My guess is that it has something to do with being a weak acid/base. Unfortunately, this doesn't make sense, because the answer is D, and D has acetic acid and NaOH, which aren't weak?

Thanks in advance :)
 
Physics news on Phys.org
it is D

0.1 mol from the 0.2 mole ethanoic acid reacts with 0.1 mol NaOH to give 0.1 mol sodium ethanoate.

0.1 mol ethanoic acid also remains.

the buffer is CH3COOH/CH3COONa
 
hmmm

Using that, couldn't e also be an answer? It has similar concentrations of an acid and base.
 
but HBr is a strong acid. it fully dissociates in solution. it is stronger than HCl. there is no equilibirum between H+ and HBr.
 
Gotcha. So the two compounds have to be weak, and one of them has to be left over in the end?
 
errmm not really. the acid has to be weak but the salt is a strong electrolyte.

e.g. CH3COOH/CH3COO-Na+

CH3COOH <------> CH3COO- + H+

CH3COO-Na+ ------> CH3COO- + Na+

when you add a small amount of acid, the H+ added will combine with the conjugate base CH3COO- from the fully ionised salt to give CH3COOH. there is no drastic change in pH.

when you add a small amount of alkali, the OH- will combine with H+ from the partially ionised acid to minimise the chnage in pH. now, H+ concentration will also decrease, but since the dissociation of the acid is an equilibrium, according to LCP, the equilibrium will shift to the right, and the H+ will be restored.

this also applies to alkaline buffers. try to guess what happens with NH3/NH4Cl?
 
Nh3 + H30+ <------> Nh4 + H20 ?
 
you have to write an equation for each species, you will get a clearer overview...

NH3 + H2O <-----> NH4+ + OH-

NH4Cl -----> NH4+ + Cl-

remeber that NH3 dissociates only partially, but the salt ionises fully.

what happens when you add some acid? or some alkali?
 
Back
Top