STPM Chemistry Form 6 Notes – Ionic Equilibrium (Part 13)

by BerryBerryTeacher

in Berry Reference (Notes)

Buffer solutions prevents pH changes of a system. This is clear from the previous part, Part 12 of Berry Berry Easy‘s notes for STPM Chemistry on Ionic Equilibrium. Do you know that, you can also create buffers? This Part 13 is focused on the make up of a buffer. typically buffers are made by mixing conjugate pairs, or through partial titration of either component in a weak reagent conjugate pair. For those who have not read Part 12, please read it first before proceeding to the following part due to the prior knowledge required.

[Tips: While it might not be overly important in STPM level to differentiate between buffering agents and buffer solutions, there are actually differences. Both regulates pH of solutions by resisting changes in pH levels. Buffering agents are the active ingredients of a buffer solution. So a buffer solution resist changes made to pH for whole system, while a buffering agent is added to solution which is already acidic or basic in order to modify/keep a new pH.]

STPM Chemistry Form 6 Notes – Ionic Equilibrium (Part 13)

Natural Universal Indicator_Purple cabbage

Natural Universal Indicator_Purple cabbage

Buffers

  • buffer is a solution where pH remain relatively constant / pH value change only very slightly after the addition of small amount of either strong acid or strong base
  • buffer are made of roughly equal mole mixture of a weak acid and its weak conjugate base in an aqueous solution.

Example:

Explain why a solution containing a strong base and its salt does not act as a buffer solution?

Solution: Addition of OH- does not shift an equilibrium toward un-ionised base, as it would with a weak base and its conjugate.

Buffer Recipes

  • buffer can be made either by mixing the conjugate pair together or
  • buffer can be made by partially titrating either component in a conjugate pair of weak reagents.

Case 1:

Half-titrate the weak acid with strong base
(Tips: a weak acid must be chosen with a pKa value close to the desired pH)

Case 2:

Half-titrate the weak conjugate base with strong acid
Buffers can be mixed by any of the method below:

  • Weak acid + the salt of the conjugate base in roughly equal mole proportions (HCOOH with HCOONa)
  • Weak base + the salt of the conjugate acid in roughly equal mole proportions (NH3 + NH4Cl)
  • Weak acid and roughly half of an equivalent of strong base (HOAc with half equivalent KOH)
  • Weak base and roughly half of an equivalent of strong acid (H3CNH2 with half equivalent HCl)

Example:

Which of the following combinations of solutes would result in the formation of a buffer solution?

A NaC2H3O2 + HC2H3O2
B NH4Cl + NH3
C HCl + NaCl
D HCl + HC2H3O2
E NaOH + HCl
F NaOH + HC2H3O2 in a 1:1 mole ratio
G NH3 + HCl in a 2:1 mole ratio
H HC2H3O2 + NaOH in a 2:1 mole ratio

Tips:

  • CH3COOH same with HC2H3O2.
  • Must has roughly equal mole mixture of a weak acid and its weak conjugate base.

Solution: The solution in G contains equimolar quantities of NH4+ and NH3. The NH4+ results from the reaction.

H+Cl- + NH3 –> NH4+ + Cl

The NH3 was present in excess.
Similar situation occurs in H

Answer: A, B, G and H.

Example:

Which of the following does not form a buffer when added to NaHCO3(s)?

A NaOH(aq)
B HCl(aq)
C H2CO3(aq)
D H2O(l)

Solution:

  • Adding ½ of an equivalent of NaOH to a sodium bicarbonate solution converts ½ of the HCO3- to its conjugate base CO32-. The mixture of the two forms a buffer, so A is correct.
  • Adding ½ of an equivalent of HCl to a sodium bicarbonate solution converts ½ of the HCO3- to its conjugate acid H2CO3. The mixture of the two forms a buffer, so B is correct.
  • Adding an equivalent of H2CO3 to a sodium bicarbonate solution results in equal portions of the HCO3- to its conjugate acid H2CO3. The mixture of the two forms a buffer, so C is correct.
  • Adding water does not make a buffer, because water is amphoteric, so it does not convert sodium bicarbonate to either its conjugate acid or conjugate base. To be a buffer, there must be both a weak acid and its weak conjugate base present in solution at the same time.

Answer: D

Example:

If 1.0 moles of a weak acid in 1.0 dm3 of water are treated with 0.4 moles of strong base, what is the pH of the solution? (Ka for the weak acid is 2.0 x 10-4)

A pH < 3.7
B pH = 3.7
C 3.7 < pH < 7
D pH > 7

Solution:

  • If the ratio of base to acid were 1:1, the pH would be equal the pKa.
  • pKa is –lg 2.0 x 10-4 = 4 – lg 2 = 4 – 0.3 = 3.7
  • To be half-titrated, it would require 0.5 moles of strong base. At the half-titration point, the pH = pKa.
  • With only 0.4 moles of base, the halfway point is not yet reached and there is excess weak acid relative to conjugate base (0.6 moles to 0.4 moles).
  • Henderson-Hasselbalch equation, the pH is less than the pKa. Therefore, A is the best choice.
  • pH is less than pKa, which is 3.7. Be aware that this question could about the [H+] as well as the pH of the solution. If a solution is half-titrated, then the Ka = [H+].
  • This can be tricky, but the answer is found from the Henderson-Hasselbalch equation. pH = pKa + lg (0.4/0.6), so pH < pKa.

Answer: A

The next part, Part 14, will be the last part in this long notes series on Ionic Equilibrium for STPM Chemistry. Before you read the next set of notes served on Berry Berry Easy, it is advisable that you reread all prior notes in the series to further reinforce your knowledge on this topic.

{ 2 comments… read them below or add one }

April 13, 2011
April 13, 2011

Previous post:

Next post: