Acids – Bases EquilibriaPart V: Buffers

Dr. C. Yau

Spring 2014

Jespersen Chap. 17 Sec 7

1

What is a buffer?

A buffer solution is one that resists large changes in pH when small amounts of H+ or OH- are added.

It contains a mix of either...

a weak acid and its conjugate base in similar concentrations (not necessarily equal)

Or

a weak base and its conjugate acid in similar concentrations. 2

Example of weak acid/conjugate baseas buffer system:

HC2H3O2 + C2H3O2-

How does this work?What happens when H+ is added?What happens when OH- is added?

Example of weak base/conjugate acid as buffer system: CH3NH2 + CH3NH3

+

What happens when H+ is added?What happens when OH- is added?Do Pract Exer 32 & 33 p. 799 3

Henderson-Hasselbalch EqnWe had earlier examined this equation. You

should review the derivation yourself. Click here.

This is a very useful equation It will be provided to you:

[base] = [A-]

[acid] = [HA]

Alternatively, you can set up an ICE table.

However this eqn allows you to take shortcuts.

[base]pH = pKa + log

[acid]

4

Example 17.6 p. 799

To study the effects of a weakly acidic medium on the rate of corrosion of a metal alloy, a student prepared a buffer solution containing both 0.11 M NaC2H3O2 and 0.090 M HC2H3O2. What is the pH of the solution.

You look up Ka of acetic acid... (you will not be reminded to do so.)

Ka = 1.8x10-5

so pKa = +4.74

5

Ans. 4.83Do Pract Exer 34, 35 p. 800

6

The Common Ion EffectThis is a commonly used expression.Know what it means: The common ion effectIn the buffer system with acetic acid and sodium acetate, the acetate ion is the “common ion”.What is the effect on the equilibrium of the dissociation of the acid having acetate supplied by another source (sodium acetate)?e.g. 0.090 M acetic acid has[H+] = 1.3x10-3M, pH = 2.89With 0.11 M acetate present, pH = 4.83(Example 17.6.) Why is pH higher?[H+] much less (pH not as low.)

7

Common Ion Effect & Le Chatelier’s Principle

Using the example of the buffer system in blood...Blood has to be in the pH range of 7.35-7.42.It utilizes the buffering effect of the conjugate

acid/base pair of H2CO3/HCO3-

Write the chem equation for the ionization of H2CO3 and discuss the effect on the pH when sodium bicarbonate is added.Write chem eqn to show how the buffer system counteracts effect of added acid and base.

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Important Aspects of Buffers• pH does not change appreciably when acid

or base is added• “Buffer capacity” means amt of acid or base

that soln can absorb w/o significant change in pH. Buffer no longer works when beyond the “buffer capacity.”

• made of weak acid/conjugate base (pH<7) or weak base/conjugate acid (pH>7)

• Most effective if [HA] ~ [A-] or [B-] ~ [HB]

• Can have buffer for other than pH 7: Pick weak acid with pKa close to desired pH.

9

Examine Ka and Kb tables...Which acid would you pick to prepare a buffer for pH = 3.3? •What do you pair with the acid?•What do you use to prepare

for pH = 4.0?•pH = 8.9?...trick question.How much do we use? Before we proceed, let’s go over some shortcuts... “simplifications”...

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Simplifications in Calculations for Buffers1) In calculating pH, initial conc ~ equil conc May use initial conc in calc instead of equil conc (In ICE table, for initial [HA], at equilibrium, equil conc = [HA] - x where x is negligible.

So.. equil conc = [HA] (the initial conc) We have been doing this all along in

simplification.2) In buffers only....we can use #mol HA or A-

instead of molar conc (e.g. [HA] or [A-].Write the equil expression of dissoc of HA.Why can we use moles instead of M?

3) Dilution of buffer soln will not change pH!Write the Henderson-Hasselbalch eqn & explain why this is so.

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Example 17.7 p. 802A soln buffered at pH 5.00 is needed in an expt. Can we use acetic acid and sodium acetate to make it?If so, how many moles of NaC2H3O2 must be added to 1.0 L of soln that contains 1.0 mol HC2H3O2 to prepare the buffer?

Ans. Add 1.8 mol NaC2H3O2.Do Pract Exer 36 & 37 p. 803

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Quantitative Analysis of pH Changes When Acid is Added to Unbuffered & Buffered Solns

Consider adding 0.020 mol HCl to 250 mL of water (unbuffered). What is the pH?

Now consider 250 mL of water containing 0.12 mol NH3 and 0.095 mol NH4Cl.Why this combination of solutes?NH3 and NH4

+ is a conjugate acid-base pair.What is the pH? Think about how you approach this problem!If we add 0.020 mol HCl to this soln, what is the change in pH?

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Summary of Calc from previous slide:

Add 0.020 mol HCl to unbuffered water,pH changes from 7.00 to 1.10 (decrease of 5.90 pH units)

Add 0.020 mol HCl to buffered soln (with NH3/NH4Cl),pH changes from 9.36 to 9.20 (decrease of 0.16 pH units)

Do p.803 Example 17.8, Pract Exer 38 & 39 p.805