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Analytical Chemistry

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Buffers

A buffered solution resists changes in pH when acids or bases

are added or when dilution occurs.

The buffer is a mixture of an acid and its conjugate base or vice

versa.

While there is a weak acid equilibrium going on, it is usually so

weak that it can be ignored. Thus most of the time it is assume

that the weak acid stays in its weak acid form (HA) and theconjugate base stays in its base form (A-).

To exert significant buffering, there must be a comparable

amount of the conjugate acid and base.2

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Mixing a weak acid and its conjugate base

If A moles of a weak acid is mixed with B moles of itsconjugate base, then the moles of acids remain close to

A and the moles of base remain close to B.

For example consider an acid with pKa = 4.00 and its

conjugate base with pKb = 10.00.

Thus calculating the fraction of dissociation for the acidin a 0.10 M solution of HA.

HA H+ + A pKa = 4.00

0.10-x x x 3

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The Ka is obtained from pKa, thus Ka = 1.0x10-4.

The concentration is obtained as follows

The fraction of dissociation is obtained as follows

The acid is only 3.1% dissociated under these conditions.

In a solution containing 0.10 mol of A dissolved in

1.00L, the extent of the reactant of Awith water is muchsmaller

A + H2O HA + OH pKb = 10.00

0.10-x x x 4

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The concentration is obtained as follows

The fraction of dissociation is obtained as follows

A is only 0.0032% dissociated under these conditions.

HA dissociates very little, and adding extra A to the

solution makes HA dissociate even less.

Also A does not react much with water, and adding

extra HA makes A

react even less. 5

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Henderson-Hasselbalch equation

The central equation for buffers is the Henderson-Hasselbalch equation, which a rearranged form of the

Ka equilibrium expression

Rearranging the upper equation gives

Thus the Henderson-Hasselbalch equation for an acid is6

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If the solution is prepared from a weak base B and itconjugate acid, then the Henderson-Hasselbalch equation is

The pKa applies to the acid BH+. Where pKa is thedissociation constant of the weak acid BH+.

The important feature of the above two equations that thebase (A- or B) appears in the numerator, and the acid is in

the denominator.

Ka is for the acids in the denominator. 7

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If [HA] = [A-] then pH = pKa.

When the ratio of [A-

]/[HA] changes by a factor of 10,then the pH changes by 1 unit.

As the concentration of the base (A-) increases, then

the pH goes up.

As the concentration of the acid (HA) increases, then

the pH goes down.

For any conjugate acid-base pair, if the pH = pKa-1

then there must be 10 times as much as HA as A-. 8

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The effect of [A-]/[HA] on pH is given in the followingtable.

Example

Sodium hypochlorite (NaOCl the active ingredient ofalmost all bleaches) was dissolved in a solution

buffered to pH 6.20. Find the ratio [OCl-

]/[HOCl]? 9

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Example

Find the pH of a 1.00 L aqueous solution prepared

with 12.34 g of tris (FM 121.135) plus 4.67 g of trishydrochloride (FM 157.569)?

Example

If we add 12.0 mL of 1.00 M HCl to the solution in

the previous example, what will be the new pH?

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The pH of buffer does not change very much when a

limited amount of strong acid or base is added.

Buffer resists the change in pH because the added

strong acid or base is consumed by B or BH+.

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