Chapter 17

Buffers

Balanced mixtures of conjugate weak acid-weak base pairs called buffers. Buffer solution containing equal mole quantity of WA (weak acid) and WB (conjugate weak base) has pH = pK a of WA. pH = pK a + log [WB]/[WA] As long as [WB]/[WA] ratio stays within the range 0.10 to 10.0 (a 100-fold change!) the pH does not change by more than 1.0 pH unit from pH = pK a (of WA). Strong acids and bases cannot exist in buffers: H + + WB  H 2 O + WA OH + WA  H 2 O + WB

Buffer Capacity

Number of moles of WA and WB in a buffer related to a quantity known as buffer capacity. Generally a balanced buffer composed of equal number of moles of WA and WB (n WA = n WB ) can absorb about 81% of this quantity (ie. 0.81 n) of either strong acid or strong base before the [WB]/[WA] ratio gets far enough out of balance to change the pH by more than 1.0 unit from pK a of WA. A buffer made from 1.0 mole of WA + 1.0 mole of WB attacked by 0.811 mol of strong acid gives: n WA = 1.0 + 0.81 = 1.81 mol, n WB = 1.0 - 0.81 = 0.19 mol; [WB]/[WA] = n WB /n WA = 0.19/1.81 = 0.105

Titration

Reaction of WB (weak base) with equal number of moles of SA (strong acid) results in creation of equal number of moles of WA (weak acid) and complete destruction of original WB.

Reaction of WA with SB (strong base) has exactly complementary (opposite) effect.

Consider (draw picture): 4 WA + 2 SB  2 WA + 2 WB + 2 H 2 O (a buffer) Now: 2 WA + 2 WB + 2 SB  4 WB + 2 H 2 O (equiv. pt.)

Titration Calculations

First step is always to convert quantities given into moles.

For complete titration (to equivalence point) moles of sample solution = moles of titrating reagent.

For partial titrations of SA or SB: moles remaining = original moles - moles titration.

To convert moles remaining to pH first convert to either [H 3 O + ] or [OH ] by dividing of SA or SB moles remaining by total volume of titrated solution.

Total volume = sample volume + titration volume.

Titration

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K

SP

Calculations (Prob. 17.47)

| CaF 2 (s)  Ca 2+ (aq)  2 F (aq)

I

 -- -x 0

M

+x 0

M

+2x

E

-- +x +2x

A

-- (a) Calculate K SP if molar solubility (x) is 1.24 x 10 -3 | Ba(IO 3 ) 2 (s)  Ba 2+ (aq)  2 IO 3 (aq)

I

 -- -x 0

M

+x 0

M

+2x

E

-- +x +2x

A

-- (c) What is the molar solubility of Ba(IO 3 ) 2 if K SP = 6.0 x 10 -10 ?

Common Ion Effect

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Dissolution of Mg(OH)

2

by Acid

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