Buffer This

• There are two common kinds of buffer solutions:1 Solutions made from a weak acid plus a soluble ionic salt of the weak acid.2 Solutions made from a weak base plus a soluble ionic salt of the weak base

1. Solutions made of weak acids plus a soluble ionic salt of the weak acid• One example of this type of buffer system is:

– The weak acid - acetic acid CH3COOH

– The soluble ionic salt - sodium acetate NaCH3COO

The Common Ion Effect and Buffers

• This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of H+ and the pH of a solution that is 0.15 M in both acetic acid sodium acetate yields:

R CH3COOH H3O+ + CH3COO-

I

C

E

One example of the type I of buffer system is:The weak acid - acetic acid CH3COOH

The soluble ionic salt - sodium acetate NaCH3COO

CH3COOH H3O+ + CH3COO-

NaCH3COO →Na+ + CH3COO-~100%

Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases

Solution [H+] pH

0.15 M CH3COOH 1.6 x 10-3 2.80

0.15 M CH3COOH &

0.15 M NaCH3COO buffer

1.8 x 10-5 4.74

[H+] is ~90 times greater in pure acetic acid than in buffer solution.Note that the pH of the buffer equals the pKa of the buffering acid.

Alternatively you might have noticed:

[base]/[acid] = 1 = 100

log 100 = 0pH = pKa + 0 = pKa = 4.74[H+] = Ka = 1.8E-5

Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases

• The general expression for the ionization of a weak monoprotic acid is:

• The generalized ionization constant expression for a weak acid is:

The Common Ion Effect and Buffers

• If we solve the expression for [H+], this relationship results:

• By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to:

The Common Ion Effect and Buffers

• The relationship developed in the previous slide is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt.

• If the salt’s cation is not univalent the relationship changes to:

The Common Ion Effect and Buffers

• Simple rearrangement of this equation and application of algebra yields the

Henderson-Hasselbach equation.

The Henderson-Hasselbach equation is one method to calculate the pHof a buffer given the concentrations of the salt and acid.

The Common Ion Effect and Buffers

• The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. • A special case exists for the Henderson-Hasselbalch equation

when [base]/[acid] = some power of 10, regardless of the actual concentrations of the acid and base, where the Henderson-Hasselbalch equation can be interpreted without the need for calculations:

[base]/[acid] = 10x

log 10x = xin general pH = pKa + x

Examples: 1. when [base] = [acid], [base]/[acid] = 1 or 100, log 1 = 0, pH = pKa, (corresponds to the midpoint in the titration of a weak acid or base)2. when [base]/[acid] = 10 or 101, log 10 = 1 then pH = pKa + 13. when [base]/[acid] = .001 or 10-2, log 10 = -2 then pH = pKa -2

• Henderson-Hasselbalch equation is valid for solutions whose concentrations are at least 100 times greater than the value of their Ka’s

Henderson-Hasselbalch - Caveats and Advantages

Buffer SolutionsThere are two common kinds of buffer solutions:

I. Commonly, solutions made from a weak acid plus a soluble ionic salt of the conjugate base of the weak acid.

II. Less common, solutions made from a weak base plus a soluble ionic salt of the conjugate acid of the weak base.

Both of the above may also be prepared by starting with a weak acid (or weak base) and add half as many moles of strong base (acid)

• This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of OH- and the pOH of the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium nitrate, NH4NO3 yeilds:

R NH3 NH4+ + OH-

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C

E

NH4NO3 → NH4 ++ NO3 -

~100%

NH3 NH4 ++ OH -

Buffer Solutions: Weak Bases Plus Salts of Their Conjugate Acids

One example of the type II of buffer system is:The weak base – ammonia NH3

The soluble ionic salt – ammonium nitrate NH4NO3

• Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

• Simple rearrangement of this equation and application of algebra yields the

Henderson-Hasselbach equation.

The Common Ion Effect and Buffers• We can derive a general relationship for buffer solutions that contain a

weak base plus a salt of a weak base similar to the acid buffer relationship.– The general ionization equation for weak bases is:

• A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect.

Solution [OH-] pH

0.15 M NH3 1.6 x 10-3 M 11.20

0.15 M NH3 &

0.15 M NH4NO3 buffer

9.0 x 10-6 M 8.95

The [OH-] in aqueous ammonia is 180 times greater than in the buffer.

The Common Ion Effect and Buffers

Buffering Action• If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is

0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl.

1 Calculate the pH of the original buffer solution.

• Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

R NH3 NH4+ + OH-

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C

E

NH4NO3 → NH4 ++ NO3 -

~100%

NH3 NH4 ++ OH -

NH3 + H+ NH4 +

Buffering Action2 Next, calculate the concentration of all species after the addition of

the gaseous HCl.– The HCl will react with some of the ammonia and change the

concentrations of the species.– This is another limiting reactant problem.

R NH3 NH4+ + OH-

I

C

E

HCl → H++ Cl -~100%

Buffering Action3 Using the concentrations of the salt and base and the Henderson-

Hassselbach equation, the pH can be calculated.

4 Finally, calculate the change in pH.

Buffering Action1. If 0.020 mole of NaOH is added to 1.00 liter of

solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH.NH4

+ + OH- NH3

R NH3 NH4+ + OH-

I

C

E

NaOH → Na++ OH -

~100%

Buffering Action

3. Finally, calculate the change in pH.

2. Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated.

Buffering Action

• Notice that the pH changes only slightly in each case.

Original SolutionOriginal

pHAcid or base

addedNew pH pH

1.00 L of solution containing 0.100 M NH3 and 0.200 M NH4Cl

8.95

0.020 mol NaOH 9.08 +0.13

0.020 mol HCl 8.81 -0.14

Preparation of Buffer Solutions• Calculate the concentration of H+ and the pH of the solution prepared by

mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions.

• Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction.

• NaOH and CH3COOH react in a 1:1 mole ratio.

• After the two solutions are mixed, Calculate total volume.

• The concentrations of the acid and base are:

• Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.

Preparation of Buffer Solutions• For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Starting with a solution that is 0.100M in aqueous

ammonia prepare 1.00L of a buffer solution that has a pH of 9.15 using ammonium chloride as the source of the soluble ionic salt of the conjugate weak acid.

• The Henderson-Hasselbalch equation is used to determine the ratio of the conjugate acid base pair

• pOH can be determined from the pH:• pKb can be looked up in a table:• [base] concentration is provided:• Solve for [acid]:

• Does this result make sense?

NH4Cl NH4 + + Cl- ~100%

NH3 NH4++ OH-

H2O