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BuffersBuffers

Buffers are solutions that undergo a minimalpH change when small quantities of a strongacid or a strong base is added.

A buffer can be made in either of two ways:

A buffer can be prepared by mixing a weak acid such as benzoic acid, HC7H5O2, and a salt containing the anion of that acid such as potassium benzoate, KC7H5O2.

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A buffer can be prepared by mixing a weak base such as, NH3, and a salt containing the cation of the base such as, NH4Br.

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HCHC77HH55OO2 2 - KC- KC77HH55OO22 Buffer Buffer

Benzoic acid being a weak acid will ionize slightly as shown below:

HC7H5O2(aq) H+(aq) + C7H5O2-(aq)

Potassium benzoate being a strong electrolyte will completely dissociate as shown below:

KC7H5O2(aq) K+(aq) + C7H5O2-(aq)→

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Benzoate, C7H5O2-, being the conjugate base

of a weak acid will hydrolyze as shown below:

C7H5O2-(aq) + H2O(l)

HC7H5O2(aq) + OH-

(aq)

The key to this buffer as with all buffers liewith the two simultaneous equilibria:

HC7H5O2(aq) H+(aq) + C7H5O2-(aq)

C7H5O2-(aq) + H2O(l)

HC7H5O2(aq) + OH-

(aq)

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There are four reacting species to consider:

HC7H5O2, H+, C7H5O2-, and OH-

When a small amount of a strong acid suchas, HCl, is added to the buffer, it willcompletely ionize.

As odd as it may sound, when you add small

amounts of a strong acid to a buffer, youproduce more weak acid as shown below:

H+(aq) + C7H5O2-(aq) → HC7H5O2(aq)

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The H+ mainly comes from the HCl that wasadded to the buffer, so the original [H+] does not change much.

When a small amount of a strong base suchas, NaOH, is added to the buffer, it willcompletely dissociate.

As odd as it may sound, when you add smallamounts of a strong base to a buffer, youproduce more weak base as shown below:

HC7H5O2(aq) + OH-(aq) → C7H5O2-(aq) + H2O(l)

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The OH- mainly comes from the NaOH thatwas added to the buffer, so the original

[OH-]does not change much.

As long as both the HC7H5O2 and the C7H5O2-

are present in the buffer to react with theaddition of OH- and H+, the pH of the bufferonly changes slightly.

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NHNH3 3 - NH- NH44Br BufferBr Buffer

Ammonia being a weak base will undergoslight hydrolysis as shown below:

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Ammonium bromide being a strongelectrolyte will completely dissociate asshown below:

NH4Br(aq) → NH4+(aq) + C7H5O2

-(aq)

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Ammonium, NH4+, being the conjugate acid

of a weak base will ionize as shown below:

NH4+(aq) H+(aq) + NH3(aq)

The key to this buffer as with all buffers liewith the two simultaneous equilibria:

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

NH4+(aq) NH3(aq) + OH-(aq)

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There are four reacting species to consider:

NH4+, OH-, H+, NH3

When a small amount of a strong acid suchas, HCl, is added to the buffer, it willcompletely ionize.

As odd as it may sound, when you add small

amounts of a strong acid to a buffer, youproduce more weak acid as shown below:

H+(aq) + NH3(aq) → NH4+(aq)

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The H+ mainly comes from the HCl that wasadded to the buffer, so the original [H+]

does not change much.

When a small amount of a strong base suchas, NaOH, is added to the buffer, it willcompletely dissociate.

As odd as it may sound, when you add small

amounts of a strong base to a buffer, youproduce more weak base as shown below:

NH4+(aq) + OH-(aq) → NH3(aq) + H2O(l)

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The OH- mainly comes from the NaOH thatwas added to the buffer, so the original

[OH-]does not change much.

As long as both the NH3 and the NH4+ are

present in the buffer to react with theaddition of OH- and H+, the pH of the bufferonly changes slightly.

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A Buffer ProblemA Buffer Problem

A buffer solution contains 0.12 mol of aceticacid, HC2H3O2, and 0.10 mol of sodiumacetate, NaC2H3O2, in 1.00 L.

(a)What is the pH of this buffer? Ka = 1.8 x 10-5 V = 1.00 L

n = 0.12 mol HC2H3O2 n = 0.10 mol NaC2H3O2

[HC2H3O2] = 0.12 M [NaC2H3O2] = 0.10 M

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HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

[ ]I 0.12 0 0.10[ ]c -x +x +x[ ]e 0.12 - x x 0.10 + x

Ka =[H+] [C2H3O2

-][HC2H3O2]

1.8 x 10-5[H+]= ×0.12 - x0.10 + x

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.

1.8 x 10-5[H+]= ×0.120.10

= 2.2 x 10-5

pH = -log[H+] = -log(2.2 x 10-5) = 4.66

(b) What is the pH of the buffer after adding 0.010 mol of NaOH?

Before looking at the equilibria of the problem,you have to determine what effect 0.010 mol ofOH- has on the stoichiometry of the problem.

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nb 0.12 mol 0.010 mol 0.10 mol

HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) +

H2O(l)na 0.11 mol 0 0.11 mol

This is a limiting reactant problem where OH-

is the limiting reactant and is totallyconsumed.

The mole ratio is 1:1:1 giving the values forna which represents the number of molesremaining after the reaction.

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The previous stoichiometry of the problem is

critical and can not be ignored.

[HC2H3O2] = [C2H3O2-] = 0.11 M

The next step is to recalculate the pH with the

new molarities.

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HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

[ ]I 0.11 0 0.11[ ]c -x +x +x[ ]e 0.11 - x x 0.11 + x

Ka =[H+] [C2H3O2

-][HC2H3O2]

1.8 x 10-5[H+]= ×0.11 + x0.11 - x

≈ 1.8 x 10-5 M

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pH = -log[H+] = -log(1.8 x 10-5) = 4.74

A sanity check is in order here!

According to the calculations the pHincreased from 4.66 to 4.74.

Does this even sound reasonable?

The answer is yes because a strong base was

added to the buffer which would cause the pH

to increase.

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The buffer should become more basic (lessacidic) as a result.

Should such a small increase (ΔpH = 0.080)result?

The answer is yes because the solution is“buffered” meaning that it will resist changesin pH.

Compare this to the change in pH of a neutralsolution (pH = 7.00) after the addition of0.010 mol OH-.

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[NaOH] = 0.010 M

NaOH(aq) → Na+(aq) + OH-(aq)

pOH = -log[OH-] = -log(0.010) = 2.000

pH + pOH = 14.00

pH = 14.00 – 2.000 = 12.00

This results in a ΔpH = 5.00.

These gyrations should clarify what is meant

by a solution being buffered.

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.

(c) What is the pH of the buffer after adding 0.010 mol of HI?

Before looking at the equilibria of the problem,you have to determine what effect 0.010 mol ofH+ has on the stoichiometry of the problem.

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nb 0.010 mol 0.10 mol 0.12 mol

H+(aq) + C2H3O2-(aq) → HC2H3O2(aq)

na 0 mol 0.09 mol 0.13 mol

This is a limiting reactant problem whereH+ is the limiting reactant and is totallyconsumed.

The mole ratio is 1:1:1 giving the values forna which represents the number of molesremaining after the reaction.

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The previous stoichiometry of the problem is

critical and can not be ignored.

[HC2H3O2] = 0.13 M

[C2H3O2-] = 0.09 M

The next step is to recalculate the pH with the

new molarities.

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HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

[ ]I 0.13 0 0.09[ ]c -x +x +x[ ]e 0.13 - x x 0.09 + x

Ka =[H+] [C2H3O2

-][HC2H3O2]

1.8 x 10-5[H+]= ×0.13 - x0.09 + x

≈1.8 x 10-5[H+] ×0.130.09

≈ 2.6 x 10-5 M

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pH = -log[H+] = -log(2.6 x 10-5) = 4.59

A sanity check is in order here!

According to the calculations the pHdecreased from 4.66 to 4.59.

Does this even sound reasonable?

The answer is yes because a strong acid was

added to the buffer which would cause the pH

to decrease.

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The buffer should become less basic (moreacidic) as a result.

Should such a small increase (ΔpH = 0.070)result?

The answer is yes because the solution is“buffered” meaning that it will resist changesin pH.

Compare this to the change in pH of a neutralsolution (pH = 7.00) after the addition of0.010 mol H+.

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[HI] = 0.010 M

HI(aq) → H+(aq) + I-(aq)

pH = -log[H+] = -log(0.010) = 2.000

This results in a ΔpH = 5.00.

These gyrations should clarify what is meant

by a solution being buffered.

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Variant of a Buffer ProblemVariant of a Buffer Problem

A sample of 1.75 L of HBr gas at 17°C and0.960 atm is bubbled through 0.600 L of0.160 M NH3 solution. Assuming that all of

theHBr dissolves and the volume of the solutionremains the same, calculate the pH of thefinal solution.

V = 1.75 L V = 0.600 L NH3

T = 17°C = 290 K [NH3] = 0.150 MP = 0.960 atm

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PV = nRT

[NH3] = n/V

0.960 atm × 1.75 L

0.0821 L•atmmol•K × 290 K

n =

n = 0.071 mol HBr

n = 0.160mol NH3

L× 0.600 L

n =0.096 mol NH3

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nb 0.071 mol 0.096 mol 0 mol

H+(aq) + NH3(aq) → NH4+

(aq) na 0 mol 0.025 mol 0.071 mol

[NH4+] = n/V

[NH4+] =

0.071 mol NH4+

0.600 L= 0.12 M

[NH3] =0.025 mol NH3

0.600 L= 0.042 M

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The question which arises many times is, do I

use the NH4+ acting as an acid or the NH3

acting as a base?

The best way to find out is to do both and then

compare the pH’s.

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NH4+(aq) H+(aq) + NH3(aq)

[ ]I 0.12 0 0.042[ ]c -x +x +x[ ]e 0.12 - x x 0.042 + x

At this point in the problem you may noticesomething disturbing. You tried looking upthe Ka of NH4

+ and couldn’t find it.

Ka =[H+] [NH3]

[NH4+]

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Or just as disturbing, you are taking an examand the problem gives the Kb for NH3.

At this point, you must calmly remember that

Ka × Kb = Kw = 1.00 × 10-14

therefore,

Do not forget parenthesis in the denominator!

Ka =1.00 x 10-14

(1.8 x 10-5)= 5.6 x 10-10

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.

Ka =[H+] [NH3]

[NH4+]

5.6 x 10-10 =x (0.042 + x)

0.12 - x≈

0.042 x0.12

[H+] = 1.6 × 10-9 M

pH = -log[H+] = -log(1.6 × 10-9) = 8.80

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NH3(aq) + H2O(l) NH4+(aq) + OH-

(aq)

[ ]I 0.042 0.12 0[ ]c -x +x

+x[ ]e 0.042 - x 0.12 + x xKb =

[OH-]

[NH3]

[NH4+]

1.8 x 10-5 = (0.12 + x) x

0.042 - x≈

0.12x0.042

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[OH-] = 6.5 x 10-6 M

pOH = -log[OH-] = -log(6.5 x 10-6) = 5.19

pH + pOH = 14.00pH = 14.00 – 5.19 = 8.81

So, in the final analysis, do you use the NH4

+ acting as an acid or the NH3 acting as a

base?

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Another Variant of a Buffer Another Variant of a Buffer ProblemProblem

Calculate the concentration of potassiumbutanate that is needed in a 0.20 M solution

ofbutanoic acid to produce a pH of 4.00.

[KC4H7O2] = ? Ka = 1.5 x 10-5

[HC4H7O2] = 0.20 M pH = 4.00

[H+] = 10-pH = 10-4.00 = 1.0 x 10-4 M

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HC4H7O2(aq) H+(aq) + C4H7O2-

(aq)

[ ]I 0.20 1.0 x 10-4 0[ ]c -x +x +x[ ]e 0.20 - x 1.0 x 10-4 + x x

Ka =[C4H7O2

-]

[HC4H7O2]

[H+]

1.5 x 10-5 = (1.0 x 10-4 + x ) x

0.20 - x≈1.0 × 10-4 x

0.20

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[H+] = [C4H7O2-] = 0.030 M

[KC4H7O2] = [C4H7O2-] = 0.030 M

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More Buffer VariantsMore Buffer Variants

(a)Calculate the mole ratio of acetic acid to potassium acetate that is necessary to

form a buffer solution with a pH = 5.28. The Ka for HC2H3O2 is 1.8 x 10-5.

pH = 5.28 Ka = 1.8 x 10-5

[H+] = 10-pH = 10-5.28 = 5.2 x 10-6 M

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HC2H3O2(aq) H+(aq) + C2H3O2-

(aq)

[ ]I y 0 0

[ ]c -x +x +x[ ]e y - x x

xKa =[C2H3O2

-]

[HC2H3O2]

[H+]

Ka

=[C2H3O2

-] [HC2H3O2] [H+]

=5.2 x 10-6

1.8 x 10-5= 0.29

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(b) How is the pH of this buffer system changed upon adding 5.0 mL of distilled water?

As shown above, the mole ratio is independent of volume because within

the same solution, the volumes cancel out.

n(HC2H3O2)/Vn(C2H3O2

-)/V= 0.29

Ka=[C2H3O2

-] [HC2H3O2][H+] × =

n(HC2H3O2)/Vn(C2H3O2

-)/V

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(c) Why should the pH of a buffer be close to

the pKa of the weak acid?

Ideally, [HC2H3O2] ≈[C2H3O2

-]. This means the concentrations in the above equation cancel, leaving [H+] = Ka.

Taking the -log of both sides of the equation yields, -log [H+] = -log Ka,

which simplifies to pH = pKa.

Ka=[C2H3O2

-] [HC2H3O2][H+] ×

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(d) What is buffer capacity?

Buffer capacity is the maximum amount

of strong acid or strong base that can be

added to a buffer before a significant pH

change results.

A buffer is destroyed when all the weak

acid or weak base is consumed.