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pK K =−log ( ) b () log K - butane.chem.uiuc.edubutane.chem.uiuc.edu/.../Worksheets/Worksheet21_Buffers_Key.pdf · 6. What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5

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Page 1: pK K =−log ( ) b () log K - butane.chem.uiuc.edubutane.chem.uiuc.edu/.../Worksheets/Worksheet21_Buffers_Key.pdf · 6. What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5

Worksheet 21 - Buffers

Until now, we’ve been primarily concerned with calculating the pH of a solution in which we initially added only one acid or base. Now we will consider solutions that initially contain both a weak acid (HA) and its conjugate base (A-). Suppose a solution contains both HF and NaF. What are the major species in solution? We know that HF is a weak acid (Ka = 7.2x10-4), so the major species resulting from HF are HF and H2O. NaF is an ionic compound, so it dissociates in water to form Na+ and F- (and of course H2O is still a major species in this solution). Altogether, the major species are HF, Na+, F- and H2O.

HF + H2O F- + H3O+ The weak acid produces some F- and H3O+ in solution. By adding NaF, we’re introducing another source of F-. LeChatelier’s principle indicates that the equilibrium should shift to the left upon addition of a product. This suggests that the [H3O+] will decrease, and the pH will increase. A solution containing both a weak acid and its conjugate base should be less acidic than a solution containing only a weak acid. A buffer solution is one which maintains an approximately constant pH when small amounts of either a strong acid or base are added. Solutions containing either weak acids and their conjugate bases, or weak bases and their conjugate acids, can be buffering solutions. Here are some important facts about buffer solutions:

• The number of moles of the conjugate pairs must be large compared to the moles of added strong acids or bases

• The ratio of the conjugate pair ([weak acid]/[conjugate base] or [weak base] / [conjugate acid]) should lie between 0.1 and 10, with optimal buffering at a 1:1 ratio

• The pH of a 1:1 ratio buffer is equal to the pKa of the weak acid or pKb of the weak base. The effective range is ± 1 from the pKa or pKb.

The Henderson-Hasselbalch equation is particularly useful for calculating the pH of buffer solutions:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

]acid[]base[logapKpH

( )( )bb

aa

KpKKpK

loglog

−=−=

Page 2: pK K =−log ( ) b () log K - butane.chem.uiuc.edubutane.chem.uiuc.edu/.../Worksheets/Worksheet21_Buffers_Key.pdf · 6. What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5

Here are some Ka and Kb values for various weak acids and bases (shown with their conjugates):

CH3COOH/CH3COO- Ka = 1.8 x 10-5 NH3/NH4

+ Kb = 1.8 x 10-5

HCO3-/CO3

2- Ka = 5.6 x 10-11 H2PO4

-/HPO42- Ka = 6.2 x 10-8

1. Which of the conjugate pairs, shown above would be appropriate for preparing a

buffer at pH = 7.0? Find pKa of each acid and pKb of each base. The one closest to 7 represents the most appropriate buffer

CH3COOH pKa = -log(1.8x10-5) = 4.74 NH3 pKb = -log(1.8x10-5) = 4.74 pKa = 14 – 4.74 = 9.26 HCO3

- pKa = -log(5.6x10-11) = 10.25 H2PO4

- pKa = -log(6.2x10-8) = 7.21 best buffer at pH = 7 2. What is the pH of a buffer formed from 50 mL of 15.0 M NH3 and 53.5 g of NH4Cl

in enough water to make 500 mL of solution? Note: It is often easier to consider buffer ICE tables in terms of moles rather than concentrations – in the Henderson-Hasselbalch equation, the volume will cancel out anyway

+

++ =××=

=×=

4

3

NH mol 1.00

NH mol 0.750

ClNH mol 1NH mol 1

ClNH 5.53ClNH mol 1

ClNH 5.53NH mol

L 1NH mol 15.0

L 050.0 mol

4

4

4

444

33

gg

NH

14.900.1750.0log26.9

]acid[]base[log =⎟

⎠⎞

⎜⎝⎛+=⎟⎟

⎞⎜⎜⎝

⎛+= apKpH

3. What is the pH of the previous solution after addition of 100 mL of 0.2 M NaOH?

(Hint: Begin by considering the major species present before addition of NaOH. With what (if anything) will the NaOH react?)

OH- is a strong base and will react completely with NH4

+ (the weak acid) NH4

+ + OH- NH3 + H2O

mol 020.0L 1 mol 0.2L 100.0 mol =×=

−− OHOH

moles NH4+ OH- NH3

Initial 1.00 0.020 0.750 Change -0.020 -0.020 +0.020 Equil. 0.980 0 0.770

16.9980.0770.0log26.9

]acid[]base[log =⎟

⎠⎞

⎜⎝⎛+=⎟⎟

⎞⎜⎜⎝

⎛+= apKpH

Page 3: pK K =−log ( ) b () log K - butane.chem.uiuc.edubutane.chem.uiuc.edu/.../Worksheets/Worksheet21_Buffers_Key.pdf · 6. What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5

Buffers can also be made by partially neutralizing a weak acid with a strong base, or a weak base with a strong acid.

4. Which of the following mixtures result in a buffering solution when equal volumes

of the two solutions are mixed?

a) 0.1 M HCl and 0.1 M NH4Cl Strong acid + weak acid no reaction; no buffer b) 0.1 M HCl and 0.1 M NH3 H+ + NH3 NH4

+ In equal proportions, the strong acid and weak base will react completely to form the conjugate weak acid; no buffer c) 0.2 M HCl and 0.1 M NH3 Excess HCl dominates this reaction, which will proceed until the NH3 is used up completely; no buffer d) 0.1 M HCl and 0.2 M NH3 Excess NH3 means that some (0.1 M) NH3 will be left after all the HCl is consumed. 0.1 M NH4

+ will be produced, so both weak acid and weak base are present in the final solution; buffer

5. What is the pH of a solution made by adding 0.40 g of NaOH to 100 ml of a 0.4 M

solution of CH3COOH? Is it a buffering solution?

COOHCH

L1COOHCH mol 0.4

L 100.0COOHCH mol

OH mol 010.0NaOH mol 1OH mol 1

NaOH 40NaOH mol 1NaOH 40.0OH mol

33

3

--

-

mol 0.040=×=

=××=g

g

moles CH3COOH OH- CH3COO-

Initial 0.040 0.010 0 Change -0.010 -0.010 +0.010 Equil. 0.030 0 0.010

26.4030.0010.0log74.4

]acid[]base[log =⎟

⎠⎞

⎜⎝⎛+=⎟⎟

⎞⎜⎜⎝

⎛+= apKpH

This is a buffer solution because both the acid and its conjugate base are present in approximately equal amounts. The pH is relatively close to the pKa value.

Page 4: pK K =−log ( ) b () log K - butane.chem.uiuc.edubutane.chem.uiuc.edu/.../Worksheets/Worksheet21_Buffers_Key.pdf · 6. What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5

6. What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5 M CHOOH (Ka = 1.8 x 10-4) to prepare a buffer with a pH of 4.0? This problem looks at buffer solutions in the reverse order. We’ll start with the pH and work through the problem in reverse to find the initial amount of NaOH

( ) ( ) 74.3108.1loglog 4 =×−=−= −aa KpK

CHOOHL1CHOOH mol 2.5L 100.0CHOOH mol mol 0.25=×=

The initial reaction has the form CHOOH + OH- CHOO- + H2O moles CHOOH OH- CHOO-

Initial 0.25 x 0 Change -x -x +x Equil. 0.25-x 0 x

( )

moles 16.0x82.2455.082.1455.0

25.082.125.0

x10

25.0xlog74.30.4

]acid[]base[log

26.0

==

=−=−−

=

⎟⎠⎞

⎜⎝⎛

−+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

xxxxx

x

x

pKpH a

Notice that the amount (moles) of CHOO- (the conjugate base) present at equilibrium is equal to the initial amount of OH- 0.16 mol OH- = 0.16 mol NaOH

mL 81L 0.081 soln NaOH volumesoln NaOH (L) volume

NaOH mol 0.16NaOH 2.0M

==

=

Caution: The 5% rule will not work for this problem because a significant amount of CHOOH reacts with the OH- to form the conjugate base.