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Common Ion Effect and Buffers

Ch. 17 in Brown LeMay

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A buffer is a substance or combination of substances capable of neutralizing limited quantities of either acids or bases.

Why is the pH of some lakes unaffected by acid rain even when they are downwind of big polluters?

The lakes are surrounded by soils which neutralize the acidic precipitation.

1. One way to neutralize acid is to add a base. Limestone (CaCO3) is a weak base.

2. Another way to neutralize either an acid or a base is to add a buffer.

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How Buffers Work A buffer has two components: A weak acid  AND a soluble salt of that acid:

HA <-> H+ + A- and LiA Li+ + A-

 Any extra H3O+ will be neutralized by the A- in the buffer.                    H3O+ + A-   <-> HA + H2O

 Any extra OH- that will be neutralized by the

acid.                         HA + OH-   <->  A- + H2O

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Buffer Solutions Resist change in pH Often made with a weak acid (or base) and

its salt Problems are worked with equilibrium

expression and RICE table. The weak acid keeps addition of a base

from changing H+ concentration. The conjugate base ensures that adding an

acid will keep original [H+] close to original--instead of increasing [H+] the added base reacts with the conjugate acid.

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What Mixtures Can Be Buffers?

A chemical buffer consists of a solution of a conjugate acid-base pair with both species present in solution at concentrations to accomplish the neutralization of added species in order to maintain pH of the solution at its original level.

Most buffers are a combination of a weak acid or base and a salt made with its conjugate: Examples are: HCOOH and NaCOOH NH3 and NH4Cl

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Can a buffer be made with a strong acid or strong base?

Yes, but it is difficult. Since the presence of both the acid and the base is

necessary for the buffer is to be able to deal with both addition and subtraction of hydrogen ions, a buffer could be made using a strong acid and salt in sufficient concentration for its conjugate to produce OH- ions to neutralize added acid.

HI + NaI in water becomes H+ + I- + Na+ + I- And--I- + H2O HI + OH-

Although this would be an acidic buffer, small amounts of H+ could be added without changing the original pH.

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BuffersThe lake: CaCO3 + H+ Ca 2+ + HCO3

- + H2O

The polyatomic ion HCO3- acts as a buffer:

HCO3- + H+ 2H2O + CO2

HCO3- + OH- H2O + CO3

2-

Another buffer is the hydrogen phosphate ion. Possible sources of this ion are Na2HPO4 and NaH2PO4.

H2PO4 -

HPO4 2-

+ H+

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How Buffers in the Blood Work

Blood must be maintained at pH = 7.35. This pH is maintained by a buffer system based on H2CO3. The blood buffer is made up from the dissolved carbon

dioxide in the plasma.

CO2 (g) + H2O     H2CO3 and H2CO3    HCO3- + H3O+

 When a base is added it reacts with the carbonic acid.

          OH- + H3CO3   HCO3- + H2O

 When an acid is added it reacts with the bicarbonate ion.

          H3O+ + HCO3-      H2CO3 + H2O

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Buffered solutions contain relatively large portions of weak acid and the corresponding weak base.

Can be HA and A- or B and BH+

When H+ added to buffer, it reacts ~ to completion with the weak base:

H+ + A- HA or H+ + B BH+

When OH- added to buffer, it reacts ~ to completion with the weak acid:

H+ + A- HA or H+ + B BH+

pH in buffered solution is determined by the ratio of the concentrations of weak acid/weak base. pH will be ~ constant as long as:

1. Buffer concentrations are large compared to concentrations of added H+ or OH-.2. The ratio of weak acid/weak base remains constant.

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Common Ion Effect/Buffer solution:Find pH of 0.50M HC2H3O2 and 0.50M NaC2H3O2

in solution. Ka acetic acid = 1.8 x 10-5

R HC2H3O2 H+ + C2H3O2 -

I 0.50 M 0 0.50 M

C -x +x +x

E 0.50 – x x 0.50 + x

Major species in solution: HC2H3O2, Na+, C2H3O2-, H2O

Dissociation of HC2H3O2 will control pH.

1.8 x 10-5 = Ka = x(.50 + x) = x(.50) =1.8 x 10-5 (.50 – x) (.50)

x = [H+] = 1.8 x 10-5 M (< 5% of 0.50 M)pH = -log[1.8 x 10-5] = 4.74

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Henderson-Hasselbalch Equation

H+ + A- HAKa = [H+][A-] [H+] = [HA](Ka)

[HA] [A-] -log[H+] = -logKa – log([HA]/[A-]

pH = pKa + log([A-]/[HA]

pH = pKa + log([base]/[acid] ]

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Practice problems

1. What is the pH of a solution after 0.1 mole of HCl is added to a buffer which contains 0.2 mole of HF and .3 mole of NaF in one liter?

2.Do problem 1 but change the HCl to 0.4 mole.

3.Do problem 1 but change HCl to 0.3 mole.

(1) 2.97, (2) 1, (3) 1.7