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Chapter 17

Additional Aspectsof AqueousEquilibria

John D. BookstaverSt. Charles Community College

Cottleville, MO

Lecture Presentation

2012 Pearson Education, Inc.

AqueousEquilibria

The Common-Ion Effect Consider a solution of acetic acid:

If acetate ion is added to the solution,Le Chtelier says the equilibrium willshift to the left.

CH3COOH( aq ) + H 2O(l ) H3O+(aq ) + CH 3COO (aq)



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AqueousEquilibria

The Common-Ion Effect

The extent of ionization of a weakelectrolyte is decreased by adding tothe solution a strong electrolyte thathas an ion in common with the weakelectrolyte.


AqueousEquilibria

The Common-Ion EffectCalculate the fluoride ion concentration andpH of a solution that is 0.20 M in HF and 0.10 M in HCl.

K a for HF is 6.8 10 4.

[H3O+] [F ]

[HF]K a = = 6.8 10 4



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AqueousEquilibria


Because HCl, a strong acid, is also present,the initial [H 3O+] is not 0, but rather 0.10 M .

[HF], M [H3O+], M [F ], M

Initially 0.20 0.10 0

ChangeAt equilibrium

HF(aq ) + H 2O(l ) H3O+(aq ) + F (aq )


AqueousEquilibria



[HF], M [H3O+], M [F ], M


Change x +x +x

At equilibrium

HF(aq ) + H 2O(l ) H3O+(aq ) + F (aq )



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AqueousEquilibria



[HF], M [H3O+], M [F ], M


Change x +x +x At equilibrium 0.20 x 0.20 0.10 + x 0.10 x

HF(aq ) + H 2O(l ) H3O+(aq ) + F (aq )


AqueousEquilibria


= x

1.4

10 3

= x

(0.10) ( x )(0.20)6.8

10 4 =

(0.20) (6.8 10 4)(0.10)



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AqueousEquilibria


Therefore, [F ] = x = 1.4 10 3

[H3O+] = 0.10 + x = 0.10 + 1.4 10 3 = 0.10 M

So,

pH = log (0.10)pH = 1.00


2012 Pearson Education, Inc.Chemistry, The Central Science , 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Sample Exercise 17.1 Calculating the pH When a CommonIon is Involved

What is the pH of a solution made by adding 0.30 mol of aceticacid and 0.30 mol of sodium acetate to enough water to make1.0 L of solution?


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Practice ExerciseCalculate the pH of a solution containing 0.085 M nitrous acid(HNO 2,K a = 4.5 10

4) and 0.10 M potassium nitrite (KNO 2).

Answer: 3.42

Sample Exercise 17.1 Calculating the pH When a Common Ionis Involved


Sample Exercise 17.2 Calculating Ion Concentrations When aCommon Ion is Involved

Practice ExerciseCalculate the formate ion concentration and pH of a solution thatis 0.050 M in formic acid (HCOOH, K a = 1.8 10 4) and 0.10

M in HNO 3.

Answer: [HCOO ] = 9.0 10 5, pH = 1.00


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AqueousEquilibria

Buffers

Buffers are solutionsof a weak conjugateacidbase pair.

They are particularlyresistant to pHchanges, even whenstrong acid or baseis added.


AqueousEquilibria

Buffers

If a small amount of hydroxide is added to anequimolar solution of HF in NaF, for example, theHF reacts with the OH to make F and water.



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AqueousEquilibria

Buffers

Similarly, if acid is added, the F reacts with it to formHF and water.


AqueousEquilibria

Buffer CalculationsConsider the equilibrium constantexpression for the dissociation of ageneric acid, HA:

[H3O+] [A ][HA]K a =

HA + H 2O H 3O+ + A



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AqueousEquilibria

Buffer Calculations

Rearranging slightly, this becomes

[A ][HA]K a = [H3O

+]

Taking the negative log of both side, we get

[A ][HA] log K a = log [H 3O+] + log

pK a

pHacid

base


AqueousEquilibria

Buffer Calculations So

pK a = pH log[base][acid]

Rearranging, this becomes

pH = p K a + log[base][acid]

This is the HendersonHasselbalchequation .



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AqueousEquilibria

HendersonHasselbalch Equation

What is the pH of a buffer that is 0.12 M in lactic acid, CH 3CH(OH)COOH, and0.10 M in sodium lactate? K a for lacticacid is 1.4 10 4.


AqueousEquilibria

pH = 3.85 + ( 0.08)

pH = 3.77

HendersonHasselbalch Equation

pH = p K a + log[base][acid]

pH = log (1.4 10 4) + log(0.10)(0.12)



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Practice ExerciseCalculate the pH of a buffer composed of 0.12 M benzoic acid and0.20 M sodium benzoate. (Refer to Appendix D.)

Answer: 4.42

Sample Exercise 17.3 Calculating the pH of a Buffer


Sample Exercise 17.4 Preparing a Buffer

How many moles of NH 4Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that theaddition of NH 4Cl does not change the volume of the solution.)


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Sample Exercise 17.4 Preparing a Buffer

Practice ExerciseCalculate the concentration of sodium benzoate that must bepresent in a 0.20 M solution of benzoic acid(C6H5COOH) to produce a pH of 4.00.

Answer: 0.13 M

AqueousEquilibria

pH Range The pH range is the range of pH values

over which a buffer system workseffectively.

It is best to choose an acid with a p K a close to the desired pH.



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AqueousEquilibria

When Strong Acids or Bases Are

Added to a BufferWhen strong acids or bases are added to a buffer, it issafe to assume that all of the strong acid or base isconsumed in the reaction.


AqueousEquilibria

Addition of Strong Acid or Baseto a Buffer

1. Determine how the neutralizationreaction affects the amounts ofthe weak acid and its conjugatebase in solution.

2. Use the HendersonHasselbalchequation to determine the new

pH of the solution.



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AqueousEquilibria

Calculating pH Changes in Buffers

A buffer is made by adding 0.300 molHC 2H3O2 and 0.300 mol NaC 2H3O2 toenough water to make 1.00 L ofsolution. The pH of the buffer is 4.74.Calculate the pH of this solution after0.020 mol of NaOH is added.


AqueousEquilibria


Before the reaction, since

mol HC 2H3O2 = mol C 2H3O2

pH = p K a = log (1.8 10 5) = 4.74



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AqueousEquilibria


The 0.020 mol NaOH will react with 0.020 mol of theacetic acid:

HC2H3O2(aq ) + OH (aq ) C2H3O2

(aq ) + H 2O(l )

HC 2H3O2 C2H3O2 OH

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction


AqueousEquilibria

Calculating pH Changes in BuffersThe 0.020 mol NaOH will react with 0.020 mol of theacetic acid:

HC2H3O2(aq ) + OH (aq ) C2H3O2 (aq ) + H 2O(l )

HC 2H3O2 C2H3O2 OH

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction 0.280 mol 0.320 mol 0.000 mol



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AqueousEquilibria


Now use the HendersonHasselbalch equation tocalculate the new pH:

pH = 4.74 + log (0.320)(0.200)

pH = 4.74 + 0.06pH

pH = 4.80



Practice Exercise

Determine (a) the pH of the original buffer described in SampleExercise 17.5 after the addition of 0.020 mol HCl and (b) the pHof the solution that would result from the addition of 0.020 molHCl to 1.000 L of pure water.

Answers: (a) 4.68, (b) 1.70

Sample Exercise 17.5 Calculating pH Changes in Buffers


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AqueousEquilibria

GOOD LUCK!

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