Aqueous Equilibria Chapter 17: Additional Aspects of Aqueous Equilibria John D. Bookstaver, St....

AqueousEquilibria

Chapter 17: Additional Aspects of Aqueous EquilibriaJohn D. Bookstaver, St. Charles Community College, St. Peters, MO,

2006, Prentice Hall, Inc.

(ppt modified for our requirements)

Chemistry, The Central Science, 10th edition AP editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

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Ch. 16 Acids-Base Equilibria

s

Ch. 17 Additional Aspects of Aqueous Equilibria

• Common Ion effect, • Buffered solutions, • Acid-Base Titrations, • Solubility Equilibria, • Factors that affect

solubility,• Precipitation and

Separation,• Qualitative Analysis for

Metallic Elements

Resources and Activities• Textbook - chapter 17 & ppt file• Online practice quiz• Lab activities• POGIL activities • Chem Guy video lecture series on

Acids-Bases (many)http://www.cosmolearning.com/video-lectures/acids-and-bases-i-properties-of-molecular-neutral-ionic-solutions/

• Chemtour videos from Norton• http://www.wwnorton.com/college/c

hemistry/gilbert2/contents/ch16/studyplan.asp

• For titrations and indicator choices:• http://www.avogadro.co.uk/chemeq

m/acidbase/titration/phcurves.htm

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Activities and Problem set for chapter 17 (due date_______)

Lab activities: Titration of Weak Acid (wet lab)

Virtual labs from

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animationsindex.htm

POGILS (5) : Introduction to Acids and Bases,

pH scale,

Acid-Base Titrations,

Weak Acid-Base Equilibria,

Buffer Solutions.

Online practice quiz ch 17 due by_____

Chapter 17 reading guide and practice problems packet

Independent work - students to view animations & interactive activities (8 in total from Norton) and write summary notes on each. These summaries are to be included in your portfolio. Some of these will be previewed in class.

Norton Animations :(Acid rain, acid-base ionization pH scale, self-ionization of water, acid strength and molecular structure, buffers, strong acid-strong base titration, titrations of weak acids)

http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch16/studyplan.asp

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Outline Common ion effect• Effect on % ionization• Effect on pH

Buffered solutions • Henderson-Hasselbalch

equation• Addition of a strong acid or a

strong base to a buffered solution

Acid-Base Titrations • Strong Acid and Strong base• Weak Acid- Strong Base• Strong Acid-Weak base• Weak Acid-Weak base• Titration of Polyprotic Acid

Solubility Equilibria

Factors that affect solubility

Precipitation and Separation

Qualitative Analysis for Metallic Elements

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The Common-Ion Effect

• Consider a solution of acetic acid:

If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

• What is the pH of a solution prepared by adding 0.30 mol of ethanoic acid (HC2H3O2) and 0.30 mol of sodium ethanoate to enough water to make a 1.0 L solution?

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)

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The Common-Ion Effect“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

What is the pH of a solution prepared by adding 0.30 mol of ethanoic acid (HC2H3O2) and 0.30 mol of sodium ethanoate to enough water to make a 1.0 L solution?

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Analyze and Plan

• Identify major species in solution & consider acidity and basicity

• Identify the impt. Equilibrium that determines pH

• Use ICE or RICE chart to tabulate concentrations

• Use equilibrium constant expression to calculate [H3O+]

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Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Ka for HF is 6.8 10−4.

[H3O+] [F−][HF]

Ka = = 6.8 10-4

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Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

[HF], M [H3O+], M [F−], M

Initially 0.20 0.10 0

Change −x +x +x

At Equilibrium 0.20 − x 0.20 0.10 + x 0.10 x

HF(aq) + H2O(l) H3O+(aq) + F−(aq)

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= 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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• 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

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Buffers:

• Solutions of a weak conjugate acid-base pair.

• They are particularly resistant to pH changes, even when strong acid or base is added.

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Buffers

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

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Buffers

If acid is added, the F− reacts to form HF and water.

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Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

[H3O+] [A−][HA]

Ka =

HA + H2O H3O+ + A−

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Buffer Calculations

Rearranging slightly, this becomes

[A−][HA]

Ka = [H3O+]

Taking the negative log of both side, we get

[A−][HA]

−log Ka = −log [H3O+] + −log

pKa

pHacid

base

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Buffer Calculations• So

pKa = pH − log[base][acid]

• Rearranging, this becomes

pH = pKa + log[base][acid]

• This is the Henderson–Hasselbalch equation.

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Henderson–Hasselbalch Equation

What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is1.4 10−4.

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Henderson–Hasselbalch Equation

pH = pKa + log[base][acid]

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

pH = 3.85 + (−0.08)

pH = 3.77

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pH Range

• The pH range is the range of pH values over which a buffer system works effectively.

• It is best to choose an acid with a pKa close to the desired pH.

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When Strong Acids or Bases Are Added to a Buffer…

…it is safe to assume that all of the strong acid or base is consumed in the reaction.

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Addition of Strong Acid or Base to a Buffer

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

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Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

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Before the reaction, since

mol HC2H3O2 = mol C2H3O2−

pH = pKa = −log (1.8 10−5) = 4.74

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

HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)

HC2H3O2 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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Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log(0.320)(0. 200)

pH = 4.74 + 0.06

pH = 4.80

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Titration

A known concentration of base (or acid) is slowly added to a solution of acid (or base).

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TitrationA pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

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Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.

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Just before and after the equivalence point, the pH increases rapidly.

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At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

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As more base is added, the increase in pH again levels off.

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Titration of a Weak Acid with a Strong Base

• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.

• The pH at the equivalence point will be >7.

• Phenolphthalein is commonly used as an indicator in these titrations.

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At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

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With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

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Titration of a Weak Base with a Strong Acid

• The pH at the equivalence point in these titrations is < 7.

• Methyl red is the indicator of choice.

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Titrations of Polyprotic Acids

In these cases there is an equivalence point for each dissociation.

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Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)

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Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42−]

where the equilibrium constant, Ksp, is called the solubility product.

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Solubility Products

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

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What is the solubility of silver chloride in g/L ?

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2

s = Ksps = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

16.6

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Factors Affecting Solubility

• The Common-Ion Effect If one of the ions in a solution equilibrium

is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.

BaSO4(s) Ba2+(aq) + SO42−(aq)

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• pH If a substance has a

basic anion, it will be more soluble in an acidic solution.

Substances with acidic cations are more soluble in basic solutions.

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Factors Affecting Solubility• Complex Ions

Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

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• Complex IonsThe formation

of these complex ions increases the solubility of these salts.

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• Amphoterism Amphoteric metal

oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

Examples of such cations are Al3+, Zn2+, and Sn2+.

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Will a Precipitate Form?

• In a solution, If Q = Ksp, the system is at equilibrium

and the solution is saturated. If Q < Ksp, more solid will dissolve until

Q = Ksp.

If Q > Ksp, the salt will precipitate until Q = Ksp.

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If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?

16.6

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

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Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.

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Qualitative Analysis of

Cations

16.11

Aqueous Equilibria Chapter 17: Additional Aspects of Aqueous Equilibria John D. Bookstaver, St....

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