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AqueousEquilibria

Chapter 15

Applications of Aqueous Equilibria

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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.

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

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“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.”

Add more of the same ion and there will be less ions of the weak one.

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• The addition of concentrated HCl to a saturated solution of NaCl will cause some solid NaCl to precipitate out of solution. The NaCl has become less soluble because the addition of additional chloride ion.

• NaCl + HCl → more NaCl due to increased Cl-

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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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• The addition of a common ion to a weak acid solution makes the solution LESS acidic.

HC2H3O2(aq) ↔ H+(aq) + C2H3O2

-(aq)

• If NaC2H3O2 is added to the system, the equilibrium shifts to undissociated HC2H3O2 raising the pH. The new pH can be calculated by putting the concentration of the anion into the Ka equation and solving for the new [H+].

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• Adding NaF to a solution of HF causes more HF to be produced. Major species are HF, Na+, F-, and H2O. Common ion is F-.

• In a 1.0M NaF and 1.0M HF solution, there is more HF in the presence of NaF.

HF(aq) ↔ H+(aq) + F-

(aq)

• Le Chatelier’s indicates that additional F- due to the NaF causes a shift to the left and thus generates more HF.

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Finding the pH

1. Always determine the major species.

2. Write the equilibrium equation and expression.

3. Determine the initial concentrations.

4. Do ICE chart and solve for x.

5. Once [H+] has been found, find pH.

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PRACTICE ONE

The equilibrium concentration of H+ in a 1.0M HF solution is 2.7 x 10-2M and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka = 7.2 x 10-4) and 1.0M NaF.

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Notice:

1.0M HF % dissociation 2.7%

versus

1.0M HF and 1.0M NaF % dissociation 0.072%

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The Common-Ion Effect

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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EQUATIONS QUIZ

• For each of the following reactions, write an equation for the reaction. Write the net ionic equation for each. Omit formulas for spectator ions or molecules in the reaction. Put a box around your final answer.

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DO NOW

• Pick handout due tomorrow.• Turn in lab – make sure the you have:

One Title page, one Prelab, one Data Table, Calculations Table for everyone, and Calculations 1-8 for everyone in that order!

• Get out notes.

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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.

• Just a case of the common ion effect.

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BUFFERS

• You are always adding a strong acid or strong base to a buffer solution.

• Buffer system (conjugate acid-base pair) acts as a net – “catches” acid or base

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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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BUFFERS

Example:

HC2H3O2 / C2H3O2- buffer system (the “net”)

• Add a strong acid:

H+ + C2H3O2- → HC2H3O2 forms a weak acid

• Add a strong base:

OH- + HC2H3O2 → C2H3O2- + H2O

forms a weak base

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BUFFERS

Example:

NH3 / NH4+ buffer system (the “net”)

• Add a strong acid: H+ + NH3 → NH4+

forms a weak acid

• Add a strong base:

OH- + NH4+ → NH3 + H2O

forms a weak base

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PRACTICE TWO

• A buffered solution contains 0.050M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Calculate the pH of the solution.

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HELPFUL TIPS• Buffered solutions are simply solutions of weak

acids and bases containing a common ion.• The pH calculations for buffered solutions

require exactly the same procedures as determining the pH of weak acid or weak base solutions learned previously.

• When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations.

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BUFFERS

• Adding a strong acid or base to a buffered solution

Requires moles

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PRACTICE THREECalculate the change in pH that occurs when 0.010mol solid NaOH is added to 1.0L of the buffered solution contains 0.050M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Compare this pH change with that which occurs when 0.010mol solid NaOH is added to 1.0L of water.

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HOW BUFFERING WORKS

• When hydroxide ions are added to the solution, the weak acid provides the source of protons. The OH- ions are not allowed to accumulate, but are replaced by A-.

OH- + HA → A- + H2O

• pH can be understood by looking at the EQ expression.

Ka = or [H+] = Ka

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HOW BUFFERING WORKS

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HOW BUFFERING WORKS

• So [H+] (and thus pH) is determined by the ratio of [HA]/[A-]. When OH- ions are added, HA converts to A- and the ratio decreases. However, is the amounts of [HA] and [A-] are LARGE, then the change in the ratio will be small.

• If [HA]/[A-] = 0.50M / 0.50M = 1.0M initially. After adding 0.010M OH-, it becomes [[HA]/[A-] = 0.49M / 0.51M = 0.96M. Not much of a change. [H+] and pH are essentially constant.

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HOW BUFFERING WORKS

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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 sides, 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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BUFFER CALCULATIONS

• For a particular buffering system, all solutions that have the same ratio of [A-] /[HA] have the same pH.

• Optimum buffering occurs when [HA] = [A-] and the pKa of the weak acid used should be as close to possible to the desired pH of the buffer system.

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HENDERSON-HASSELBACH

• The equation needs to be used cautiously. • It is sometimes used as a quick, easy

equation in which to plug in numbers. • A Ka or Kb problem requires a greater

understanding of the factors involved and can ALWAYS be used instead of the HH equation.

• However, at the halfway point (as in a titration), the HH is very useful.

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PRACTICE FOUR

What is the pH of a buffer that is 0.75 M lactic acid, HC3H5O3, and 0.25 M in sodium lactate? Ka for lactic acid is 1.4 10−4.

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HENDERSON–HASSELBALCH EQUATION

pH = pKa + log[base][acid]

pH = −log (1.4 10−4) + log(0.25)(0.75)

pH = 3.85 + (−.477)

pH = 3.37

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HINTS

1. Determine the major species involved.

2. If a chemical reaction occurs, write the equation and solve stoichiometry.

3. Write the EQ equation.

4. Set up the equilibrium expression (Ka or Kb) of the HH equation.

5. Solve.

6. Check the logic of the answer.

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PRACTICE FIVE

A buffered solution contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl. Calculate the pH of this solution.

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PRACTICE SIX

Calculate the pH of the solution that results when 0.10mol gaseous HCl is added to 1.0Lof the buffered solution of contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl.

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BUFFERING CAPACITY

• This is the amount of acid or base that can be absorbed by a buffer system without a significant change in pH.

• In order to have a large buffer capacity, a solution should have large concentrations of both buffer components.

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PRACTICE SEVENCalculate the change in pH that occurs when 0.010mol gaseous HCl is added to 1.0L of each of the following solutions (Ka for acetic acid = 1.8 x 10-5):

• Solution A: 5.00M HC2H3O2 and 5.00M NaC2H 3O2

• Solution B: 0.050M HC2H3O2 and 0.050M NaC2H 3O2

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HINT

• We see that the pH of a buffered solution depends on the ratio of the [base] to [acid] (or [acid] to [base]).

• Big concentration difference = large pH change

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PRACTICE EIGHTA chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt:

a. chloroacetic acid Ka = 1.35 × 10-3

b. propanoic acid Ka = 1.3 × 10-5

c. benzoic acid Ka = 6.4 × 10-5

d. hypochlorus acid Ka = 3.5 × 10-8

Calculate the ratio of [HA] / [A-] required for each system to yield a pH of 4.30. Which system works best?

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TITRATIONS and pH CURVES

• Only when the acid AND base are both strong is the pH at the equivalence point 7.

• Any other conditions and you get to do an equilibrium problem. It is really a stoichiometry problem with a limiting reactant. The “excess” is responsible for the pH

• Weak acid + strong base equivalence pt. > pH 7

• Strong acid + weak base equivalence pt. < pH 7

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END PT VS EQUIVALENCE PT

• There is a distinction between the equivalence point and the end point.

• The end point is when the indicator changes color.

• If you’ve made a careful choice of indicators, the equivalence point, when the number of moles of acid = number of moles of base, will be achieved at the same time.

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VOCABULARY

• Titrant – solution of know concentration (usually in the buret). The titrant is added to a solution of unknown concentration until the substance being analyzed is just consumes (stoichiometric point or equivalence point).

• Titration or pH Curve – plot of pH as a function of the amount of titrant added.

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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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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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STRONG ACID-STRONG BASE

H+(aq) + OH-

(aq) → H2O(l)

• To compute H+, we have to know how much H+ remains at that point in the titration.

• New unit: millimole, mmol – titrations usually involve small quantities.

• This means Molarity = = . • So a 1.0M = =

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Example - For the titration of 50.0 mL of 0.200M HNO3 with 0.100M NaOH, calculate the pH of the solution at the following selected points of the titration: A. 0.0 mL of 0.100M NaOH has been added.

B. 10.0 mL of 0.100M NaOH has been added.

C. 20.0 mL of 0.100M NaOH has been added.

D. 50.0 mL of 0.100M NaOH has been added.

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a. 0.0 mL of 0.100M NaOH has been added to 0.200M HNO3.

• Major Species: H+, NO3-, H2O

HNO3 = strong acid

• pH = -log[H+] = -log (0.200) = 0.699

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b. 10.0 mL of 0.100M NaOH has been added.• Major Species: H+, NO3

-, H2O, Na+, OH-

• 1.00mmol (10.0mL x 0.100M) OH- reacts with 1.00mmol H+.

H+ + OH- → H2O

• Before: H+ = 10.0mmol; OH- = 1.00mmol• After: H+ = 10.0mmol – 1.00mmol = 9.0mmol;

OH- = 1.00mmol – 1.00mmol = 0.00mmol• So after, Major Species: H+, NO3

-, H2O, Na+. Determine pH with H+ remaining.

• pH = -log[H+] = -log () = -log[0.15] = 0.82

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c. 20.0 mL (total as opposed to additional) of 0.100M NaOH has been added.

• Major Species: H+, NO3-, H2O, Na+, OH-

• 2.00mmol (20.0mL x 0.100M) OH- reacts with 2.00mmol H+. H+ + OH- → H2O


OH- = 2.00mmol – 2.00mmol = 0.00mmol

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• So after, Major Species: H+, NO3-, H2O,

Na+. Determine pH with H+ remaining.• pH = -log[H+] = -log () = -log[0.11] =

0.94

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d. 50.0 mL (total) of 0.100M NaOH has been added.




OH- = 5.00mmol – 5.00mmol = 0.00mmol

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Na+. Determine pH with H+ remaining.• pH = -log[H+] = -log () = -log[0.050] =

1.30

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e. 100.0 mL (total) of 0.100M NaOH has been added.




OH- = 10.0mmol – 10.0mmol = 0.0mmol

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• So after, Major Species: NO3-, H2O,

Na+. • This is the EQUIVALENCE POINT (or

stoichiometric point).• pH = 7.00, neutral

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f. 150.0 mL (total) of 0.100M NaOH has been added.




OH- = 15.0mmol – 10.0mmol = 5.00mmol

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Na+. Determine pH with H+ remaining.• pOH = -log[OH-] = -log () = -log[0.025] =

1.60, so pH = 12.40

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g. 200.0 mL (total) of 0.100M NaOH has been added.




OH- = 20.0mmol – 10.0mmol = 10.0mmol

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Na+. Determine pH with H+ remaining.• pOH = -log[OH-] = -log () = -log[0.040] =

1.40, so pH = 12.60.

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The results of a-g are plotted. The pH changes gradually until the titration is close to the equivalence point when there is a dramatic change. Why is this?

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Characteristics:• At equivalence point, pH = 7.• Before the equivalence point, [H+] and

thus pH can be calculated by dividing mmol H+ by total volume of solution.

• After the equivalence point, [OH-] and thus pOH and then pH can be calculated by dividing mmol OH- by total volume of solution.

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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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WEAK ACID – STRONG BASE

• We have to do a series of buffer problems like we did earlier.

• Remember that although the acid is weak, it reacts to completion with the OH- ion, a very strong base.

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• A two-step procedure:A stoichiometry problem: The rxn of OH_

with the weak acid is assumed to run to completion, and concentrations of the acid remaining and conjugate base formed are determined.

An equilibrium problem: The position of the weak acid equilibrium is determined and pH is calculated.

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• At halfway to the equivalence point, pH = pKa.

• At the equivalence point, a basic salt is present and pH > 7

• After the equivalence point, the strong base will be the dominant species and a simple pH calculation can be done after stoichiometry.

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Example - For the titration of 50.0 mL of 0.10M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10M NaOH, calculate the pH of the solution at the following selected points of the titration:

a. 0.0 mL of 0.10M NaOH has been added.

b. 10.0 mL of 0.10M NaOH has been added.

c. 25.0 mL (total as opposed to additional) of 0.100M NaOH has been added.



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a. 0.0 mL of 0.10M NaOH has been added. • Major Species: HC2H3O2, H2O

• HC2H3O2 → C2H3O2

- + H+

• Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

HC2H3O2 → C2H3O2

- + H+

I 0.10 0 0

C -x +x +x

E 0.10-x x x

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Ka = 1.8 x 10-5 = = x2 / 0.10

x2 = 1.8 x 10-6 x = 0.0013

pH = -log[H+] = -log (0.0013) = 2.89

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b. 10.0 mL of 0.10M NaOH has been added.• Major Species: HC2H3O2, H2O, Na+, OH-

• HC2H3O2 + OH- → C2H3O2

- + H2O

• Before: OH- = 1.0mmol; HC2H3O2 = 5.0mmol; C2H3O2

- = 0.0mmol

• After: OH- = 1.0mmol – 1.0mmol = 0.0mmol; HC2H3O2 = 5.0mmol – 1.0mmol = 4.0mmol; C2H3O2

- = 0.0mmol + 1.0mmol = 1.0mmol formed.

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So after, Major Species: HC2H3O2, C2H3O2-,

H2O, Na+.

Determine pH with HC2H3O2 equilibrium

[Initial]: HC2H3O2 =

[Initial]: C2H3O2- =

[Initial]: H+ ≈ 0

Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

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HC2H3O2 → C2H3O2

- + H+

I 0.067 0.017 0

C -x +x +x

E 0.067 - x 0.017 + x x• Ka = 1.8 x 10-5 = = x(0.017)/0.067 = 0.25x

• x = 7.2 x 10-5

• pH = -log[H+] = -log[7.2 x 10-5] = 4.14

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Henderson Hasselbach Equation

pH = pKa + log

= 4.74 + -0.60

= 4.14

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c. 25.0 mL (total as opposed to additional) of 0.100M NaOH has been added.• Major Species: HC2H3O2, H2O, Na+, OH-

• HC2H3O2 + OH- → C2H3O2

- + H2O

• Before: OH- = 2.5mmol;

HC2H3O2 = 5.0mmol; C2H3O2- = 0.0mmol



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So after, Major Species: HC2H3O2, C2H3O2-,

H2O, Na+.

Determine pH with HC2H3O2 equilibrium



[Initial]: H+ ≈ 0

Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

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HC2H3O2 → C2H3O2

- + H+

I 0.033 0.033 0

C -x +x +x

E 0.033 - x 0.033 + x xKa = 1.8 x 10-5 = = x(0.033)/0.033 = x

x = 1.8 x 10-5

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

This is halfway to the equivalence point. Half of the HC2H3O2 has been converted. [HC2H3O2]0 = [C2H3O2

-]0. Ka = [H+] and pH = pKa.

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pH = pKa + log

= 4.74 + 0.0

= 4.74

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• Major Species: HC2H3O2, H2O, Na+, OH-

• HC2H3O2 + OH- → C2H3O2

- + H2O


- = 0.0mmol



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So after, Major Species: HC2H3O2, C2H3O2-, H2O,

Na+.



[Initial]: H+ ≈ 0

Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5

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HC2H3O2 → C2H3O2

- + H+

I 0.011 0.044 0

C -x +x +x

E 0.011 - x 0.044 + x x• Ka = 1.8 x 10-5 = = x(0.044)/0.011 = 4.0x

• x = 4.5 x 10-6

• pH = -log[H+] = -log[4.5 x 10-6] = 5.35

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pH = pKa + log

= 4.74 + 0.60

= 5.34

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• Major Species: HC2H3O2, H2O, Na+, OH-

• HC2H3O2 + OH- → C2H3O2

- + H2O


- = 0.0mmol


- = 0.0mmol + 5.0mmol = 5.0mmol formed. EQUIVALENCE POINT!

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So after, Major Species: C2H3O2-, H2O, Na+.

[Initial]: HC2H3O2 = 0


[Initial]: OH- ≈ 0

Kb = 1.0 x 10-14/1.8 x10-5 = 5.6 x 10-10

Kb = [C2H3O2-][H+] / [HC2H3O2] = 5.6 x 10-10

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C2H3O2- → HC2H3O2

+ OH-

I 0.050 0 0

C -x +x +x

E 0.050 - x x x

Kb = 5.6 x 10-10 = = x2 / 0.050

x2 = 2.8 x 10-11 x = 5.3 x 10-6

pOH = -log[OH-] = -log[5.3 x 10-6] = 5.28,

so pH = 8.72

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f. 60.0 mL (total) of 0.100M NaOH has been added.

• Major Species: HC2H3O2, C2H3O2-, H2O, Na+, OH-

• HC2H3O2 + OH- → C2H3O2

- + H2O


- = 0.0mmol



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After, Major Species: C2H3O2-, H2O, Na+, OH-.


[Initial]: C2H3O2- = this is weak

[Initial]: OH- =

pOH = -log[OH-] = -log[9.1 x 10-3] = 2.04, so pH = 11.96

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g. 75.0 mL (total) of 0.100M NaOH has been added.

• Major Species: HC2H3O2, C2H3O2-, H2O, Na+,

OH-

• HC2H3O2 + OH- → C2H3O2

- + H2O


- = 0.0mmol



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After, Major Species: C2H3O2-, H2O, Na+, OH-.


[Initial]: C2H3O2- = this is weak

[Initial]: OH- =

pOH = -log[OH-] = -log[2.0 x 10-2] = 1.70,

so pH = 12.30

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WEAK ACID – STRONG BASENote the difference between the two curves on the left and the curve for a strong acid and strong base titration. The difference is noted to the right. Why is the shape the same after the equivalence point?

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• Remember that the equivalence point is defined by stoichiometry NOT by pH.

• When enough titrant has been added to react exactly with all the acid or base being titrated is the equivalence point.

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PRACTICE NINE

• HCN is a very weak acid (Ka = 6.2 x 10-10) when dissolved in water. If a 50.0mL sample of 0.100M HCn is titrated with 0.100M NaOH, calculate the pH of the solution

A. After 8.00mL of 0.100M NaOH has been added.

B. At the halfway point of the titration.

C. At the equivalence point of the titration.

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PRACTICE NINE

• HCN is a very weak acid (Ka = 6.2 x 10-10) when dissolved in water. If a 50.0mL sample of 0.100M HCN is titrated with 0.100M NaOH, calculate the pH of the solution

A. After 8.00mL of 0.100M NaOH has been added. pH = 8.48 HH pH = 8.49

B. At the halfway point of the titration.

pH = 9.21

C. At the equivalence point of the titration.

pH = 10.95

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• It took the same amount of NaOH in both the example and Practice Nine to reach the equivalence point. It is the AMT of the acid, not its strength that determines the equivalence point.

• The pH value at the equivalence point IS affected by acid strength.

• The strength of a weak acid has a significant effect on the shape of its pH curve. The weaker the acid, the greater the pH at the equivalence point. As the acid becomes weaker, the vertical region (shaded to the left) becomes shorter.

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The pH Curves for the Titrations of 50.0-mL Samples pf 0.10 M Acids with Various Ka Values with 0.10 M NaOH

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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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TITRATION CURVE

FIVE POINTS OF INTEREST ALONG A TITRATION CURVE for weak acids/bases:

A. The pH before the titration begins. Treat as usual, the acid or base in the flask determines the pH. If weak, an ICE chart is in order.

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TITRATION CURVEFIVE POINTS OF INTEREST ALONG A TITRATION CURVE:

B. The pH on the way to the equivalence point. You are in the “land of buffer” as soon as the first drop from the buret makes a splash and reacts to form the salt. Whatever is in the buret is the “added” part. Use to solve for the hydrogen ion concentration and subsequently the pH. Either the acid or the base [whichever is in the buret] starts at ZERO. Do the stoichiometry and then an ICE chart.

AqueousEquilibria

TITRATION CURVE

FIVE POINTS OF INTEREST ALONG A TITRATION CURVE:

C. The pH at the midpoint of the titration (½ equivalence point): once the midpoint is reached, [H+] = Ka since ½ of the acid or base has been neutralized, AND the resulting solution in the beaker is composed of the half that remains AND the salt. That means that pH = pKa.

AqueousEquilibria

TITRATION CURVE


D. The pH at the equivalence point.—you are simply calculating the pH of the salt, all the acid or base is now neutralized [to salt + water!]. Write the hydrolysis reaction for your ICE chart.

AqueousEquilibria

TITRATION CURVE


E. The pH beyond the equivalence point—it is stoichiometry again with a limiting reactant. Calculate the molarity of the EXCESS and solve for either pH directly (excess H+) or pOH (excess OH−) and subtract it from 14.00 to arrive at pH. Be sure to track the total volume when calculating the molarity!

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WEAK BASE – STRONG ACID

• We have to do a series of buffer problems like we did earlier.

• Remember that although the base is weak, it reacts to completion with the H+ ion, a very strong acid.

• At the equivalence point, an acidic salt is present and pH < 7

• After the equivalence point, the strong acid will be the dominant species and a simple pH calculation can be done after stoichiometry.

AqueousEquilibria

PRACTICE TEN

• For the titration of 100.0 mL of 0.050M NH3 (Kb = 1.8 x 10-5) with 0.10M HCl, calculate the pH of the solution at the following selected points of the titration:

a. 0.0 mL of 0.10M HCl has been added.

b. Before the equivalence point. 10.0mL

c. At the equivalence point 50.0mL

d. Beyond the equivalence point 60.0mL

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SUMMARY

AqueousEquilibria

SUMMARY1. Buffered solutions contain relatively large

concentrations of a weak acid and the corresponding weak base. They can involve a weak acid, HA, and the conjugate base, A-, or a weak base, B and the conjugate acid, BH+.

2. When H+ is added to a buffered solution, it reacts essentially to completion with the weak acid present. H+ + A- → HA or H+ + B → BH+

AqueousEquilibria

SUMMARYC. When OH- is added to a buffered solution, it

reacts essentially to completion with the weak acid present. OH- + HA → A- + H2O or OH- + BH+ → B + H2O

D. The pH of the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or B and BH+) are large compared with the amounts of H+ and OH- added.

AqueousEquilibria

TITRATIONS OF POLYPROTIC ACIDS

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

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ACID BASE INDICATORS• Marks the end point of a titration by changing

color. • The end point is NOT the equivalence point• With careful selection of an indicator, it can be.• We want the indicator end point and the titration

equivalence point to be as close as possible – choose wisely.

• Since strong acid-strong base titrations have a large vertical area, color changes will be sharp and a wide range of indicators can be used.

AqueousEquilibria

ACID BASE INDICATORS

• With weak acids and bases, we must be more careful in our choice of indicator.

• Indicators are usually weak acids, HIn. They have one color in their acidic form, HIn, and another color in their basic form, In-.

• Common indicator = phenolphthaleincolorless in its HIn formpink in its In-

form.

It changes color in the pH range of 8-10.

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ACID BASE INDICATORS• Usually 1/10 of the initial form of the indicator must

be changed to the other form before a new color is apparent.

• Equations used to determine the pH at which an indicator will change color:For titration of an ACID: pH = pKa + log1/10 = pKa – 1

For titration of a BASE: pH = pKa + log10/1 = pKa + 1

• The useful range of an indicator is usually pKa + 1.

• When choosing an indicator, determine the pH at the equivalence point of the titration and then choose an indicator with a pKa close to that.

AqueousEquilibria

Figure 15.8, page 715

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INDICATOR USE

The pH curve for the titration of 100.0 mL of 0.10 M of HCl with 0.10 M NaOH

AqueousEquilibria

INDICATOR USE

The pH curve for the titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH; Phenolphthalein will give an end pt very close to the equivalence pt of the titration

AqueousEquilibria

PRACTICE ELEVENBromothymol Blue, an indicator with a Ka of 1.0 x10-7, is yellow in its HIn form and blue in its In- form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible?

AqueousEquilibria

SOLUBILITY EQ

• Saturated solutions of salts are another type of chemical equilibria.

• Saturated solution of AgCl:

AgCl(s) ↔ Ag+(aq) + Cl-(aq).

• The SOLUBILITY PRODUCT expression:

Ksp = [Ag+][Cl-].

The AgCl(s) is left out of the expression

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SOLUBILITY EQ• The Ksp for AgCl = 1.6 x 10-10.

• If the product of [Ag+][Cl-] is < 1.6 x 10-

10, then the solution is unsaturated and no solid would be present.

• If the product were 1.6 x 10-10, the product is exactly saturdated and no solid would be present.

• If the product > 1.6 x 10-10, the solution is saturated and a solid (precipitate) would form.

AqueousEquilibria

SOLUBILITY EQ

• 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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SOLUBILITY EQ

Calculating Concentrations

For Ag2CO3:

Ag2CO3 (s) ↔ 2Ag+(aq) + CO3

-2(aq)

Ksp = [Ag+]2[CO3-2], Ksp = 8.1 x 10-12

We can determine the concentrations of 2Ag+

(aq) and CO3-2

(aq) using the equation Ksp = [Ag+]2[CO3

-2] = 8.1 x 10-12.

AqueousEquilibria

SOLUBILITY EQ

Initial [Ag+]2 = 0 Initial [CO3-2] = 0

EQ [Ag+]2 = 2x EQ [CO3-2] = x

Ksp = [Ag+]2[CO3-2] = 8.1 x 10-12

Ksp = (x)(2x)2 = 4x3 = 8.1 x 10-12

x3 = 2.0 x 10-12

x = 1.3 x 10-4

EQ [Ag+]2 = 1.3 x 10-4M

EQ [CO3-2] = 2.6 x 10-4M

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PRACTICE TWELVEThe Ksp value for copper(II) iodate, Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at 25°C.

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PRACTICE THIRTEENCopper(I) bromide has a measured solubility of 2.0 x 10-

4M at 25°C. Calculate the Ksp value.

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PRACTICE FOURTEENCalculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15M at 25°C.

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COMMON ION EFFECT

What happens if we dissolve the substance into something other than pure water?

What happens when the solution contains a common ion?

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PRACTICE FIFTEENWhat is the molar solubility of solid calcium fluoride, CaF2 in a 0.025M solution of sodium fluoride?

Ksp = 4.0 x 10-11

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PRECIPITATE AND QUAL ANALYSIS

• This is now considering the reverse of the above - forming a solid from solution.

• The product of the initial concentrations of the ions (raised to the power of their coefficients) is called the ION PRODUCT or Q.

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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.

AqueousEquilibria

PRACTICE SIXTEENA solution is prepared by adding 750.0mL of 4.00 x 10-3M Ce(NO3)3 to 300.0mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 with a Ksp = 1.9 x 10-10 precipitate from this solution? Justify your answer.

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SELECTIVE PRECIPITATION OF IONS

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

AqueousEquilibria