1 Chapter 15 Aqueous Equilibrium AP Chemistry Unit 12

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Chapter 15Aqueous Equilibrium

AP Chemistry Unit 12

The Common Ion Effect• Consider a solution of acetic acid.• If we add acetate ion as a second solute (i.e., sodium

acetate), the pH of the solution increases:

LeChâtelier’s principle: What happens to [H3O+] when the

equilibrium shifts to the left?

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.”

The Common-Ion Effect

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

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)

The Common-Ion Effect

= x

1.4 10−3 = x

(0.10) (x)(0.20)6.8 10−4 =

(0.20) (6.8 10−4)(0.10)

The Common-Ion Effect

• 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

Example 15.16

Calculate the pH of an aqueous solution that is both 1.00 M CH3COOH and 1.00 M CH3COONa.

Buffers

• Buffers are solutions of a weak conjugate acid-base pair.

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

Buffer Solutions

• The acid component of the buffer neutralizes small added amounts of OH–, forming the weaker conjugate base which does not affect pH much:

HA + OH– H2O + A–

• The base component neutralizes small added amounts of H3O+, forming the weaker conjugate acid which does not affect pH much.

A– + H3O+ H2O + HA

• Pure water does not buffer at all …

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.

Buffers

Similarly, if acid is added, the F− reacts with it to form HF and water.

Buffer Calculations

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

[H3O+] [A−][HA]

Ka =

HA + H2O H3O+ + A−

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

Buffer Calculations

• SopKa = pH − log

[base][acid]

• Rearranging, this becomes

pH = pKa + log[base][acid]

• This is the Henderson–Hasselbalch equation.• To use this equation, the ratio [conjugate

base]/[weak acid] must have a value between 0.10–10 and both concentrations must exceed Ka by a factor of 100 or more.

Buffer Capacity andBuffer Range

• There is a limit to the ability of a buffer solution to neutralize added acid or base.

• This buffer capacity is reached before either buffer component has been consumed.

• In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize.

• As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal or nearly so.

• Therefore, a buffer is most effective when the desired pH of the buffer is very near pKa of the weak acid of the buffer.

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.

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.

Example 15.17

A buffer solution is 0.24 M NH3 and 0.20 M NH4Cl. (a) What is the pH of this buffer? (b) If 0.0050 mol NaOH is added to 0.500 L of this solution, what will be the pH?

Example 15.18

What concentration of acetate ion in 0.500 M CH3COOH(aq) produces a buffer solution with pH = 5.00?

Acid–Base Indicators

• An acid–base indicator is a weak acid or base.• The acid form (HA) of the indicator has one color, the

conjugate base (A–) has a different color. One of the “colors” may be colorless.

• In an acidic solution, [H3O+] is high. Because H3O+ is a common ion, it suppresses the ionization of the indicator acid, and we see the color of HA.

• In a basic solution, [OH–] is high, and it reacts with HA, forming the color of A–.

• Acid–base indicators are often used for applications in which a precise pH reading isn’t necessary.

Different indicators have different values of Ka, so they exhibit color changes at different values of pH

Example 15.19 A Conceptual Example Explain the series of color changes of thymol blue indicator

produced by the actions pictured in Figure 15.14:

(a) A few drops of thymol blue are added to HCl(aq).

(b) Some sodium acetate is added to solution (a).

(c) A small quantity of sodium hydroxide is added to solution (b).

(d) An additional, larger quantity of sodium hydroxide is added to solution (c).

Neutralization Reactions

• At the equivalence point in an acid–base titration, the acid and base have been brought together in precise stoichiometric proportions.

• The endpoint is the point in the titration at which the indicator changes color.

• Ideally, the indicator is selected so that the endpoint and the equivalence point are very close together.

• The endpoint and the equivalence point for a neutralization titration can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.

Titration Curve, Strong Acid with Strong Base

Bromphenol blue, bromthymol blue, and

phenolphthalein all change color at very

nearly 20.0 mL

At about what volume would we see a color

change if we used methyl violet as the indicator?

Example 15.20Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH:

H3O+ + Cl– + Na+ + OH– Na+ + Cl– + 2 H2O

(a) before the addition of any NaOH

(b) after the addition of 10.00 mL of 0.500 M NaOH

(c) after the addition of 20.00 mL of 0.500 M NaOH

(d) after the addition of 20.20 mL of 0.500 M NaOH

Titration Curve, Weak Acid with Strong BaseThe equivalence-point pH is NOT 7.00 here.

Why not??

Bromphenol blue was ok for the strong acid/strong

base titration, but it changes color far too early

to be useful here.

Titration of a Weak Acid with a Strong Base

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.

Titration of a Weak Acid with a Strong Base

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

Example 15.21Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH:

CH3COOH + Na+ + OH– Na+ + CH3COO– + H2O

(a) before the addition of any NaOH

(b) after the addition of 8.00 mL of 0.500 M NaOH

(c) after the addition of 10.00 mL of 0.500 M NaOH

(d) after the addition of 20.00 mL of 0.100 M NaOH

(e) after the addition of 21.00 mL of 0.100 M NaOH

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.

Titrations of Polyprotic Acids

When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

Example 15.22 A Conceptual Example

This titration curve shown in Figure 15.18 involves 1.0 M solutions of an acid and a base. Identify the type of titration it represents.

• Solubility product constant, Ksp: the equilibrium constant expression for the dissolving of a slightly soluble solid.

The Solubility Product Constant, Ksp

Ksp = [ Ba2+ ][ SO42–]

BaSO4(s) Ba2+(aq) + SO42–(aq)

• Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble”)

• An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution:

Example 16.1Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO4, and (b) chromium(III) hydroxide, Cr(OH)3.

• Ksp is an equilibrium constant

• Molar solubility is the number of moles of compound that will dissolve per liter of solution.

• Molar solubility is related to the value of Ksp, but molar solubility and Ksp are not the same thing.

• In fact, “smaller Ksp” doesn’t always mean “lower molar solubility.”

• Solubility depends on both Ksp and the form of the equilibrium constant expression.

Ksp and Molar Solubility

Example 16.2At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag2CO3 per liter of solution. Calculate Ksp for Ag2CO3 at 20 °C. The balanced equation is

Ag2CO3(s) 2 Ag+(aq) + CO32–(aq) Ksp = ?

Example 16.3From the Ksp value for silver sulfate, calculate its molar solubility at 25 °C.

Ag2SO4(s) 2 Ag+(aq) + SO42–(aq)

Ksp = 1.4 x 10–5 at 25 °C

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)

Example 16.5 Calculate the molar solubility of Ag2SO4 in 1.00 M Na2SO4(aq).

Factors Affecting Solubility

• 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.

Example 16.11 What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x 10–5 M, that is, pH = 9.00?

Mg(OH)2(s) Mg2+(aq) + 2 OH–(aq) Ksp = 1.8 x 10–11

Example 16.12 A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH)2(s) is most soluble:(a) 1.00 M NH3

(b) 1.00 M NH3 /1.00 M NH4+

(c) 1.00 M NH4Cl.

Factors Affecting Solubility• Complex Ions

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

Factors Affecting Solubility

• Complex Ions– The formation

of these complex ions increases the solubility of these salts.

Factors Affecting Solubility

• 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+.

• Qip can then be compared to Ksp.

• Precipitation should occur if Qip > Ksp.

• Precipitation cannot occur if Qip < Ksp.

• A solution is just saturated if Qip = Ksp.

• In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered.

• Qip is the ion product reaction quotient and is based on initial conditions of the reaction.

Qip and Qc: new look, same great taste!

Will Precipitation Occur?

Example 16.6If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3, will a precipitate form?

Ag2CrO4(s) 2 Ag+(aq) + CrO42–(aq) Ksp = 1.1 x 10–12

Example 16.7 A Conceptual ExamplePictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2. What is the solid that first appears? Explain why it then disappears.

Example 16.8

If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form?

MgF2(s) Mg2+(aq) + 2 F–(aq) Ksp = 3.7 x 10–8

Selective Precipitation of Ions

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