Applications of Aqueous EquilibriaChapter 15

Solutions—9 out of 75 multiple choice

Free Response—almost every year

Common Ion Effect

Common ion—an ion that is produced by more than one species in a solutionShifts equilibrium because concentration of common ion is increased

Example

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

ExampleNitrous acid (HNO2) and sodium nitrite (NaNO2)

Ka = [H+][NO2] / [HNO2]Acid will dissociate less because the solution has nitrite ion present already

Sample ProblemCalculate the [H+] and percent dissociation of a solution that is 1.00 M in HF and 1.00 M in NaF. The Ka of HF is 7.2 x 10-4.

BuffersSolutions that resist change in pH when either H+ or OH- are addedContains a weak acid & its salt or a weak base & its saltTakes advantage of common ion effect

Key Points on Buffered Solutions

1.They are weak acids or bases containing a common ion.

2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

Sample ProblemA buffer solution contains .50 M acetic acid and .50 M sodium acetate. What is the pH of the solution? (The Ka for acetic acid is 1.8 x 10-5.)

Sample ProblemCalculate the pH of a solution that is .75 M in lactic acid and .25 M in sodium lactate. (Ka for lactic acid = 1.4 x 10-4.)

Henderson-Hasselbalch

EquationUseful for calculating pH when the [A]/[HA] ratios are known.

Sample ProblemWhat is the pH of a buffer solution that contains .50 M acetic acid and .50 M sodium acetate? Ka for acetic acid is 1.8 x 10-5.

Sample ProblemCalculate the pH of a solution that is .75 M in lactic acid and .25 M in sodium lactate. (Ka for lactic acid = 1.4 x 10-4.)

Sample problemWhat is the pH of a buffer solution that contains .25 M NH3 (Kb = 1.8 x 10-5) and .40 M NH4Cl?

pH Change in Buffered Solutions

Buffers contain relatively large amounts of weak acid and corresponding base.

Added H+ reacts to completion with the weak base.

Added OH- reacts to completion with weak acid

pH Change in Buffered Solutions

The pH is determined by the ratio of the concentrations of the weak acid and weak base.Think H-H equation.

Sample ProblemHow will the pH change when .010 mol of NaOH is added to 1.0 L of the acetic acid/sodium acetate buffer system? (Original pH was 4.74.)

Sample ProblemCalculate the pH change when .010 mol of gaseous HCl is added to 1.0 L of a buffer solution containing 5.00 M HC2H3O2 and 5.0 M NaC2H3O2.

Ka = 1.8 x 10-5

Buffering Capacity

the amount of H+ or OH the buffer can absorb without a significant change in pH

Titration (pH) Curve

A plot of pH of the solution being analyzed versus the amount of titrant added.

Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.

Strong Acid/Strong Base

Easiest type—acid & base dissociate completely and react completelyInitial—pH depends only on concentrationAdd acid/base—do stoichiometry

Strong Acid/Strong Base

pH depends on amount of substance remaining after reaction

Sample ProblemYou are titrating 50.00 mL of a .50 M NaOH solution with 1.0 M HCl.What is the pH of the base solution before any acid has been added?

Sample ProblemWhat is the pH of the base solution after 10.00 mL of acid have been added?

Sample ProblemWhat is the pH of the solution after 12.5 mL of acid have been added?

Sample ProblemWhat is the pH of the solution after 20.00 mL of acid have been added?

Sample ProblemWhat is the pH of the solution after 25.00 mL of acid have been added?

Sample ProblemWhat is the pH of the solution after 30.00 mL of acid have been added?

Weak Acid/Strong Base

Initailly—find pH by doing a weak acid problemAfter adding base—buffer problem—Use H-H equationAt midpoint—pH = pKa

Weak Acid/Strong Base

At equivalence point—find pH by doing a weak base problem for conj. baseAfter end point—pH depends on amount of excess base

Sample ProblemTitration of 50.0 mL of .10 M acetic acid (Ka = 1.8 x 10-5) with .10 M NaOH.What is the initial pH of the acetic acid solution?

Sample ProblemWhat is the pH after 15.0 mL of base have been added?

Sample ProblemWhat is the pH after 25.0 mL of base have been added?

Sample ProblemWhat is the pH after 35.0 mL of base have been added?

Sample ProblemWhat is the pH at the endpoint?

Sample ProblemWhat is the pH after 60.0 mL of base have been added?

Weak Base/Strong Acid

Initailly—find pH by doing a weak base problemAfter adding acid—buffer problem—Use H-H equationAt midpoint—pH = pKa of conjugate acid

Weak Base/Strong Acid

At equivalence point—find pH by doing a weak acid problem for conj. acidAfter end point—pH depends on amount of excess acid

Sample ProblemCalculate the pH at each of the following points for the titration of 100.0 mL of .050 M NH3 with .10 M HCl.

Sample ProblemBefore any acid is addedAfter 20 mL of acid At the equivalence pointAfter 40 mL of acid At the end pointAfter 60 mL of acid

Acid-Base IndicatorA substance, usually a weak acid,

that changes color at different pH’smarks the end point of a titration

by changing colorThe equivalence pt.

(stoichiometry) is not necessarily the same as the end pt. (color change).

IndicatorsSince they are weak acids (HIn), they will have a Ka value.Assume Ka = 1.0 x 10-8 for HIn.HIn<==> H+ + In-

(red) (blue)What would happen in acid? In base?

Calculating pH for Color Change

Ka = 1.0 x 10-8 for HIn.HIn<==> H+ + In-

(red) (blue)Assume 1/10 of the substance must change color before it is detectable.

Color change (end point) range is pKa + or –1.

Choosing an IndicatorCalculate pH at endpointChoose an indicator that changes in that range.

Solubility Equilibrium

Same as other equilibrium expressions.Describes the amount of solute that will dissolve for slightly soluble materials.

Ksp ExpressionsWrite equations for complete dissociation.Remember pure solids and liquids are not included.Account for all ions.

Solubility ProductFor solids dissolving to form aqueous solutions.

Bi2S3(s) 2Bi3+(aq) + 3S2(aq)

Ksp = solubility product constant Ksp = [Bi3+]2[S2]3

Solubility Product vs. Solubility

Solubility—s—amount of substance that dissolves at a given temp.—variable if common ion is presentSolubility Product—Ksp—product of dissolved ions in solution—constant at given temp.Remember, pure solids & liquids don’t matter.

Sample ProblemCalculate Ksp for copper (I) bromide if the molar solubility is 2.0 x 10-4 mol/L at 25*C

Practice ProblemThe solubility for bismuth sulfide (Bi2S3) is 1.0 x 10-

15 mol/L. Calculate Ksp.

Answer: 1.1 x 10-73

Sample ProblemCopper (II) iodate has a Ksp value of 1.4 x 10-7 at 25.0o C. What is its solubility at this temperature?

Copper (II) iodate has a Ksp value of 1.4 x 10-7 at 25.0* C. What is its solubility at this temperature if it is dissolving in a solution that is .125 M in sodium iodate?