Aqueous Equilibria

Mar 31-7:41 AM

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Common Ion Effect

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

We have found the concentration of ions in solutions where only a weak acid or weak base have been present.

In this chapter we will look situations where you have a both a weak acid or base and its conjugate.

Which way will equilibrium shift if the strong electrolyte NaC2H3O2 is added to this solution?

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 electrolye.

Shifts right reducing H+

Mar 22-7:33 AM

2

What is the pH of a 0.30M solution of HC2H3O2? The Ka of acetic acid is 1.8x10-5.

What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate (NaC2H3O2) to enough water to make 1.0L of solution?

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

HC2H3O2 H+ C2H3O2-

2.64

4.74

Mar 22-8:37 AM

3

Calculate the pH of a solution containing 0.085 M nitrous acid,(HNO2; 4.5x10-4), and 0.10 M potassium nitrite (KNO2).

3.42

Mar 23-6:05 PM

4

Calculate the fluoride concentration and pH of a solutionthat is 0.20M in HF and 0.10M HCl. Ka = 6.8x10-4

HF(aq) H+(aq) + F-(aq)

HF H+ F-

1.4x10-3M = [F-] pH = 1.00

Mar 22-8:47 AM

5

Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCHO2; Ka = 1.8x10-4) and0.10M in HNO3.

[CHO2-] = 9.0x10-5; pH = 1.00

Mar 24-5:08 AM

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Buffered Solutions

Solutions which contain a weak conjugate acid base paircan resist changes in pH and known as buffers.

How do buffers resist changes in pH.Use the following reactions in your explanation:NH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)HC2H3O2(aq) + H2O(l) H3O

+(aq) + C2H3O2-(aq)

Buffered solutions are generally made by mixing a weak acidor weak base with a salt of that acid or base.

Take a solution of HC2H3O2 and add NaC2H3O2.Take a solution of NH3 and add NH4Cl.

Mar 22-8:50 AM

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For a reaction: HF(aq) H+(aq) + F-(aq)

Ka = [H+][F-] [HF]

Solved for [H+] = Ka[HF] [F-]

What two factors influence the [H+] and the pH

Mar 24-5:34 AM

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Buffer Capacity and pHBuffer capacity is the amount of acid or base that the buffer can neutralize before the pH begins to change to an appreciable degree.

The higher the concentrations of the conjugate acid basepair the more resistant to pH changes the buffer is.

[H+] = Ka[HX] [X-]

How do you find pH if you knowthe [H+]? How would you do thatfor the pH for a buffer? Enjoy the algebra?

pH = pKa + log [base][acid]

Henderson Hasselbach Equation

What about bases?

Mar 24-5:48 AM

9

What is the pH of a buffer that is 0.12M in lactic acid (HC3H5O3)and 0.10M in sodium lactate? For lactic acid the Ka=1.4x10-4

HC3H5O3 C3H5O3-H+

3.77

Mar 24-6:49 AM

10

Calculate the pH of a buffer composed of 0.12 benzoic acidand 0.20M sodium benzoate. The Ka of benzoic acid is 6.5x10-5.

4.42How many moles of NH4Cl must be added to 2.0L of 0.10M NH3 to form a buffer whose pH is 9.00? Assume that the NH4Cl does not change the volume of the solution.Kb = 1.8x10-5

0.36 mol NH4Cl

Mar 24-6:55 AM

11

Calculate the concentration of sodium benzoate that mustbe present in a 0.20M solution of benzoic acid (HC7H5O2)to produce a pH of 4.00.

0.13MAdding Strong Acids or Bases to Buffers

HX(aq) + H2O(l) H3O+(aq) + X-(aq)

What happens to the buffer below when you add astrong acid or base to it?

Mar 24-7:06 AM

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Adding Strong Acids or Bases to Buffers

1. Consider the acid-base neutralization reaction, and determine its effect on the [HX] and [X-]. This stage ofthe calculation is a stoichiometry calculation.

2. Use Ka and the new concentrations of [HX] and [X-] fromstep 1 to calculate [H+].This second stage of the procedureis a standard equilibrium calculation and is most easily done using the Henderson-Hasselbach equation.

Mar 24-7:13 AM

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A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 molNaC2H3O2 to enough water to makde 1.00L of solution. The pHof the buffer is 4.74.a) Calculate the pH of this solution after 0.020 mol of NaOH is added.b) Calculate the pH of this solution after 0.020 mol of NaOH is added to 1.00L of pure water.

a) 4.80

b) 12.30

Mar 24-7:23 AM

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A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 molNaC2H3O2 to enough water to makde 1.00L of solution. The pHof the buffer is 4.74.a) Calculate the pH of this solution after 0.020 mol of HCl is added.b) Calculate the pH of this solution after 0.020 mol of HCl is added to 1.00L of pure water.

4.68

1.70

A

B

Mar 24-9:49 PM

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Acid Base TitrationsIn a titration, a solution containing a known concentrationof base is slowly added to an acid (or the acid is added tothe base).

Acid base indicators can be used to signal the equivalencepoint.A pH meter can be used to monitor pH to generate a titration curve.

Three main types of titrations

1) Strong Acid - Strong Base2) Weak Acid - Strong Base3) Polyprotic Acid - Strong Base

Mar 24-9:55 PM

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Strong Acid Strong Base Titrations1) Initial pH: The pH of the solutionbefore the addition of any base isdetermined by the initial concentrationof the strong acid.

0.100M NaOH added to 50mL of 0.100M HCl

Initial pH = 1.002) Between the initial pH and theequivalence point: As base is added,the pH increases slowly at first and the rapidly as the equivalence point isreached. The pH of the solution isdetermined by the concentration of acid that has not yet been neutralized.23, 29, 31, 33, 37, 41,43, 45, 47, 49

Mar 24-10:09 PM

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Strong Acid Strong Base Titrations

3) The equivalence point: At the equivalence point an equal number of moles of acid and base have reacted. The pH is 7 because a neutral salt has been produced.

4) After the equivalence point: The pH of thesolution after the equivalence point is determined by the concentration of the excess base in the solution.

Mar 24-10:11 PM

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When choosing pH Indicators you need to look forone that will change color near the equivalence point.

Why is it not important for it to change exactly atthe equivalence point?

The point where thepH indicator changescolor is called theend point.

Mar 25-7:22 AM

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Calculate the pH when the following quantities of 0.100M NaOHsolution have been added to 50.0mL of 0.100M HCl solution:a) 49.0mLb) 51.0mL

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H+(aq) + OH-(aq) H2O(l)BeforeReaction

AfterReaction

Remember volumes are additive!!

a) 3.00b) 11.00

Mar 25-7:07 AM

20

Calculate the pH when the following quantities of 0.10M HNO3have been added to 25.0mL of 0.10M KOH:a) 24.9mLb) 25.1mL

10.30 3.70A B

Mar 25-7:17 AM

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What do you think the graph of a titration of a solution of a strongbase with a strong acid would look like?

pH 7

Volume of acid added

Mar 25-7:22 AM

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Weak Acid-Strong Base Titrations

The shape of the titration curve is similar to that of astrong acid strong base curve.

The main difference is the position of the equivalencepoint.

50.0mL of 0.100M acetic acidwith 0.100M NaOH

1) The initial pH: The pH is just the pHof the weak acid.

2) Between the initial pH andthe equivalence point: To determine the pH in this range you must first consider the neutralization reaction. Thencalculate the pH of the resultingbuffer.

Mar 25-7:34 AM

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2) Between the initial pH and the equivalence point: To determine the pH in this range you must first consider the neutralization reaction. Then calculate the pH of the resulting buffer.

3) The equivalence point: At this pointall of the acid has been neutralized butthere is anion of the weak acid willaffect the pH. As a weak base, the anionwill react with water and cause the pHto go up. The pH should be calculatedusing the Kb of the anion.

The equivalence point of aweak acid strong basetitration is always above 7.

Mar 25-7:42 AM

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4. After the equivalence point: In this region, the [OH-] producedby the anion of the weak acid is negligible compared to the [OH-] from the excess. The pH will be calculated using the excess [OH-] left over after the acid has been neutralized.

Calculate the pH of the solution formed when 45.0mL of 0.100MNaOH is added to 50.0mL of 0.100M HC2H3O2 (Ka=1.8x10-5)

Where is this at on the titration curve?

5.70

Mar 25-7:54 AM

25

Calculate the pH in the solution formed by adding 10.0mL of 0.050MNaOH to 40.0mL of 0.0250M benzoic acid (HC7H5O2, Ka=6.3x10-5).

4.20

Calculate the pH in the solution formed by adding 10.0mL of 0.100M HCl to 20.0mL of 0.100M NH3.

9.26

Mar 25-8:13 AM

26

Calculate the pH at the equivalence point in the titration of 50mLof 0.100M HC2H3O2 with 0.100M NaOH.

What has happened at the equivalence point?

How many mL of NaOH will be needed to reach this point?

What is the Kb for C2H3O2-?

What is the pH for this solution?

8.72

Mar 25-8:22 AM

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Weak acid-strong base titrations differ from strong acid-strong basetitrations in three ways:

1. The solution of the weak acid has a higher initial pHthan a solution of a strong acid of the same concentration.2. The pH change at the rapid rise portion of the curvenear the equivalence point is smaller for the weak acidthan it is for the strong acid.3. The pH at the equivalence point is above 7.00 for theweak acid-strong base titration.

As the Ka becomes smaller the pH changeat the equivalence point becomes lessmarked because the acid is weaker.

The pH at the equivalence point increasesas Ka decreases.

This makes choosing a proper indicator is much more important.

Mar 25-8:03 AM

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Titration curve of a weak base with strong acid(blue line) versus a strong base with a strongacid (red line)

Titration curve of a polyprotic acid versus a strong baseWhy are there multiple equivalence points?What are the major species present at the other two equivalence points?

Mar 25-8:41 AM

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Solubility Equilibria

Mar 24-5:15 AM

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Solubility Product Constant, Ksp

Saturated solution = A solution in which undissolved solute and dissolved solute are in equilibrium.

Unsaturated solution = A solution containing less solute than a saturated solution.

Supersaturated solution = A solution containing more solute than a saturated solution.

BaSO4(s) Ba2+(aq) + SO42-(aq)

Ksp is equal to the products of the concentration of the ions involvedin the equilibrium, each raised to the power of its coefficient in the equilibrium expression.

Ksp = [Ba2+][SO42-]

Mar 25-8:49 AM

31

What is the expression for the solubility product constantfor the following compounds:

Calcium flouride

Barium carbonate

Silver sulfateSolubility versus Solubility Product Constant

Solubility is the quantity of a solute that dissolves to form a saturated solution.This is usually expressed as grams of solute per liter of solution (g/L) or molarsolubility (mol/L) which is the number of moles of solute that dissolve in a liter of solution.

Solubility product constant is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution.

Mar 25-5:38 PM

32

The solubility of a substance can change as the concentrations of other solutes change.

What is the solubility expression for Mg(OH)2?

How do you think the solubility of Mg(OH)2 would be affectedby changes in pH?

What would happen to the solubility if the pH was lowered? Why?

What would happen to the solubility if the pH was raised? Why?

The solubility of a compound may change but the solubility product constant for the same compound does not change except when temperature changes.

Mar 27-8:39 AM

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The Ksp can be used to calculate solubility under a variety of conditions.This works best for ions with low charges (+1 and -1) but can applied to ions with largercharges to give a rough estimate of for the solubility.

Solid silver chromate is added to pure water at 25℃. Some of the solid remainsundissolved at the bottom of the flask. The mixture is stirred for several days toensure the equilibrium is achieved between the undissolved Ag2CrO4(s) and thesolution. Analysis of the equilibrated solution shows that its silver ion concentrationis 1.3x10-4M. Assuming that Ag2CrO4 completely in water and that there are no otherimportant equilibria involving the Ag+ or CrO4

-2 ions in the solution. Calculate the Kspfor this compound.

1. Write the solubility reaction for silver chromate.

2. You know the concentration of one of the ions.Use that to find the concentration of the other ion.

3. Substitute the concentrations into the equlibrium expression.

1.1x10-12

Mar 27-9:10 AM

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A saturated solution of Mg(OH)2 in contact with undissolved solid is prepared at 25℃.The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no simultaneous equilibria involving the Mg2+ orOH- ions in the solution, calculate Ksp for this compound.

1.6x10-12

The Ksp for CaF2 is 3.9x10-11 at 25℃. Assuming that CaF2 dissociates completelyupon dissolving and that there are no other important equilibria affecting its solubility,calculate the solubility of CaF2 in gram per liter.

1 - Write the solubility expression for calcium floride.

I

C

E

CaF2(s) Ca+2(aq) + 2F-(aq)

2 - Fill in ice chart. Do you put anything into the column for the solid?

3 - Substitute values into equilibrium expression and solve for x (molar solubility)

Pul

lP

ull

Mar 27-10:52 AM

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9.28x10-6mol/L

The Ksp for LaF3 is 2x10-19. What is the solubility of LaF3 in waterin moles per liter?

Factors That Affect Solubility

Common Ion EffectThe solubility of a slightly soluble salt is decreased by thepresence of a second solute that furnishes a common ion.

CaF2(s) Ca2+(aq) + 2F-(aq)

What happens for the solubility of calcium flouride if sodium flourideis added?

Mar 28-12:42 PM

36

Calculate the molar solubility of CaF2 (Ksp = 3.9x10-11) at 25 C in a solution that isa) 0.010M in Ca(NO3)2b) 0.010M in NaF

I

C

E

CaF2(s) Ca2+(aq) + 2F-(aq)

a) 3.1x10-5M

b) 3.9x10-7M

Mar 27-11:20 AM

37

The value for Ksp for manganese(II) hydroxide, Mn(OH)2, is 1.6x10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020M NaOH.

Solubility and pHIf a substance has a anion that is basic its solubility willbe affected by changes in pH.

Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp=1.8x10-11

What is the concentration of Mg2+ when the pH is 10.52? 9.50?How did the change in pH affect the solubility? Why?

Mar 28-4:00 PM

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The solubility of calcium flouride is also affected by changes in pH. Flouride is a weak base that will react with any free hydrogen.

CaF2(s) Ca2+(aq) + 2F-(aq)

F-(aq) + H+(aq) HF(aq)

What is the reaction for the overall process?

The solubility of slightly soluble salts containing basic anions increasesas [H+] increases (as pH is lowered). Salts with anions of negligible basicity are unaffected by pH changes. Which anions fall into this category?

Mar 28-4:53 PM

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Which of the following substances will be more soluble in acidic solutionthan in basic solution:

a) Ni(OH)2(s)b) CaCO3(s)c) BaF2(s)d) AgCl(s)

Write the net ionic equation for the reaction of the following copper(II) compounds with acid:a) CuSb) Cu(N3)2

CuS(s) + 2H+(aq) Cu2+(aq) + H2S(aq)

Cu(N3)2(s) + 2H+ Cu2+(aq) + 2HN3

Mar 28-5:00 PM

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Formation of Complex IonsMetal ions can act as Lewis acids or electron pair acceptors toward molecules such as water that can behave as Lewis bases.

What do Lewis bases need to have?AgCl(s) Ag+(aq) + Cl-(aq)

Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)

AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq)

Ag(NH3)2+ is a complex ion. The larger the equilbrium constant for the

formation of the complex ion the more stable it is.

What is the equilibrium expression for this reaction?

The solubility ofmetal salts go upin the prescenceof Lewis basessuch as NH3, CN-, or OH-.

Mar 28-5:12 PM

41

Calculate the concentration of Ag+ present in solution at equilibriumwhen concentrated ammonia is added to a 0.010M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20M.

I

C

E

Ag(NH3)2+(aq) Ag+(aq) + 2NH3(aq)

Kf = 1.7x107

[Ag+] = 5.9x10-8

Mar 28-5:32 PM

42

Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol

of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. The Kf of Cr(OH)4

- is 8x1029.

1.x10-16M

AmphoterismSome metal hydroxides and oxides are insoluble in water butdissolve in acidic and basic solutions. The happens becausethese hydroxides and oxides are amphoteric. Some examplesinclude the hydroxides and oxides of Al3+, Cr3+, Zn2+, and Sn2+.

Aluminum example:

Mar 28-5:49 PM

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Precipitation and Separation of IonsRemember that equilibrium can be achieved from either direction.The equilibrium that exists between BaSO4(s), Ba2+(aq), and SO4

2-(aq) can be achieved starting with the solid barium sulfate of with a mixture of solutions that contain the two ions such as barium chloride and sodium sulfate. To form the precipitate the proper concentrationsare needed.

BaSO4(s) Ba2+(aq) + SO42-(aq)

The Return of QIf Q > Ksp, precipitation occurs until Q = KspIf Q = Ksp, equilibrium exists (saturated solution)If Q < Ksp, solid dissolves until Q = Ksp

Q = [Ba2+][SO42-]

Will a precipitate form when 0.10L of 8.0x10-3M Pb(NO3)2 is added to0.40L of 5.0x10-3M Na2SO4? Ksp = 6.3x10-7.

YES NO

YES

Mar 28-5:59 PM

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Will a precipitate form when 0.050L of 2.0x10-2M NaF is mixedwith 0.010L of 1.0x10-2M Ca(NO3)2? Ksp = 3.9x10-11.

YES

Ions can be separated from each other bases on the solubilitiesof their salts.

A solution contains 1.0x10-2M Ag+ and 2.0x10-2 M Pb2+. WhenCl- is added to the solution both AgCl (Ksp = 1.8x10-10) and PbCl2 (Ksp=1.7x10-5) precipitate from the solution. What concentrationof Cl- is needed to begin the precipitation of each salt? Which saltprecipitates first?

1.8x10-8M for the silver2.9x10-2M for the lead

Mar 28-6:44 PM

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A solution consists of 0.050M Mg2+ and 0.020M Cu2+. Which ionwill precipitate first as OH- is added to the solution? What concentration of OH- is necessary to begin the precipitation of each cation? (Ksp = 1.8x10-11 for Mg(OH)2 and 2.2x10-20 for Cu(OH)2)

1.0x10-9M for the copper1.9x10-5M for the magnesium

Mar 28-6:58 PM

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Apr 15-8:33 AM

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