chemistry 102(001) fall 2012

17-1CHEM 102, Fall 2010, LA TECH

CTH 328 10:00-11:15 am

Instructor: Dr. Upali Siriwardane

e-mail: [email protected]

Office: CTH 311 Phone 257-4941

Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu, Th, F 8:00 - 10:00am..

Exams: 10:00-11:15 am, CTH 328.

September 27, 2012 (Test 1): Chapter 13

October 18, 2012 (Test 2): Chapter 14 &15

November 13, 2012 (Test 3): Chapter 16 &18

Optional Comprehensive Final Exam: November 15, 2012 :

Chapters 13, 14, 15, 16, 17, and 18

Chemistry 102(001) Fall 2012


Chapter 16.Aditional Aqueous Equilibria17.1 Buffer Solutions17.2 Acid-Base Titrations17.3 Acid Rain17.4 Solubility Equilibria and the Solubility Product Constant,

Ksp

17.5 Factors Affecting Solubility / Precipitation: Will It Occur?


Reaction of a basic anion or acidic cation with water is an ordinary Brønsted-Lowry acid-

base reaction.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

NH4+

(aq) + H2O(l) NH3 (aq) + H3O+

(aq)

This type of reaction is given a special name.

Hydrolysis

The reaction of an anion with water to produce the conjugate acid and OH-.

The reaction of a cation with water to produce the conjugate base and H3O+

.

Hydrolysis


Acid-Base Properties of Typical Ions


What salt solutions would be acidic, basic and neutral?1) strong acid + strong base =

neutral 2) weak acid + strong base =

basic 3) strong acid + weak base =

acidic 4) weak acid + weak base =

neutral, basic or an acidic solution

depending on the relative strengths of the acid and the base.


What pH? Neutral, basic or acidic?

•a)NaCl • neutral•b) NaC2H3O2

• basic•c) NaHSO4 • acidic•d) NH4Cl• acidic


1) If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic?

a) Solid sodium carbonate-(Na2CO3):

b) Sodium chloride- (NaCl):

c) Sodium acetate- (NaC2H3O2):

d) Ammonium sulfate-((NH4)2SO4):


How do you calculate pH of a salt solution?Find out the pH, acidic or basic?If acidic it should be a salt of weak baseIf basic it should be a salt of weak acidif acidic calculate Ka from Ka= Kw/Kb

if basic calculate Kb from Kb= Kw/Ka

Do a calculation similar to pH of a weak acid or base


What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5)Find out the pH, acidic

if acidic calculate Ka from Ka= Kw/Kb

Ka= Kw/Kb = 1 x 10-14 /1.8 x 10-5)

Ka= 5.56. X 10-10

Do a calculation similar to pH of a weak acid


Continued NH4

+ + H2O H 3+O + NH3

[NH4+] [H3

+O ] [NH3 ]Ini. Con. 0.5 M 0.0 M 0.00 MChange -x x xEq. Con. 0.5 - x x x [H 3+O ] [NH3 ] Ka(NH4

+) = -------------------- = [NH 4+] x2

---------------- ; appro.:0.5 - x . 0.5 (0.5 - x)


x2 Ka(NH4

+) = ----------- = 5.56 x 10 -10

0. 5 x2

= 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10

x= 2.78 x 10 -10 = 1.66 x 10-5

[H+ ] = x = 1.66 x 10-5 MpH = -log [H+ ] = - log 1.66 x 10-5

pH = 4.77pH of 0.5 M NH4Cl solution is 4.77

(acidic)

Continued


2) What is the pH of a 0.05 M aqueous NH4Cl solution? (Kb (NH3) = 1.8 x 10-5)

a) equilibrium reaction for the hydrolysis of salt:b) Ka for the conjugate acid NH4

+:c) ICE set-up:I:___________________________________________C:__________________________________________E___________________________________________d) Calculation of x, [H3O]+:e) pH of the solution:


3) What is the pH of a 0.05 M aqueous NaC2H3O2 solution? (Ka (HC2H3O2) = 1.8 x 10-5)

a) equilibrium reaction for the hydrolysis of salt:b) Kb for the conjugate base C2H3O2

-:c) ICE set-up:I:___________________________________________C:__________________________________________E___________________________________________d) Calculation of x, [OH]-:e) pOH and pH of the solution:


Acid-Base Chemistryof Some Antacids


4) A 50.00-mL sample of 0.100 M KOH is being titrated with 0.100 M HNO3. Calculate the pH of the solution after 52.00 mL of HNO3 is added.

a) acid base reaction:

b) moles of KOH: c) moles of HNO3:

d) [H3O]+:

e) pH of the solution:


Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

This type of reaction is given a special name.

Hydrolysis

The reaction of an anion with water to produce the conjugate acid and OH-.

The reaction of a cation with water to produce the conjugate base and H3O+.

Hydrolysis


Common Ion Effect

Weak acid and salt solutionsE.g. HC2H3O2 and NaC2H3O2

Weak base and salt solutionsE.g. NH3 and NH4Cl. H2O + C2H3O2

- OH- + HC2H3O2 (common ion)

H2O + NH4 + H3+O + NH3

(common ion)


Solutions that resist pH change when small amounts of acid or base are added.

Two types

Mixture of weak acid and its salt

Mixture of weak base and its salt

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Add OH- Add H3O+

shift to right shift to left

Based on the common ion effect.

Buffers


The pH of a buffer does not depend on the absolute amount of the conjugate acid-base

pair. It is based on the ratio of the two.

Henderson-Hasselbalch equation.

Easily derived from the Ka or Kb expression.

Starting with an acid

pH = pKa + log

Starting with a base

pH = 14 - ( pKb + log )[HA]

[A-]

[A-]

[HA]

Buffers


Henderson-Hasselbalch Equation

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

[H3O+] [A-] Ka = ---------------- [HA]

[H3O+] = Ka ([HA]/[A-])

pH = pKa + log([A-]/[HA])

when the [A-] = [HA]pH = pKa


Calcualtion of pH of BuffersHenderson Hesselbach Equation

[ACID]pH = pKa - log --------- [BASE] [BASE]pH = pKa + log --------- [ACID]


Control of blood pH

Oxygen is transported primarily by hemoglobin in the red blood cells.

CO2 is transported both in plasma and the red blood cells.

CO2 (aq) + H2O H2CO3 (aq)

H+

(aq) + HCO3-(aq)

The bicarbonate

buffer is essential

for controlling

blood pH

Buffers and blood


Buffer Capacity

Refers to the ability of the buffer to retard changes in pH when small amounts of acid or base are added

The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity


Adding an Acid or Baseto a Buffer


Buffer Systems


5) For the (buffer effect) of HC2H3O2/NaC2H3O2

a) Acid dissociation reaction:

b) Salt hydrolysis reaction:

c) Common ions in both equilibria:

d) Which way salt hydrolysis equilibrium move adding

H3O+:

e) Which way salt hydrolysis equilibrium move adding

OH-:


6) Describe the (buffer effect) of NH3/NH4Cl

a) Buffer type: (weak acid or base)/soluble salt):

b) Base dissociation reaction:

c) Salt hydrolysis reaction:

d) common ions in both equilibria:

e) Which way salt hydrolysis equilibrium move adding

H3O+:

f) Which way salt hydrolysis equilibrium move adding

OH-:


7) What is the pH of a solution that is 0.2 M in acetic acid (Ka = 1.8 x 10-5) and 0.2 M in sodium acetate?

a) Is it a acid, base, salt or buffer solution?

b) Henderson-Hesselbalch equation:

c) pKa: d)

e) pH of the solution:


Titrations ofAcids and Bases

TitrationAnalyteTitrant

analyte + titrant => products


Acid-base indicators are highly colored weak acids or bases.

HIn In- + H+

color 1 color 2

They may have more than one color transition.

Example. Thymol blue

Red - Yellow - Blue

One of the forms may be colorless - phenolphthalein (colorless to pink)

Indicators


Acid-Base Indicator

HIn + H2O H3O+ + In-

acid base color color

[H3O+][In-]Ka =

[HIn]They may have more than one color transition.

Example. Thymol blue Red - Yellow - Blue

Weak acid that changes color with changes in pH


What is an Indicator?Indicator is an weak acid with different Ka, colors to

the acid and its conjugate base. E.g. phenolphthalein

Hin H+ + In-

colorless pinkAcidic colorlessBasic pink


Selection of an indicator for a titration

a) strong acid/strong base b) weak acid/strong base c) strong acid/weak base d) weak acid/weak base

Calculate the pH of the solution at he equivalence point or end point


pH and Color of Indicators


Red Cabbage as IndicatorC O H

O

N N N

CH3

CH3

(aq)

C O-

O

N N N

CH3

CH3

(aq) + H3O+ (aq)

yellow

red

+ H2O (l)


Acid-base indicators are weak acids that undergo a color change at a

known pH.

phenolphthalein

pH

Indicator examples


8) If 50 ml of a 0.01 M HCl solution is titrated with a 0.01 M NaOH solution, what will be the concentration of salt (NaCl) the pH at the endpoint?

a) NaCl solution acidic, basic or neutral?

b) Concentration of [NaCl]:

c) pH of the solution?

d) Suitable indicator for the titration:


Titration Apparatus

Burette delivering base to a flask containing an acid.

The pink color in the flask is due to the

phenolphthalein indicator.


Endpoint vs. Equivalence Point

Endpointpoint where there is a physical change, such

as color change, with the indicator

Equivalence Point# moles titrant = # moles analyte

#molestitrant=(V M)titrant

#molesanalyte=(V M)analyte


9) If 50 mL of a 0.01 M HCl solution is titrated with a 0.01 M NH3 (Kb = 1.8 x 10-5) solution, what will be

a) The initial pH (0.01 M NH3):

b) Concentration of NH4Cl at the endpoint:

c) pH at the endpoint:

d) Suitable indicator for the titration:


10) If 50 ml of a 0.01 M HC2H3O2 solution is titrated with a 0.01 M NaOH solution, what will be the

a) Molarity of NaC2H3O2 at the endpoint:

b) The pH at the endpoint:

c) What indicator would be most suitable for this titration:


Polyprotic Acids


Organic or Carboxylic Acids

H C

H

H

C

H

H

C

H

H

C

O

O H

nonacidic hydrogens

butanoic acid

acidic hydrogen

CH 3 C

O

acetic acid

OH

electron-attractingoxygen atom

acidic hydrogenCH 3 C

O

OH

-

CH 3 C

O

O-

acetate ion


FCH2CO2H (strongest acid) > ClCH2CO2H > BrCH2CO2H (weakest acid).

Acid Ka pKa

HCOOH (formic acid) 1.78 X 10-43 0.75CH3COOH (acetic acid) 1.74 X 10-54 0.76CH3CH2COOH (propanoic acid)1.38 x 10-5 4.86

Organic or Carboxylic Acids


Acid-Base in the Kitchenvinegar - acetic acidlemon juice (citrus juice) - citric acidbaking soda - NaHCO3

milk - lactic acidbaking powder - H2PO4

- & HCO3-


Household Cleaners

CH 3CH2CH 2CH2CH 2CH2CH 2CH 2CH 2CH 2CH2CH 2CH2CH 2 SO3-Na+

Oil-soluble part(hydrophobic)

Water-soluble part(hydrophilic)

A Typical Synthetic Detergent Molecule

CH 3(CH 2)4COO(CH 2)2O( CH2CH 2O) 2CH 2CH 2OH

esterlink

(hydro-philic)

etherlink

etherlink

(hydrophilic)

hydrocarbonchain

(hydro-phobic)

alcohol group(hydrophilic)

A nonionic detergent


Dishwashing Detergent


Acid-Base Indicator Behavioracid color shows when

[In-] 1

£ [HIn] 10

[H3O+

][In-] 1

= [H3O+

] = Ka [HIn] 10

base color shows when

[In-] 1 ³ [HIn] 10

[H3O+][In-] = 10 [H3O+] = Ka [HIn]


Indicator pH Range

acid color shows when

pH + 1 = pKa

and base color shows when

pH - 1 = pKa

Color change range is

pKa = pH 1 or pH = pKa 1


Titration curves

Acid-base titration curveA plot of the pH against the amount of acid or base

added during a titration.Plots of this type are useful for visualizing a titration.

It also can be used to show where an indicator undergoes its color change.


Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH


EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.at 0.00 mL of NaOH added, initial point

[H3O+

] = [HCl] = 0.1000 M

pH = 1.0000


EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH.

at 15.00 mL of NaOH added

Va Ma > Vb Mb thus

(Va Ma) - (Vb Mb)

[H3O+] = (Va + Vb)

((35.00mL) (0.1000M)) - ((15.00mL) (0.1000M))

= (35.00 + 15.00)mL

= 4.000 10-2

M pH = 1.3979




Va Ma = Vb Mb , equivalence point

at equivalence point of a

strong acid - strong base titration

pH º 7.0000




Vb Mb > Va Ma , post equvalence point

(Vb Mb) - (Va Ma)

[OH-] = (Va + Vb)

((50.00mL) (0.1000M)) - ((35.00mL) (0.1000M))

=

(35.00 + 50.00)mL

= 1.765 10-2

M pOH = 1.7533

pH = 14.00 - 1.7533 = 12.25


0

5

10

15

0 10 20 30 40 50

Volume of Base Added

pH equivalence point x

Titration of Strong Acid with Strong Base


Titration of Weak Acid with Strong Base


Effect of Acid Strength on Titration Curve


Titration of Weak Base with Strong Acid


Titration of Diprotic Weak Acid with Strong Base


pH range of Indicatorslitmus (5.0-8.0)bromothymole blue (6.0-7.6) methyl red (4.8-6.0)thymol blue (8.0-9.6) phenolphthalein (8.2-10.0) thymolphthalein (9.4-10.6)


Acid Rainacid rain is defined as rain with a pH < 5.6pH = 5.6 for rain in equilibrium with

atmospheric carbon dioxide


Sulfuric Acid from Sulfur burning

SO2

S + O2 => SO2

SO3

2 SO2 + O2 => 2 SO3

Sulfuric AcidSO3 + H2O => H2SO4


Nitric Acid2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)


How Acid Precipitation Forms


Acid Precipitation in U.S.


Solubility Productsolubility-product - the product of the solubilitiessolubility-product constant => Ksp

constant that is equal to the solubilities of the ions produced when a substance dissolves


Solubility Product Constant

In General:AxBy xA+y + yB-x

[A+y

]x [B-x

]yK =

[AxBy]

[AxBy] K = Ksp = [A+y

]x

[B-x

]y

Ksp = [A+y

]x [B

-x]y

For silver sulfate

Ag2SO4 2 Ag+

+ SO4-2

Ksp = [Ag+

]2[SO4

-2]


Solubility Product Constant Values


Dissolving Slightly Soluble Salts Using AcidsInsoluble salts containing anions of

Bronsted-Lowry bases can be dissolved in solutions of low pH


Calcium CarbonateDissolved in Acid

Limestone Dissolving in Ground WaterCaCO3(S) + H2O + CO2 => Ca+2

(aq) + 2 HCO3-(aq)

Stalactite and Stalagmite FormationCa+2

(aq) + 2 HCO3-(aq) => CaCO3(S) + H2O + CO2


The Common Ion Effectcommon ionsecond source which is completely dissociatedIn the presence of a second source of the ion,

there will be less dissolved than in its absencecommon ion effecta salt will be less soluble if one of its constitutent

ions is already present in the solution


EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?

AgCl Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.82 10-10M2

let x = molar solubility = [Ag+] = [Cl-](x)(x) = Ksp = [Ag+][Cl-] = 1.82 10-10M2

x = 1.35 10-5M


EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?

AgCl Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.82 10-10M2

let x = molar solubility = [Ag+][Cl-] = 1.0 M

Ksp = [Ag+][Cl-] = (x)(1.0M) = 1.82 10-10M2

x = 1.82 10-10M


Formation of Complexesligand - Lewis basecomplexes - product of Lewis acid-base reaction

Ag+(aq) + 2 NH3(aq) [Ag(NH3)2(aq)]+

Ag+(aq) + Cl-

(aq) AgCl(s)

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


Sodium Thiosulfate Dissolves Silver Bromide


11) For a saturated solution of a) AgCl in water:

i) Solubility equilibrium reaction:

ii) Ksp expression:


11) For a saturated solution of b) CaF2 in water:i) Solubility equilibrium reaction:

ii) Ksp expression:

c) Fe2S3 in wateri) Solubility equilibrium reaction:

ii) Ksp expression:


12) Which of following has the highest molar solubility (mole/L)?

Salt Ksp

a) CaCO3 5 × 10-9

b) PbCO 3 1.4 × 10-13

c) Li2CO3 2 × 10-3

d) NiCO3 1.2 × 10-7


Formation Constants


Amphoterism


Reactant Quotient, Q

Reactant Quotient, Qion product of the initial concentrationsame form as solubility product constant

Q < Ksp - no precipitate forms an unsaturate solutionQ > Ksp - precipitate may form to restore

condition of saturated solutionQ = Ksp - no precipitate forms, saturated solution


Will Precipitation Occur?


13) For Li2CO3, Ksp is 2 × 10-3 M3. What is the concentration of Li+ in a saturated solution of Li2CO3?


14) Chemical analysis gave [Pb2+] = 0.012 M, and [Br-] = 0.024 M in a solution. From a table, you find Ksp for PbBr2 has a value of 4 x 10-5 M3.

a) Solubility equilibrium reaction:

b) Qsp:

c) Ksp:d) Qsp < Ksp, Qsp = Ksp or Qsp > Ksp?e) Is the solution saturated, oversaturated or

unsaturated?


Kidney StonesKidney stones are normally composed of:

calcium oxalatecalcium phosphate

magnesium ammonium phsphate


Calcium OxalatePrecipitate formed from calcium ions from food rich

in calcium, dairy products, and oxalate ions from fruits and vegetables

Ca+2 + C2O4-2 CaC2O4


PRECIPITATION REACTIONS


Analysis of Silver Group

These salts formed are insoluble, they do dissolve to some SLIGHT extent.

AgCl(s) Ag+(aq) + Cl-(aq)When equilibrium has been

established, no more AgCl dissolves and the solution is SATURATED.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2



AgCl(s) Ag+(aq) + Cl-(aq)When solution is SATURATED, expt.

shows that [Ag+] = 1.67 x 10-5 M.(SOLUBILITY) of AgCl.What is [Cl-]?

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2



AgCl(s) Ag+(aq) + Cl-(aq)Saturated solution has

[Ag+] = [Cl-] = 1.67 x 10-5 MUse this to calculate Ksp

Ksp = [Ag+] [Cl-] = (1.67 x 10-5)(1.67 x 10-5) = 2.79 x 10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2


Because this is the product of “solubilities”, we call it

Ksp = solubility product constant See Table in the Text

Ksp = solubility product


Lead(II) ChloridePbCl2(s) Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5


Consider PbI2 dissolving in waterPbI2(s) Pb2+(aq) + 2 I-(aq)Calculate Ksp if solubility = 0.00130 MSolution1. Solubility = [Pb2+]

= 1.30 x 10-3 M [I-] = 2 x [Pb2+] = 2.60 x 10-3 M2. Ksp = [Pb2+] [I-]2

= [Pb2+] {2 • [Pb2+]}2

= 4 [Pb2+]3 = 4 (solubility)3

Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

Solubility of Lead(II) Iodide


Precipitating an Insoluble SaltHg2Cl2(s) Hg2

2+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

If [Hg22+] = 0.010 M, what [Cl-] is req’d to just begin

the precipitation of Hg2Cl2?

(maximum [Cl-] in 0.010 M Hg22+ without forming

Hg2Cl2?)



2+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Recognize that

Ksp = product of maximum ion concs.

Precip. begins when product of ion Concs. EXCEEDS the Ksp.



2+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Solution

[Cl-] that can exist when [Hg22+] = 0.010 M,

If this conc. of Cl- is just exceeded, Hg2Cl2 begins to precipitate.

[Cl ] = Ksp0.010

= 1.1 x 10-8 M



2+(aq) + 2 Cl-(aq)Ksp = 1.1 x 10-18

Now raise [Cl-] to 1.0 M. What is the value of [Hg2

2+] at this point?Solution[Hg2

2+] = Ksp / [Cl-]2

= Ksp / (1.0)2 = 1.1 x 10-18 MThe concentration of Hg2

2+ has been reduced by 1016 !


Separating Metal Ions Cu2+, Ag+, Pb2+

Ksp ValuesAgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrO4 1.8 x 10-14

Cu2+

Ag+

Pb2+

Cl-

Insoluble

PbCl2 AgClSoluble

CuCl2

Heat

Insoluble

AgCl

Soluble

PbCl2

CrO4-2

Insoluble

PbCrO4


Separating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO42-

to precipitate Ag2CrO4 (red) and PbCrO4 (yellow). Which precipitates first?

Ksp for Ag2CrO4 = 9.0 x 10-12

Ksp for PbCrO4 = 1.8 x 10-14

SolutionThe substance whose Ksp is first exceeded

precipitates first. The ion requiring the smaller amount of CrO4

2- ppts. first.



[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]

= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M[CrO4

2-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4 precipitates first.

Solution

Calculate [CrO42-

] required by each ion.


How much Pb2+ remains in solution when Ag+ begins to precipitate?

SolutionWe know that [CrO4

2-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4. What is the Pb2+ conc. at this point?[Pb2+] = Ksp / [CrO4

2-] = 1.8 x 10-14 / 2.3 x 10-8 M = 7.8 x 10-7 MLead ion has dropped from 0.020 M to < 10-6 M



Common Ion EffectAdding an Ion “Common” to an Equilibrium

PbCl2(s) Pb2+

(aq) + 2Cl-(aq)

NaCl Na+

(aq) + Cl- (aq)


Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

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

Solutiona) Solubility in pure water = [Ba2+] = [SO4

2-] = xKsp = [Ba2+] [SO4

2-] = x2

x = (Ksp)1/2 = 1.1 x 10-5 M

The Common Ion Effect



Ksp = 1.1 x 10-10

Solutionb) Now dissolve BaSO4 in water already containing

0.010 M Ba2+. Which way will the “common ion” shift the

equilibrium? ___ Will solubility of BaSO4 be less than or greater than

in pure water?___




Solution [Ba2+] [SO4

2-]initialchangeequilib.



Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10


Solution [Ba2+] [SO4

2-]initial 0.010 0change + y + yequilib. 0.010 + y y



Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)

Because y << x (1.1 x 10-5 M) 0.010 + y 0.010. Therefore,

Ksp = 1.1 x 10-10 = (0.010)(y)y = 1.1 x 10-8 M =

solubility in presence of added Ba2+ ion.

Le Chatelier’s Principle is followed!


chemistry 102(001) fall 2012

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