Solubility Products, Solubility Products, KKsp, and Ion Products sp, and Ion Products

QQsp sp

Ch 15, Part 4Ch 15, Part 4

The Other Half of the The Other Half of the PrankPrank

Take-Home Test PostedTake-Home Test Posted

Due Wed NightDue Wed Night

heterogeneous heterogeneous equilibriumequilibrium

solubility products, solubility products, KKspsp: : Equilibrium Expression for Equilibrium Expression for KKsp sp

and and QQsp sp

Equilibrium Expression for Ksp and Qsp

AgCl(s) = Ag+(aq) + Cl-(aq) [Ag+] [Cl-]

CaCO3(s) = Ca2+(aq) + CO32-(aq) [Ca2+] [CO3

-]

Li2CO3(s) = 2 Li+(aq) + CO32-(aq) [Li+]2 [CO3

-]

KKspsp and Q and Qspsp

KKspsp is the equilibrium expression used is the equilibrium expression used when the solution is saturatedwhen the solution is saturated

When the solution is NOT saturated, QWhen the solution is NOT saturated, Qspsp is usually used to describe the is usually used to describe the equilibriumequilibrium– QQspsp is called the ION PRODUCT QUOTIENT is called the ION PRODUCT QUOTIENT

– Equilibrium expression is the sameEquilibrium expression is the same– If solution is not saturated, precipitate will If solution is not saturated, precipitate will

not form.not form.

Comparing QComparing Qspsp and K and Kspsp

Qsp < Ksp Unsaturated solution

Qsp = Ksp Saturate solution

Qsp > Ksp Oversaturate solution

Molar Solubilities and Solubility Molar Solubilities and Solubility Products Products

Solubility products, Solubility products, KKsp, of salts are indirect indication of sp, of salts are indirect indication of their solubilities expressed in mol/L (called their solubilities expressed in mol/L (called molar molar solubilitysolubility). ). – However, the solubility products are more useful than molar However, the solubility products are more useful than molar

solubility. solubility. – The molar solubilities are affected when there are common ions The molar solubilities are affected when there are common ions

present in the solution. present in the solution. – We need to employ the solubility products to estimate the molar We need to employ the solubility products to estimate the molar

solubilities in these cases. solubilities in these cases. – When a salt is dissolved in pure water, solubility products and When a salt is dissolved in pure water, solubility products and

molar solubilities are related. This is illustrated using calcium molar solubilities are related. This is illustrated using calcium carbonate. If carbonate. If xx is the concentration of Ca is the concentration of Ca2+2+ (= [CO (= [CO33

2-2-]) in the ]) in the saturated solution, then saturated solution, then

KKsp = sp = xx2 2 – Because Ksp = [CaBecause Ksp = [Ca2+2+] [CO] [CO33

2-2-]]

Example 1Example 1

The The KKsp for AgCl is 1.8e-10 Msp for AgCl is 1.8e-10 M22. . – What is the molar solubility of AgCl What is the molar solubility of AgCl

in pure water? in pure water? SolutionSolution

– Let Let xx be the molar solubility, then be the molar solubility, then – AgCl = AgAgCl = Ag++ + Cl + Cl--

xx xx

– xx = (1.8e-10 M = (1.8e-10 M22))1/21/2 = 1.3e-5 M = 1.3e-5 M

Molar SolubilityMolar Solubility The solubility product, The solubility product, KKsp is a better sp is a better

indicator than the usual solubility indicator than the usual solubility specification of g per 100 mL of solvent or specification of g per 100 mL of solvent or moles per unit volume of solvent. moles per unit volume of solvent.

For the AgCl case, when the cation For the AgCl case, when the cation concentration is not the same as the anion concentration is not the same as the anion concentration ([Ag+] =/= [Cl-]) solubility concentration ([Ag+] =/= [Cl-]) solubility of AgCl can not be defined in terms of of AgCl can not be defined in terms of moles per L. moles per L.

Example 1Example 1– In this case, the system can be In this case, the system can be

divided into three zones. The divided into three zones. The condition [Ag+] [Cl-] = condition [Ag+] [Cl-] = KKsp, is sp, is represented by a line which represented by a line which divides the plane into two divides the plane into two zones. zones.

– When [Ag+] [Cl-] < When [Ag+] [Cl-] < KKsp, no sp, no precipitate will be formed. precipitate will be formed.

– When [Ag+] [Cl-] > When [Ag+] [Cl-] > KKsp, a sp, a precipitate will be formed. precipitate will be formed.

– When AgCl and NaCl dissolve in When AgCl and NaCl dissolve in a solution, both salts give Cl- a solution, both salts give Cl- ions. The effect of [Cl-] on the ions. The effect of [Cl-] on the solubility of AgCl is called the solubility of AgCl is called the Common ion effectCommon ion effect

Example 2Example 2 The The KKsp for Agsp for Ag22CrOCrO44 is 9e-12 M is 9e-12 M33. .

– What is the molar solubility of AgWhat is the molar solubility of Ag22CrOCrO44 in pure water? in pure water? SolutionSolution

Let Let xx be the molar solubility of Ag be the molar solubility of Ag22CrOCrO44, then , then – AgAg22CrOCrO44 = 2 Ag = 2 Ag+ + + CrO+ CrO44

2- 2-

22xx xx

– (2 (2 xx))22(x) = (x) = KKsp sp – xx = {(9e-12)/4} = {(9e-12)/4}(1/3) (1/3) = 1.3e-4 M = 1.3e-4 M – [Ag+] = 2.6e-4 M and the molar solubility is 1.3e-4 M. [Ag+] = 2.6e-4 M and the molar solubility is 1.3e-4 M.

A diagram similar to AgCl can be drawn but A diagram similar to AgCl can be drawn but shape of the curve is differentshape of the curve is different

Example 3Example 3 What is the pH in a saturated solution of What is the pH in a saturated solution of

Ca(OH)Ca(OH)22??KKsp = 5.5e-6 M3 for Ca(OH)sp = 5.5e-6 M3 for Ca(OH)22. .

Let [CaLet [Ca2+2+] = ] = xx, then [OH, then [OH--] = 2 ] = 2 xx. The . The equilibrium and concentration are equilibrium and concentration are represented below: represented below:

Ca(OH)2 = CaCa(OH)2 = Ca2+2+ + 2 OH + 2 OH-- xx 2 2 xx

xx = {5.5e-6/4} = {5.5e-6/4}(1/3)(1/3) [OH [OH--] = 2 ] = 2 xx..Note the stoichiometry of equilibrium.Note the stoichiometry of equilibrium.

Answer 12.35Answer 12.35

Example 4Example 4

At 760◦C, Kc = 33.3 for the reaction– PCl5(g) ⇀↽ PCl3(g) + Cl2(g)

– If a mixture that consists of 0.54 mol PCl3 and 0.85 mol Cl2 is placed in a 9 L reaction vessel and heated to 760◦C, what is the equilibrium composition of PCl5?

Initially, – [PCl3] = 0.54 mol/9 L = 0.06 mol/L

– [Cl2] = 0.85 mol/9 L = 0.0944444 mol/L

Example 4Example 4

33.3 x = x2 − 0.154444 x + 0.005666670 = x2 − 33.4544 + 0.00566667

Since 33.4543 is larger than the initial con-centrations, it can be discarded.Thus [PCl5] = 0.000169385 mol/L

AP Question #1AP Question #1

Propanoic acid, HCPropanoic acid, HC33HH55OO22, ionizes in , ionizes in water according to the equation below.water according to the equation below.

– HCHC33HH55OO22(aq) ↔ C(aq) ↔ C22HH55OO22--(aq) + H(aq) + H++(aq)(aq)

– Ka = 1.34 x 10Ka = 1.34 x 10-5-5

A.A. Write the equilibrium expressionWrite the equilibrium expression

AP Question #1AP Question #1

Propanoic acid, HCPropanoic acid, HC33HH55OO22, ionizes in , ionizes in water according to the equation below.water according to the equation below.

– HCHC33HH55OO22(aq) ↔ C(aq) ↔ C22HH55OO22-(aq) + H+(aq)-(aq) + H+(aq)

– Ka = 1.34 x 10-5Ka = 1.34 x 10-5

A.A. Write the equilibrium expressionWrite the equilibrium expression

][

]][[

253

253

OHHC

HOHCKa

AP Example #1bAP Example #1bHC3H5O2(aq) ↔ C2H5O2

-(aq) + H+

(aq)

0.265 0 0

-x +x +x

0.265-x X x

726.2)00188.0log(]log[

00188.0][][

)265.0(1034.1

][

]][[

253

25

253

253

HpH

MHOHCx

x

xx

OHHC

HOHCKa

AP Example #1AP Example #1

B.B. Calculate the pH of a 0.265 M solution of Calculate the pH of a 0.265 M solution of propanoic acidpropanoic acid

A 0.496 g sample of sodium propanate, A 0.496 g sample of sodium propanate, NaC3H5O2, is added to 50.0 mL sample of a NaC3H5O2, is added to 50.0 mL sample of a 0.265 M solution of propanoic acid. 0.265 M solution of propanoic acid. Assuming no change in volume of the Assuming no change in volume of the solution occurs, calculate each of the solution occurs, calculate each of the following:following:

– The concentration of the propanate ion, The concentration of the propanate ion, C3H5O2- (aq), in the solution.C3H5O2- (aq), in the solution.

– The concentration of the H+ (aq) in the solution.The concentration of the H+ (aq) in the solution.

AP Example #1CAP Example #1C

253253

253253

253253

253253

25.131

0.50

1

265.0::

103.00.50

17.5][

17.51

1000

0.96

1496.0

OHmmolHCmL

xL

OHmolHCmolesMV

MmL

OHmmolCOHC

OHmmolCmol

mmolx

OHgNaC

OHmolNaCx

OHgNaC

HC3H5O2(aq) ↔ C2H5O2-(aq) + H+

(aq)

13.25 5.17 0

-x +x +x

13.25-x 5.17+x x

AP Example #1cAP Example #1c

MxH

OHHC

OHCpKapH

KapKa

pH 5

253

253

1043.3][10

464.425.13

17.5log873.4

][

][log

873.4)log(

AP Example #2AP Example #2

The methanote ion, HCO2- (aq) reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation.HCO2

- (aq) + H2O (l) HCO2H (aq) + OH- (aq)

1) Given that [OH-] is 4.18 x 10-6 M in a 0.309 M solution of sodium methanoate, calculate each of the following

I)The value of Kb for the methanoate ion, HCO2- (aq)

II)The value of Ka for methanoic acid, HCO2H

4

1114

1166

2

2

2

62

1077.1

)1065.5)((100.1

1065.5309.0

)1018.4)(1018.4(

][

]][[

309.0][

1018.4][][

xK

xKxKxKK

xxx

HCO

OHHHCOK

MHCO

xOHHHCO

a

awba

b

AP Example #3 (AP Example #3 (2003)2003) 1.1. CC66HH55NHNH22(aq)(aq) + H + H22OO(l)(l) C C66HH55NHNH33

++(aq)(aq) + + OHOH––(aq) (aq) Aniline, a weak base, reacts with Aniline, a weak base, reacts with water according to the reaction represented water according to the reaction represented above.above.(a)(a) Write the equilibrium constant expression, Write the equilibrium constant expression,

KKbb, for the reaction represented above., for the reaction represented above.(b)(b) A sample of aniline is dissolved in water to A sample of aniline is dissolved in water to

produce 25.0 mL of 0.10 produce 25.0 mL of 0.10 MM solution. The pH of the solution. The pH of the solution is 8.82. Calculate the equilibrium solution is 8.82. Calculate the equilibrium constant, Kconstant, Kbb, for this reaction., for this reaction.

(c)(c) The solution prepared in part (b) is titrated The solution prepared in part (b) is titrated with 0.10 with 0.10 MM HCl. Calculate the pH of the solution HCl. Calculate the pH of the solution when 5.0 mL of the acid has been titrated.when 5.0 mL of the acid has been titrated.

(d)(d) Calculate the pH at the equivalence point of Calculate the pH at the equivalence point of the titration in part (c).the titration in part (c).

(e)(e) The pThe pKKaa values for several indicators are values for several indicators are given below. Which of the indicators listed is most given below. Which of the indicators listed is most suitable for this titration? Justify your answer.suitable for this titration? Justify your answer.

AP Example #3 (2003)AP Example #3 (2003)

Indicator pKa

Erythrosine 3

Litmus 7

Thymolphthalein

10

(e)(e) The pThe pKKa values for several indicators are a values for several indicators are given below. Which of the indicators listed is most given below. Which of the indicators listed is most suitable for this titration? Justify your answer.suitable for this titration? Justify your answer.

AP Example #3 AnswersAP Example #3 Answers

C6H5NH3 OH

C6H5NH2

6.6110 6 2

0.10 – 6.6110 6

(a) Kb =

(b) pOH = 14 – pH = 14 – 8.82 = 5.18-log[OH–] = 5.18; [OH–] = 6.6110–6 M[OH–] = [C6H5NH3+]

Kb = = 4.410–10

AP Example #3 (Answers)AP Example #3 (Answers)

X X 0.50 mmol

30.0 mL

20.0 mmol30.0 mL

110 14

1.810 9

X 2

(0.050 X )

(c) 25 mL = 2.5 mmol C6H5NH2

5 mL = 0.5 mmol H+ added2.0 mmol base remains in 30.0 mL solution

4.410–10 =

X = 1.8010–9 = [OH–][H+] =

(d) when neutralized, there are 2.5 mmol of C6H5NH3+ in 50.0 mL of solution, giving

a [C6H5NH3+] = 0.050 M his cation will partially ionize according to the following

equilibrium:C6H5NH3+(aq) C6H5NH2(aq) + H+(aq)

at equilibrium, [C6H5NH2] = [H+] = X

[C6H5NH3+] = (0.050–X) = Ka = 2.310-5

X = 1.0610–3 = [H+]pH = –log[H+] = 2.98

= 5.6*10–6; pH = 5.26

AP Example #3 (Answers)AP Example #3 (Answers)

(e)(e) erythrosine; the indicator will erythrosine; the indicator will change color when the pH is near its change color when the pH is near its ppKKa, since the equivalence point is a, since the equivalence point is near pH 3, the indicator must have a near pH 3, the indicator must have a ppKKa near this value.a near this value.

AP Example #4 (2004)AP Example #4 (2004)1. Answer the following questions relating to the 1. Answer the following questions relating to the

solubilities of two silver compounds, Ag2CrO4 solubilities of two silver compounds, Ag2CrO4 and Ag3PO4.and Ag3PO4.

Silver chromate dissociates in water according to Silver chromate dissociates in water according to the equation shown below.the equation shown below.Ag2CrO4(Ag2CrO4(ss) ) 2 Ag+( 2 Ag+(aqaq) + CrO42–() + CrO42–(aqaq) ) Ksp Ksp = 2.6 × 10–12 at 25°C= 2.6 × 10–12 at 25°C

a)a) Write the equilibrium-constant expression for the Write the equilibrium-constant expression for the dissolving of Ag2CrO4(dissolving of Ag2CrO4(ss).).

b)b) Calculate the concentration, in mol L-1, of Ag+Calculate the concentration, in mol L-1, of Ag+((aqaq) in a saturated solution of Ag2CrO4 at 25°C.) in a saturated solution of Ag2CrO4 at 25°C.

c)c) Calculate the maximum mass, in grams, of Calculate the maximum mass, in grams, of Ag2CrO4 that can dissolve in 100. mL of water at Ag2CrO4 that can dissolve in 100. mL of water at 25°C.25°C.

AP Example #4 (2004)AP Example #4 (2004)1. Answer the following questions relating to the solubilities 1. Answer the following questions relating to the solubilities

of two silver compounds, Agof two silver compounds, Ag22CrOCrO44 and Ag and Ag33POPO44..Silver chromate dissociates in water according to the Silver chromate dissociates in water according to the

equation shown below.equation shown below.AgAg22CrOCrO44((ss) ) 2 Ag 2 Ag++((aqaq) + CrO) + CrO44

2–2–((aqaq) ) Ksp Ksp = 2.6 × 10–12 at 25°C= 2.6 × 10–12 at 25°Cd)d) A 0.100 mol sample of solid AgNO3 is added to a 1.00 L A 0.100 mol sample of solid AgNO3 is added to a 1.00 L

saturated solution of Agsaturated solution of Ag22CrOCrO44 . Assuming no volume change, . Assuming no volume change, does [CrOdoes [CrO44

2–2–] increase, decrease, or remain the same? Justify ] increase, decrease, or remain the same? Justify your answer.your answer.

In a saturated solution of AgIn a saturated solution of Ag33POPO44 at 25°C, the concentration at 25°C, the concentration of Agof Ag++((aqaq) is 5.3 × 10) is 5.3 × 10–5–5 M. M. The equilibrium constant The equilibrium constant expression for the dissolving of Agexpression for the dissolving of Ag33POPO44((ss) in water is ) in water is shown below.shown below.Ksp Ksp = [Ag= [Ag++]]33 [PO [PO44

3-3-]]d)d) Write the balanced equation for the dissolving of AgWrite the balanced equation for the dissolving of Ag33POPO44 in water. in water.

Calculate the value of Calculate the value of Ksp Ksp for Agfor Ag33POPO44 at 25°C. at 25°C.e)e) A 1.00 L sample of saturated AgA 1.00 L sample of saturated Ag33POPO44 solution is allowed to solution is allowed to

evaporate at 25°C to a final volume of 500. mL. What is [Agevaporate at 25°C to a final volume of 500. mL. What is [Ag++] in ] in the solution? Justify your answer.the solution? Justify your answer.

AP Example #4 (Answers)AP Example #4 (Answers)

a) Write the equilibrium constant a) Write the equilibrium constant expression for the dissolving of expression for the dissolving of Ag2CrO4.Ag2CrO4.

Ksp = [AgKsp = [Ag2+2+]]22[CrO[CrO442-2-]]

b) Calculate the concentration in b) Calculate the concentration in mol L-1, of Agmol L-1, of Ag++ in a saturated in a saturated solution of Agsolution of Ag22CrOCrO44 at 25˚C. at 25˚C.

Ksp [Ag]2[CrO42 ]

2.6x10 12 [2s]2[s] 4s3

s 8.7x10 5

[Ag] 2s 1.7x10 4 M

AP Example #4 (Answers)AP Example #4 (Answers)

c) Calculate the maximum mass in c) Calculate the maximum mass in grams of Aggrams of Ag22CrOCrO44 that can dissolve in that can dissolve in 100. mL of water at 25˚C.100. mL of water at 25˚C.

gmol

gx

Lx

L

CrOmolAgx0029.0

1

332

1

100.0

1

107.8 425

d) A 0.100 mol sample of solid AgNO3 is added to 1.00 L saturated solution of Ag2CrO4. Assuming no volume change, does [CrO4

2-] increase, decrease, or remain the same? Justify your answer.

Common ion effect. Adding AgNO3 is adding Ag+ to the product side. This will shift the equilibrium to the left (LeChatliers Principle). Thus the concentration of the chromate will go down. You could justify this with math but since they don’t ask it then don’t.

AP Example #5 (2001)AP Example #5 (2001) 1.1. Answer the following questions Answer the following questions

relating to the solubility of the chlorides relating to the solubility of the chlorides of silver and lead.of silver and lead.(a)(a) At 10At 10C, 8.9 C, 8.9 10-5 g of AgCl 10-5 g of AgCl(s)(s) will will

dissolve in 100. mL of water.dissolve in 100. mL of water.(i)(i) Write the equation for the Write the equation for the

dissociation of AgCldissociation of AgCl(s)(s) in water. in water.

(ii) Calculate the solubility, in mol L–1, of AgCl(ii) Calculate the solubility, in mol L–1, of AgCl(s)(s) in water at 10in water at 10C.C.

(iii) Calculate the value of the solubility-product (iii) Calculate the value of the solubility-product constant, constant, KspKsp for AgCl for AgCl(s)(s) at 10 at 10C.C.

AP Example #5 (2001)AP Example #5 (2001)(b)(b) At 25At 25C, the value of C, the value of KspKsp for PbCl2 for PbCl2(s)(s) is is

1.6 1.6 10-5 and the value of 10-5 and the value of KspKsp for AgCl for AgCl(s)(s) is is 1.8 1.8 10-10. 10-10.

(i)(i) If 60.0 mL of 0.0400 If 60.0 mL of 0.0400 MM NaCl NaCl(aq)(aq) is is added to 60.0 mL of 0.0300 added to 60.0 mL of 0.0300 MM Pb(NO3)2 Pb(NO3)2(aq)(aq), will a , will a precipitate form? Assume that volumes are additive. precipitate form? Assume that volumes are additive. Show calculations to support your answer.Show calculations to support your answer.

(ii)(ii) Calculate the equilibrium value of Calculate the equilibrium value of [Pb2+[Pb2+(aq)(aq)] in 1.00 L of saturated PbCl2 solution to ] in 1.00 L of saturated PbCl2 solution to which 0.250 mole of NaClwhich 0.250 mole of NaCl(s)(s) has been added. Assume has been added. Assume that no volume change occurs.that no volume change occurs.

(iii)(iii) If 0.100 If 0.100 MM NaCl NaCl(aq)(aq) is added slowly to a is added slowly to a beaker containing both 0.120 beaker containing both 0.120 MM AgNO3 AgNO3(aq)(aq) and 0.150 and 0.150 MM Pb(NO3)2 Pb(NO3)2(aq)(aq) at 25 at 25C, which will precipitate first, C, which will precipitate first, AgClAgCl(s)(s) or PbCl2 or PbCl2(s)(s)? Show calculations to support your ? Show calculations to support your answer.answer.

AP Example #5 (2001)AP Example #5 (2001)(a)(a) At 10° C, 8.9 x10-5 g of AgCl(s) will At 10° C, 8.9 x10-5 g of AgCl(s) will

dissolve in 100 mL of water.dissolve in 100 mL of water.(i)(i) Write the equation for the dissociation of Write the equation for the dissociation of

AgCl(s) in water.AgCl(s) in water.

AgCl(s) <===> AgAgCl(s) <===> Ag++ + Cl + Cl--

(ii)(ii) Calculate the solubility, in mol/L, of Calculate the solubility, in mol/L, of AgCl(s) in water at 10° C.AgCl(s) in water at 10° C.

(8.9 x10-5 g AgCl)(mol AgCl/169 g AgCl) = 5.27 (8.9 x10-5 g AgCl)(mol AgCl/169 g AgCl) = 5.27 x10-7 molx10-7 mol

5.27 x10-7 mol/0.1 L = 5.27 x10-6 M5.27 x10-7 mol/0.1 L = 5.27 x10-6 M(iii)(iii) Calculate the value of the solubility Calculate the value of the solubility

product constant, Ksp, for AgCl(s) at 10° C.product constant, Ksp, for AgCl(s) at 10° C.Ksp = [Ag +][Cl -] = [5.27 x10-6]2 = 2.78 Ksp = [Ag +][Cl -] = [5.27 x10-6]2 = 2.78

x10-11x10-11

AP Example #5 (2001)AP Example #5 (2001) (b)(b) At 25° C, the value of Ksp for PbClAt 25° C, the value of Ksp for PbCl22(s) is (s) is

1.6 x101.6 x10-5-5 and the value for AgCl(s) is 1.8 x10 and the value for AgCl(s) is 1.8 x10--

1010.. (i)(i) If 60.0 mL of 0.0400 M NaCl(aq) is If 60.0 mL of 0.0400 M NaCl(aq) is

added to 60.0 mL of 0.0300 M Pb(NOadded to 60.0 mL of 0.0300 M Pb(NO33))22(aq), (aq), will a precipitate form? Assume that will a precipitate form? Assume that volumes are additive. Show calculations to volumes are additive. Show calculations to support your answer.support your answer.– (0.06 L)(0.04 M) = 2.4 x10(0.06 L)(0.04 M) = 2.4 x10-3-3 mol Cl mol Cl--/0.120 L - = /0.120 L - =

0.02 mol M Cl0.02 mol M Cl--

– (0.06 L)(0.03 M) = 1.8 x10(0.06 L)(0.03 M) = 1.8 x10-3-3 mol Pb mol Pb2+2+/0.120 L = /0.120 L = 0.015 M Pb0.015 M Pb2+2+

– [Pb[Pb2+2+][Cl][Cl--]]22 = Q = [0.015][0.02]2 = 6.0 x10 = Q = [0.015][0.02]2 = 6.0 x10-6-6

– No pptNo ppt 1.6 x101.6 x10-5-5 > 6.0 x10 > 6.0 x10-6-6

AP Example #5 (2001)AP Example #5 (2001)(ii)(ii) Calculate the equilibrium value of Calculate the equilibrium value of

[Pb[Pb22(aq)(aq)] in 1.00 L of saturated PbCl] in 1.00 L of saturated PbCl22 solution to solution to

which 0.2050 mole of NaClwhich 0.2050 mole of NaCl(s)(s) has been added. has been added. Assume that no volume change occurs.Assume that no volume change occurs.1.6 x101.6 x10-5-5 = [Pb = [Pb2+2+][0.25]][0.25]22 = 2.56 x10 = 2.56 x10-4-4

(iii)(iii) If 0.100 M is added slowly to a beaker If 0.100 M is added slowly to a beaker containing both 0.120 M AgNOcontaining both 0.120 M AgNO3(aq)3(aq) and 0.150 and 0.150 M Pb(NOM Pb(NO33))22 at 25° C, which will precipitate at 25° C, which will precipitate first, AgClfirst, AgCl(s)(s) or PbCl or PbCl2(s)2(s)? Show calculations to ? Show calculations to support your answer.support your answer.QQAgClAgCl = [0.12][0.1]2 = 1.2 x10 = [0.12][0.1]2 = 1.2 x10-3-3

QQPbCl2PbCl2 = [0.15][0.1] = [0.15][0.1]22 = 1.5 x10 = 1.5 x10-3-3

AP Example #6AP Example #6

5. Answer the questions below that relate to the five aqueous solutions at 25C shown above.

(a) Which solution has the highest boiling point? Explain.(b) Which solution has the highest pH? Explain.(c) Identify a pair of the solutions that would produce a precipitate when mixed together. Write the formula of the precipitate.(d) Which solution could be used to oxidize the Cl–(aq) ion? Identify the product of the oxidation.(e) Which solution would be the least effective conductor of electricity? Explain.

AP Example #6AP Example #6(a)(a) Which solution has the highest boiling Which solution has the highest boiling

point? Explain.point? Explain.Pb(NO3)2:Pb(NO3)2: greatest number of moles of particlesgreatest number of moles of particles

(b)(b) Which solution has the highest pH? Which solution has the highest pH? ExplainExplainC2H3O2 - + H2O <==> HC2H3O2 + OH -C2H3O2 - + H2O <==> HC2H3O2 + OH -

(c)(c) Identify a pair of solutions that would Identify a pair of solutions that would produce a precipitate when mixed together. produce a precipitate when mixed together. Write the formula of the precipitate.Write the formula of the precipitate.Pb2+ + Cl - ===> PbCl2Pb2+ + Cl - ===> PbCl2

(d)(d) Which solution could be used to oxidize Which solution could be used to oxidize the Cl -(aq) ion? Identify the product of the the Cl -(aq) ion? Identify the product of the solution.solution.Cl - + MnO4 ===> Cl2 + Mn 2+Cl - + MnO4 ===> Cl2 + Mn 2+

(e)(e) Which solution would be the least Which solution would be the least effective conductor of electricity? Explain.effective conductor of electricity? Explain.C2H5OH:C2H5OH: stays in solution as a molecular speciesstays in solution as a molecular species