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Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

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Page 1: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Acid-Base and Solubility Equilibria

• Common-ion effect

• Buffer solutions

• Acid-base titration

• Solubility equilibria

• Complex ion formation

• Qualitative analysis

Page 2: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

I. Common ion effect• Common ion: an ion that is produced from more than one solute• Common ion effect: shift in equilibrium caused by the addition

of a compound having an ion in common with the dissolved substance.

Example: Consider CH3COOH (weak acid, partial ionizes)

Add CH3COONa (strong electrolyte, completely ionizes)

CH3COOH (aq) H+ (aq) + CH3COO- (aq) (*)

CH3COONa (s) Na+ (aq) + CH3COO- (aq)common ion

Le Chatelier's principle: equilibrium (*) shift

This shift to the left will also result in a decrease of the [H+]

The ionization of a weak electrolyte is decreased by adding a strong electrolyte (i.e. a salt) that produces an common ion

Page 3: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Ex:What is the the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa at 25oC? (Ka=1.8 x10-5).

CH3COOH (aq) H+ (aq) + CH3COO– (aq)

Initial (M)Change (M)

Equilibrium (M)

0.20 0.00-x +x

0.20 - x

0.30+x

x 0.30+ x

Ka =x(0.30+x)0.20 - x

= 1.8 x 10-5

Ka 0.30x0.20

= 1.8 x 10-5

Assume 0.20 – x 0.20 0.30 + x 0.30

x = 1.2 x 10-5 M

[H+] = [CH3COO– ] = 1.2 x 10-5 M

pH = -log [H+] = 4.92

Page 4: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

II. Buffer solution

A. A solution that resist pH change when acids or base are added.

B. In general, a buffer is made of

– A weak acid and its salt

– A weak base and its salt

HF+NaF

NH3+NH4Cl

Add strong acid

H+ (aq) + HCOO- (aq) HCOOH (aq)

Add strong base

OH- (aq) + HCOOH (aq) HCOO- (aq) + H2O (l)

Consider an equal molar mixture of HCOOH + HCOONa

? How does buffer work ?

Page 5: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Ex. Which of the following can be used to prepare for buffer solutions? (a) NaF/HF (b) NaBr/HBr, (c)Na2SO3/NaHSO3,

and (d) CH3COOH and NaOH.

(a) HF is a weak acid and F- is its conjugate base

buffer solution

(b) HBr is a strong acidnot a buffer solution

(c) SO32- is a weak base and HSO3

- is it conjugate acidbuffer solution

(d) CH3COOH is a weak base and NaOH is a strong base

buffer solution

CH3COOH + NaOH --> CH3COONa + H2O

Page 6: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

C. pH of a Buffer solutionConsider a mixture of a weak acid (HA) and its salt (XA)

HA (aq) H+ (aq) + A- (aq)

XA (s) X+ (aq) + A- (aq)

Ka =[H+][A-]

[HA][H+] =

Ka [HA]

[A-]

-log [H+] = -log Ka - log[HA]

[A-]= - log Ka + log [A-]

[HA]

pH = pKa + log[conjugate base]

[acid]Henderson-Hasselbalch

equation

pKa = -log Ka As pKa , the acid strength______.

Page 7: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Ex: What is the the pH of a solution containing 0.20 M CH3COOH and 0.30 M CH3COONa at 25oC? (Ka=1.8 x10-5)? What is the pH after the addition of (a) 20.0 mL of 0.050 M NaOH or (b) 20.0 mL of 0.050 M HCl to 80.0 mL of this buffer solution?

Henderson-Hasselbalch eq. pH = pKa + log[conjugate base]

[acid]

pH = -log(1.8 x10-5)+ log0.300.20

= 4.74+ 0.18 = 4.92

(a) Addition of 20.0 mL of 0.050 M NaOH (0.001 mol NaOH)NaOH will react with the acid in the buffer

CH3COOH (aq) + OH- (aq) CH3COO– (aq) + H2O (l)start (moles)

end (moles)0.016 0.001 0.024

final volume = 80.0 mL+ 20.0 mL=100.0 mL

= 4.96 pH = 4.74 + log[0.25][0.15]

[CH3COOH] = =0.15 M0.015

0.10

[CH3COO–] = =0.25 M0.025

0.10

0.015 0.00 0.025

Page 8: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Continued

Henderson-Hasselbalch eq. pH = pKa + log[conjugate base]

[acid]pKa = 4.74

(b) Addition of 20.0 mL of 0.050 M HCl (0.001 mol HCl)

HCl will react with the base in the buffer

CH3COO– (aq) + H+ (aq) CH3COOH (aq)

start (moles)

end (moles)0.024 0.001 0.016

0.023 0.0 0.017

final volume = 80.0 mL+ 20.0 mL=100.0 mL

= 4.87 pH = 4.74 + log[0.23][0.17]

[CH3COOH] = =0.17 M0.017

0.10

[CH3COO–] = =0.23 M0.023

0.10

Page 9: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

D. Preparation of buffer• Select acid/base pair with pKa that is closest to the desired pH.

• Adjust ratio of base/acid to have the desired pH.

pH = pKa + log[conjugate base]

[acid]

Ex. How to prepare a buffer with pH = 3.85?

[H+] =10-3.85=1.4 x10-4

HCOOH seems to be a good choice, Ka=1.7 x10-4

pH = 3.85 = -log(1.7 x10-4) + log[HCOO–][HCOOH]

[HCOO–][HCOOH]

log =0.080[HCOO–][HCOOH]

=1.2

The buffer can be prepared by mixing 1.0 M HCOOH and 1.2 M HCOONa

Page 10: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis
Page 11: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

E. Buffer capacity

• Quantity of acid or base that can be added before pH changes significantly

– Buffers work best when the ratio of [base]/[acid] is 1.

– The greater the concentrations of both [base] and [acid], the ________ buffering capacity

Ex. Which of the following has the greater buffering capacity?

(a) 1.0 L of 1.0 M NaF + 1.0 M HF solution, or

(b) 1.0 L of 0.1 M NaF + 0.1M HF solution

greater

Page 12: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

III. Titration

• Acid-base titration: to determine the concentration of unknown solution

• Standard solution (titrant): a solution of known concentration

• Equivalence point : the point at which the reaction is complete

• Indicator : a substance changes color at/near the equivalence point.

Page 13: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

A. Strong acid-strong base titration

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

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

Titration curve: plot of pH as a function of mL

Page 14: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q. A 40.0 mL sample of 0.200 M HCl is titrated with 0.200 M NaOH. Calculate the pH of the soution after the following volume of NaOH have been added. (a) 35.0 mL, (b) 39.0 mL, (c) 41.0 mL, and (d) 45.0 mL.

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

(a) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0350 L x 0.200 M = 0.00700 molexcess HCl: 0.00800 mol - 0.00700 mol = 0.00100 molFinal volume = (40.0 + 35.0)mL =75.0 mL[H+]=0.00100 mol/ 0.0750 L = 0.0133 MpH= -log[H+]= 1.875

(b) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0390 L x 0.200 M = 0.00780 molexcess HCl: 0.00800 mol - 0.00780 mol = 0.00020 molFinal volume = (40.0 + 39.0)mL =79.0 mL[H+]=0.00020 mol/ 0.0790 L = 0.0025 MpH= -log[H+]= 2.60

Initial pH = 0.699

Page 15: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

continued

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

(c) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0410 L x 0.200 M = 0.00820 molexcess NaOH: 0.00820 mol - 0.00800 mol = 0.00020 molFinal volume = (40.0 + 41.0)mL =81.0 mL[OH–]=0.00020 mol/ 0.0810 L = 0.0025 MpOH= -log[OH–]= 2.61 pH=14.0 -pOH = 11.39

(d) mol HCl: 0.0400 L x 0.200 M = 0.00800 mol mol NaOH: 0.0450 L x 0.200 M = 0.00900 molexcess NaOH: 0.00900 mol - 0.00800 mol = 0.00100 molFinal volume = (40.0 + 45.0)mL =85.0 mL[OH–]=0.00100 mol/ 0.0850 L = 0.0118 MpOH= -log[OH–]= 1.929 pH=14.0 -pOH = 12.071

Page 16: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

B. Weak acid-strong base titration

CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

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

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

At equivalence point (pH > 7):

Buffering region

Page 17: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q. Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)end (moles)

0.01 0.010.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x 0.05 x 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Page 18: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

B. Weak acid-strong base titration curve

1. Initial pH of a weak acid2. Before equivalence point: buffering region3. At the equivalence point: a weak base4. After the equivalence point: excess strong base raises pH

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100

NaOH added (mL)

pH

Ka=2 x10-6

Ka=2 x10-8

Ka=2 x10-4

pH=pKa

equivalence point

50 mL of 0.1 M weak acid is titrated with 0.1 M NaOH

Page 19: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

C. Strong acid-weak base titration

HCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

At equivalence point (pH < 7):

H+ (aq) + NH3 (aq) NH4Cl (aq)

Buffering region

Page 20: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

IV. Acid-base indicator

HIn (aq) H+ (aq) + In- (aq)

10[HIn]

[In-]Color of acid (HIn) predominates

0.1[HIn]

[In-]Color of conjugate base (In-) predominates

Page 21: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

The titration curve of a strong acid with a strong base

Page 22: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q: Which indicator(s) would you use for a titration of HCOOH with NaOH ?

Weak acid titrated with strong base.At equivalence point, will have conjugate base of weak acid.

At equivalence point, pH > 7

Use cresol red or phenolphthalein

Page 23: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

V. Solubility equilibriaA. Solubility product constant, Ksp

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

CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp = [Ca2+][F-]2

Ag2SO4 (s) 2Ag+ (aq) + SO42- (aq) Ksp = [Ag+]2[SO4

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Page 24: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

B. Solubility and KSP

a. Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solutionQ < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

b. Determine Ksp from solubility or solubility from Ksp

Molar solubility (mol/L): moles of solute in 1 L of a saturated solution

Solubility (g/L): grams of solute dissolved in 1 L of a saturated solution.

Page 25: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q. What is the solubility of silver chloride in g/L ? Ksp=1.6x10-10

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

Ksp = [Ag+][Cl-]

Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2 s = Ksp s = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl =

1.3 x 10-5 mol AgCl1 L soln

143.35 g AgCl1 mol AgCl

x = 1.9 x 10-3 g/L

Page 26: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q.The solubility of SrF2 is found to be 1.1 x 10-2 g in 100 mL of aqueous solution. Determine the value of Ksp

of SrF2.

SrF2 (s) Sr2+ (aq) + 2 F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+2s

s 2s

Ksp = [Sr2+][F–]2 = s(2s)2

s: Solubility of SrF2 in M =

-s

1.1 x10-2 g SrF2/(125.6 g/mol)0.100 L soln

= 4 s3

= 8.74 x 10–4 M

= 2.7 x 10-9

Page 27: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis
Page 28: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form? (Ksp = 8.0 x10-6)

Ca(OH)2 (s) Ca2+ (aq) + 2 OH- (aq)

Which ions are present in solution?

Only possible precipitate is _________(solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M x 1.00 L/(1.00 L +0.00200 L)=0.100 M

[OH-]0 = 0.00200 L x 0.200 M /(1.00L + 0.00200 L)=4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

Na+, OH-, Ca2+, Cl-.

Ca(OH)2

Page 29: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

C. Factors affecting solubility

1. Common ion effect: common ion decreases the solubility.

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

If NaF is added, equilibrium shifts __________.left

Ex. What is the molar solubility of CaF2 in (a) pure water and (b) 0.010 M NaF? (Ksp=3.9 x10-11)

Ksp = 3.9 x 10-11 =s(2s)2 = 4s3

s = 2.1 x 10-4

NaF (s) Na+ (aq) + F- (aq) [F-] = 0.010 M

[F-] = 0.010 + 2s 0.010Ksp = (0.010)2 x s

s = 3.9 x 10-7

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

(b)

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

Initial (M)

Change (M)

Equilibrium (M)

0.00 0.010

s 0.010+2s

-s +s +2s

Page 30: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

2. Effect of pH

• Insoluble bases dissolve in acidic solutions• Insoluble acids dissolve in basic solutions

removeadd

In acidic solution, solubility_________.

Why?

increases

CaF2(s) +2 H+ (aq) Ca2+ (aq) + 2 HF (aq)

Mg(OH)2(s) Mg2+ (aq) + 2OH- (aq)Ex.

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

F–(aq) + H+ (aq) HF (aq)

Ex. Which salt will be more soluble in acidic solution than in pure water (a) ZnCO3, (b) AgCN, and (c) BiI3.

(a) and (b) because CO32– and CN– react with H+

Page 31: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Q. What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.010 M?

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

Ksp = [Ag+][Cl-]

Ksp = 1.6 x 10-10

AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13

Ksp = [Ag+][Br-]

[Ag+] = Ksp

[Br-]7.7 x 10-13

0.010= = 7.7 x 10-11 M

[Ag+] = Ksp

[Cl-]1.6 x 10-10

0.010= = 1.6 x 10-8 M

7.7 x 10-11 M < [Ag+] < 1.6 x 10-8 M

Page 32: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

VI. Complex ion equilibria and solubility

1. Complex ion: an ion containing a central metal cation (Lewis acid) bonded to one or more molecules or ions (Lewis base).

2. Solubility of slightly soluble salts can be increased by complex ion formation AgCl (s) Ag+ (aq) + Cl– (aq)

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

By adding NH3, solubility of AgCl increases due to

3. The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation.

Kf = =1.7 x107[Ag(NH3)2

+ ]

[Ag+][NH3]2

Kf

stability of complex

Page 33: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis
Page 34: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

VII. Qualitative analysis

Page 35: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Solution containsNa+, K+, NH ions

4+

Solution containing ionsof remaining groups

Solution containing ionsof all cation groups

Solution containing ionsof remaining groups

Group 2 precipitatesCuS, CdS, SnS, Bi2S3

Group 1 precipitatesAgCl, Hg2Cl2, PbCl2

Solution containing ionsof remaining groups

Filtration+HCl

Filtration

+H2S

Group 4 precipitatesBaCO3, CaCO3, SrCO3

Filtration

+Na2CO3

Group 3 precipitatesCoS, FeS, MnS, NiSZnS, Al(OH)3,Cr(OH)3

Filtration+NaOH

Flow chart for separation of cations

Page 36: Acid-Base and Solubility Equilibria Common-ion effect Buffer solutions Acid-base titration Solubility equilibria Complex ion formation Qualitative analysis

Flame test of cations

lithium sodium potassium copper