Upload
amberly-roberta-nicholson
View
229
Download
3
Tags:
Embed Size (px)
Citation preview
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
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
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 ?
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
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______.
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
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
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
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
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.
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
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
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
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
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
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
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
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
The titration curve of a strong acid with a strong base
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
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
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.
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
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
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
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
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+
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
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
VII. 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
Flame test of cations
lithium sodium potassium copper