acid-base equilibria
DESCRIPTION
Acid-Base Equilibria. Chapter 16. The presence of a common ion suppresses the ionization of a weak acid or a weak base. CH 3 COONa ( s ) Na + ( aq ) + CH 3 COO - ( aq ). common ion. CH 3 COOH ( aq ) H + ( aq ) + CH 3 COO - ( aq ). - PowerPoint PPT PresentationTRANSCRIPT
Acid-Base Equilibria
Chapter 16
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
common ion
16.2
A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
16.3
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
Consider an equal molar mixture of CH3COOH and CH3COONa
Adding more acid creates a shift left IF enough acetate ions are present
Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) KF is a weak acid and F- is its conjugate basebuffer solution
(b) HCl is a strong acidnot a buffer solution
(c) CO32- is a weak base and HCO3
- is it conjugate acidbuffer solution
16.3
What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30 0.00
-x +x
0.30 - x
0.52
+x
x 0.52 + x
16.2
Mixture of weak acid and conjugate base!
KKaa for HCOOH = 1.8 x 10 for HCOOH = 1.8 x 10 -4-4
[H+] [HCOO-]Ka = [HCOOH]
x = 1.038 X 10 -4
pH = 3.98
OR…… Use the Henderson-Hasselbach equationConsider mixture of salt NaA and weak acid HA.
HA (aq) H+ (aq) + A- (aq)
NaA (s) Na+ (aq) + A- (aq)Ka =
[H+][A-][HA]
[H+] =Ka [HA]
[A-]
-log [H+] = -log Ka - log [HA][A-]
-log [H+] = -log Ka + log [A-][HA]
pH = pKa + log [A-][HA]
pKa = -log Ka
Henderson-Hasselbach equation
16.2
pH = pKa + log [conjugate base][acid]
What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30 0.00
-x +x
0.30 - x
0.52
+x
x 0.52 + x
Common ion effect0.30 – x 0.300.52 + x 0.52
pH = pKa + log [HCOO-][HCOOH]
HCOOH pKa = 3.77
pH = 3.77 + log [0.52][0.30]
= 4.01
16.2
Mixture of weak acid and conjugate base!
HCl H+ + Cl-
HCl + CH3COO- CH3COOH + Cl-
16.3
pH= 9.18
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Initial
End
0.30 0.36 0
0.30 - x 0.36 + x x
16.3
[NH4+] [OH-]
[NH3]Kb =
Change - x + x + x
= 1.8 X 10-5
(.36 + x)(x)
(.30 – x)
1.8 X 10-5 =
1.8 X 10-5 0.36x
0.30
x = 1.5 X 10-5 pOH = 4.82
pH = 9.20
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
start (M)
end (M)
0.29 0.01 0.24
0.28 0.0 0.25
final volume = 80.0 mL + 20.0 mL = 100 mL
16.3
NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M
OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M
NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M
Ka= = 5.6 X 10-10[H+] [NH3]
[NH4+]
= 5.6 X 10-10[H+] 0.25
0.28
[H+] = 6.27 X 10 -10
= 9.20
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) H+ (aq) + NH3 (aq)
pH = pKa + log [NH3][NH4
+]pKa = 9.25 pH = 9.25 + log [0.30]
[0.36]= 9.17
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
start (M)
end (M)
0.29 0.01 0.24
0.28 0.0 0.25
pH = 9.25 + log [0.25][0.28]
final volume = 80.0 mL + 20.0 mL = 100 mL
16.3
Chemistry In Action: Maintaining the pH of Blood
16.3
TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at the endpoint (hopefully close to the equivalence point)
Slowly add baseto unknown acid
UNTIL
The indicatorchanges color
(pink) 4.7
Strong Acid-Strong Base TitrationsNaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)
OH- (aq) + H+ (aq) H2O (l)
16.4
100% ionization!
No equilibrium
Weak Acid-Strong Base TitrationsCH3COOH (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):
16.4
Strong Acid-Weak Base TitrationsHCl (aq) + NH3 (aq) NH4Cl (aq)
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)
At equivalence point (pH < 7):
16.4
H+ (aq) + NH3 (aq) NH4Cl (aq)
Acid-Base Indicators
16.5
pH16.5
The titration curve of a strong acid with a strong base.
16.5
Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base.At equivalence point, will have conjugate base of weak acid.At equivalence point, pH > 7Use cresol red or phenolphthalein
16.5
Finding the Equivalence Point(calculation method)
• Strong Acid vs. Strong Base– 100 % ionized! pH = 7 No equilibrium!
• Weak Acid vs. Strong Base– Acid is neutralized; Need Kb for conjugate base
equilibrium• Strong Acid vs. Weak Base
– Base is neutralized; Need Ka for conjugate acid equilibrium
• Weak Acid vs. Weak Base– Depends on the strength of both; could be
conjugate acid, conjugate base, or pH 7
Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of 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 0.01
0.0 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
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.
Co2+ (aq) + 4Cl- (aq) CoCl4 (aq)2-
Co(H2O)62+ CoCl4
2-
16.10
16.10
Complex Ion Formation
• These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions).
• As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)4
2+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2
+.• Memorize the common ligands.
Common LigandsLigands Names used in the ion
H2O aqua
NH3 ammine
OH- hydroxyCl- chloroBr- bromoCN- cyanoSCN- thiocyanato (bonded through
sulphur) isothiocyanato (bonded through nitrogen)
Names• Names: ligand first, then cation
Examples:– tetraamminecopper(II) ion: Cu(NH3)4
2+
– diamminesilver(I) ion: Ag(NH3)2+.
– tetrahydroxyzinc(II) ion: Zn(OH)4 2-
• The charge is the sum of the parts (2+) + 4(-1)= -2.
When Complexes Form• Aluminum also forms complex ions as do some post transitions
metals. Ex: Al(H2O)63+
• Transitional metals, such as Iron, Zinc and Chromium, can form complex ions.
• The odd complex ion, FeSCN2+, shows up once in a while• Acid-base reactions may change NH3 into NH4
+ (or vice versa) which will alter its ability to act as a ligand.
• Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)4
2+, forms.• Keywords such as "excess" and "concentrated" of any solution
may indicate complex ions. AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.
Coordination Number
• Total number of bonds from the ligands to the metal atom.
• Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.
Some Coordination Complexes
molecular formula
Lewis base/ligand
Lewis acid donor atom
coordination number
Ag(NH3)2+ NH3 Ag+ N 2
[Zn(CN)4]2- CN- Zn2+ C 4
[Ni(CN)4]2- CN- Ni2+ C 4
[PtCl6] 2- Cl- Pt4+ Cl 6
[Ni(NH3)6]2+ NH3 Ni2+ N 6