27
1 6-4-2011
1
6-4-2011
2
TitrationsDefinition: Volumetric determination of
the amount of an acid or base by addition of a standard acid or base until neutralization has occurred.
2) A burette (buret) is usually used to add the titrant & can be read to ±0.01 mL
3) a plot of pH vs volume of titrant is called a pH curve or a titration curve.
3
4) End Point can be determined by either an indicator or by use of a pH meter.
5) Can easily determine Ka since pH = pKa half way to the end point for a weak acid, at this point [HA]=[A-], or mmol HA = mmol A-.
6) A quick examination of the pH curve will tell if titrating an acid or base as well as if they are strong or weak.
4
Overview: Titrations
5
What will we study?
Three types of titrations:1. Strong acid with strong base or vice
versa2. Weak acid with strong base3. Weak base with strong acid
One of the reactants must be strong. You can not titrate weak acids with weak bases
6
pH Determination while Titrating a Strong Acid like HCl with a strong baseBefore Titration Starts: calculate pH from
M of HCl
During Titration & Before End Point: Calculate HCl remaining; then find pH from remaining HCl.
At End Point: Only NaCl left in water;
pH = 7
After End Point: Calculate Excess NaOH; Find pH from excess NaOH.
7
Assume we have 50 mL of 0.1 M HNO3 to which 10, 30, 50, 70 and 90 mL of 0.1 M NaOH is added:
HNO3 + NaOH NaCl + H2O
10 mL0.1*500.1*10
Initial51
Final40
Conc4/60 M0[H+] = 4/60 M, pH = 1.18
30 mL0.1*500.1*30
Initial53
Final20
Conc2/80 M[H+] = 2/80 M, pH = 1.60
8
Assume we have 50 mL of 0.1M HNO3 to which 10, 30, 50, and 70 mL of NaOH is added:
HNO3 + NaOH NaCl + H2O
50 mL0.1*500.1*50
Initial55
Final00
ConcEquivalence point, pH = 7
70 mL0.1*500.1*70
Initial57
Final02
Conc02/120 M[OH-] = 2/120 M,
pOH = 1.78, pH = 12.22
9
Find the pH of a 25.0 mL solution of 0.100 M HCl after addition of 0, 10, 20, 25, 35, and 45 mL of 0.100 M NaOH.
After addition of 0 mL NaOHWe have HCl solution only:[H+] = 0.100 M since HCl is a strong acidpH = 1.00
After addition of 10 mL NaOH
H+ + OH- H2Ommol H+ remaining = initial mmol H+ - mmol NaOH added
mmol H+ remaining = 0.100*25.0 – 0.100*10 = 1.50[H+] = {mmol/mL} = 1.5/35 = 0.043 and pH = 1.37
10
After addition of 20 mL NaOH
mmol H+ remaining = initial mmol H+ - mmol NaOH added
mmol H+ remaining = 0.100*25.0 – 0.100*20 = 0.50
[H+] = {mmol/mL} = 0.50/45 = 0.011 and pH = 1.95
After addition of 25 mL NaOH
mmol H+ remaining = initial mmol H+ - mmol NaOH added
mmol H+ remaining = 0.100*25.0 – 0.100*25 = 0
This is the equivalence point
[H+] = [OH-] = 10-7 M
pH = 7
11
After addition of 35 mL NaOH
mmol OH- excess = mmol NaOH added - mmol H+
mmol OH- excess = = 0.100*35.0 – 0.100*25 = 1.0
[OH-] = {mmol/mL} = 1.0/60 = 0.017
pOH = 1.78, and pH = 14 – 1.78 = 12.22
After addition of 45 mL NaOH
mmol OH- excess = mmol NaOH added - mmol H+
mmol OH- excess = = 0.100*45.0 – 0.100*25 = 2.0
[OH-] = {mmol/mL} = 2.0/70 = 0.029
pOH = 1.54, and pH = 14 – 1.54 = 12.46
12
25.0 mL of 0.100 M HCl with 0.100 M NaOH (Strong Acid with a Strong Base). Note: Large change in pH near end point and pH at equivalence point = 7.00
13
pH Determination while Titrating a Weak Acid like HA with a strong base
Before Titration Starts: Initial pH from M of weak acid (HA)
During Titration & Before End Point: Find remaining HA and A- formed; Calculate pH for the buffer containing HA & A-.
At End Point: Only have A-. Calculate pH from reaction of A- with H2O using Kb.
After End Point: Calculate Excess NaOH; then find pH directly from excess OH- .
14
If one reactant is weak, two steps are needed:
1. Quantitative step: Find conc. Of remaining acid, the formed conjugate base, and OH- after reaction of weak acid with strong base.
2. Equilibrium step: use the appropriate equilibrium constant expression to find [H+] or [OH-] and then calculate pH.
15
Assume we have 50 mL of 0.1M HA (weak) to which 10, 30, 50, 70 and 90 mL of NaOH is added:
HA + OH- A- + H2O
10 mL0.1*500.1*100
Initial510
Final401
Conc4/60 M01/60Buffer
30 mL0.1*500.1*300
Initial530
Final203
Conc2/80 M03/80Buffer
16
Assume we have 50 mL of 0.1M HA (weak) to which 10, 30, 50, 70 and 90 mL of NaOH is added:
HA + OH- A- + H2O
50 mL0.1*500.1*500
Initial550
Final005
Conc005/100 MSalt, use Kb
70 mL0.1*500.1*700
Initial570
Final025
Conc02/120 M5/120 MpOH from base, OH-
17
Weak Acid-Strong Base Titrations
Find the pH of a 25 mL solution of 0.10 M HOAc (Ka = 1.8*20-5), after addition of 0, 5, 15, 25, 30, and 40 mL of 0.10 M NaOH.
After addition of 0.0 mL NaOH
We have HOAc only, in solution. This is a weak acid which you have studied how to calculate its pH
HOAc H+ + OAc-
Initial 0.10 0 0
Change -x +x +x
Equilibrium 0.10 – x x x
18
Ka = x2/(0.10 – x), assume 0.10>>x since Ka is very small.
1.8*10-5 = x2/0.1
X = 1.34*10-3
RE = (1.34*10-3 /0.1)*100 = 1.3%, OK
X = [H+] = 1.34*10-3M
pH = 2.87
19
After addition of 5.0 mL NaOHHOAc + OH- OAC- + H2O
mmol HOAc left = initial mmol HOAc – mmol NaoH
mmol HOAc left = 0.10*25 – 0.10*5 = 2.0mmol OAc- formed = 0.5[HOAc] = 2.0/30 M, and [OAC-] = 0.5/30 M (This is a buffer solution)
HOAc H+ + OAc-
Initial 2.0/30 0 0.5/30Change -x +x +xEquil 2.0/30 – x x 0.5/30 + x
20
Ka = x(0.5/30 +x)/(2.0/30 – x), assume 0.5/30>>x since Ka is very small (buffer solution).
1.8*10-5 = (0.5/30*x)/(2.0/30)
X = 7.2*10-5
Since it is a buffer, no need to calculate RE
X = [H+] = 7.2*10-5 M
pH = 4.14
21
After addition of 15.0 mL NaOHHOAc + OH- OAC- + H2O
mmol HOAc left = initial mmol HOAc – mmol NaoH
mmol HOAc left = 0.10*25 – 0.10*15 = 1.0mmol OAc- formed = 1.5[HOAc] = 1.0/30 M, and [OAC-] = 1.5/30 M (This is a buffer solution)
HOAc H+ + OAc-
Initial 1.0/40 0 1.5/40Change -x +x +xEquil 1.0/40 – x x 1.5/40 + x
22
Ka = x(1.5/40 +x)/(1.0/40 – x), assume 1.0/40>>x since Ka is very small.
1.8*10-5 = (1.5/40*x)/(1.0/40)
X = 1.2*10-5
Since it is a buffer, no need to calculate RE
X = [H+] = 1.2*10-5 M
pH = 4.92
23
After addition of 25.0 mL NaOHHOAc + OH- OAC- + H2O
mmol HOAc left = initial mmol HOAc – mmol NaoH
mmol HOAc left = 0.10*25 – 0.10*25 = 0.0This is the equivalence point (all HOAc is
consumed)mmol OAc- formed = 2.5[OAC-] = 2.5/50 = 0.05 M
H2O + OAc OH- + HOAcInitial 0.05 0 0Change -x +x +xEquil 0.05 – x x x
24
Kb = x2/(0.05 – x), assume 0.05>>x since Kb is very small.
(10-14/1.8*10-5) = x2/0.05X = 5.3*10-6
RE = (5.3*10-6 /0.05)*100 = 0.01%, OKX = [OH-] = 5.3*10-6 MpOH = 5.28pH = 14 – pOH = 14 – 5.28 = 8.72 Note that the pH at equivalence point is
basic in the titration of weak acids.
25
After addition of 30.0 mL NaOH
HOAc + OH- OAC- + H2OInitial 2.5 3.0 0Final 0 0.5 2.5mmol OH- excess = mmol NaoH added - mmol
HOAc Now the solution has [OAc-] = 2.5/55 M and
[OH-] = 0.5/55 M. The OH- from the base is much greater than that coming from OAc-
since Kb is very small.[OH-] = 9.1*10-3 M, and pOH = 2.04pH = 14 – 2.04 = 11.96
26
After addition of 40.0 mL NaOH
HOAc + OH- OAC- + H2OInitial 2.5 4.0 0Final 0 1.5 2.5mmol OH- excess = mmol NaoH added - mmol
HOAc Now the solution has [OAc-] = 2.5/65 M and
[OH-] = 1.5/65 M. The OH- from the base is much greater than that coming from OAc-
since Kb is very small.[OH-] = 2.3*10-2 M, and pOH = 1.64pH = 14 – 1.64 = 12.36
27
Summary of Weak Acid-Strong Base TitrationsCH3COOH (aq) CH3COO- (aq) + H+ (l)
CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)
CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)
At equivalence point (pH > 7):
