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Buffers/Titration Aqueous Equilibria - I
Slide 1 / 113
Review hydrolysis of salts
Slide 2 / 113
1 The hydrolysis of a salt of a weak base and a strong acid should give a solution that is __________.
A weakly basic
B neutral
C strongly basic
D weakly acidic
Slide 3 / 113
2 A solution of one of the following is acidic. The compound is _________.
A NaCl
B CH3COONa
C NH4Cl
D NH3
Slide 4 / 113
3 Which salt would undergo hydrolyzes to form an acidic solution?
A KCl
B NaCl
C NH4Cl
D LiCl
Slide 5 / 113
4 Which substance when dissolved in water will produce a solution with a pH greater than 7?
A CH3COOH
B NaCl
C NaC2H3O2
D HCl
Slide 6 / 113
5 A basic solution would result from the hydrolysis of one of the ions in this compound. The compound is__________.
A NaNO3
B NH4Cl
C CH3COONa
D CaCl2
Slide 7 / 113
6 A water solution of which compound will turn blue litmus red?
A K2CO3
B NH4Cl
C NaOH
D NaCl
Slide 8 / 113
The Common-Ion Effect
· Consider an aqueous solution of acetic acid:
· If acetate ion is added to the solution, Le Châtelier 's Principle predicts that the equilibrium will shift to the left.
CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO-(aq)
Slide 9 / 113
The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”
In the following Sample Problem, compare the pH of a 0.20 M solution of HF with the pH of the same solution after some NaF is added to it.
Slide 10 / 113
SAMPLE PROBLEM #1a) Calculate the pH of a 0.20 M HFsolution. The Ka for HF is 6.8 ´ 10−4.
The Common-Ion Effect
HF(aq) H2O(l) H3O+(aq) F−(aq) Initial 0.20 M 0 0
Change -x +x +x
At Equilibrium 0.20-x x x
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
Slide 11 / 113
6.8 x 10-4 = x2
(0.20)
Ka = [H3O+][F-][HF] = 6.8 x 10-4
x2 = (0.20) (6.8 ´ 10−4)
x = 0.012
So, [H+] = [F-] = 0.012
and pH = 1.93
The Common-Ion Effect
Slide 12 / 113
The Common-Ion Effect
SAMPLE PROBLEM #1 (con't)b) Calculate the pH of a for the same 0.20 M HF solution that also contains 0.10 M NaF.
Ka for HF is 6.8 ´ 10−4.
Ka = [H3O+][F-][HF] = 6.8 x 10-4
Slide 13 / 113
The Common-Ion Effect
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
b) Calculate the pH of a for the same 0.20 M HF solution that also contains 0.10 M NaF.
· We use the same equation and the same Ka expression, and set up a similar ICE chart. · But because NaF is a soluble salt, it completely dissociates, so F- is not initially zero.
HF(aq) H2O(l) H3O+(aq) F−(aq) Initial 0.20 M 0 0.10 M
Change
at Equilibrium
Slide 14 / 113
HF(aq) H2O(l) H3O+(aq) F−(aq) Initial 0.20 M 0 0.10 M
Change -x +x +x
At Equilibrium 0.20-x x 0.10 + x
The Common-Ion Effect
Slide 15 / 113
The Common-Ion EffectTherefore,
x = [H3O+] = 1.4 x 10−3
and
pH = −log (1.4 x 10−3)
pH = 2.87
Slide 16 / 113
The Common-Ion Effect
1.4 ´ 10−3 = x
6.8 x 10-4 = (0.10)(x)(0.20)
(0.20)(6.8 x 10-4)(0.10)
= x
Ka =
Remember what the "x" is that you are solving for!
Slide 17 / 113
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
· Consider the two solutions we just examined in Sample Problem #1.
· Compare the following:
· How do these results support Le Chatelier's Principle?
The Common-Ion Effect
Solution Final [H3O+] pH
HF 0.12
HF and NaF 1.4 x 10-3
Slide 18 / 113
The Ka for acetic acid, CH3COOH or HC2H3O2, is 1.8 x 10-5.
SAMPLE PROBLEM #2
a) Calculate the pH of a 0.30 M acetic acid, CH3COOH.
b) Calculate the pH of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate.
The Common-Ion Effect
Slide 19 / 113
The Ka for acetic acid, CH3COOH, is 1.8 x 10-5.
SAMPLE PROBLEM #2 - Answers
a) Calculate the pH of a 0.30 M acetic acid solution.
The Common-Ion Effect
x2
1.8 x 10-5 =
Ka = [H3O+][C2H3O2-]
[HC2H3O2] = 1.8 x 10-5
(0.30)
x = [H3O+] = ______________
and pH = ___________
Slide 20 / 113
The Ka for acetic acid, CH3COOH or HC2H3O2, is 1.8 x 10-5.
SAMPLE PROBLEM #2 - Answers (con't)b) Calculate the pH of a solution containing 0.30 M acetic acid and 0.10 M sodium acetate.We use the same Ka expression, except now the acetate ion concentration is not the same as [H3O+].
The Common-Ion Effect
[H3O+] (0.10)1.8 x 10-5 =
Ka = [H3O+][C2H3O2-]
[HC2H3O2] = 1.8 x 10-5
(0.30)
So now, [H3O+] = ______________
and pH = ___________
Slide 21 / 113
HC2H3O2 (aq) + H2O (l) ↔ C2H3O2 -
(aq) + H3O+(aq)
· Consider the two solutions we just examined in Sample Problem #2.
· Compare the following:
· How do these results support Le Chatelier's Principle?
The Common-Ion Effect
Solution [OH-] pH
HC2H3O2
HC2H3O2 and NaC2H3O2
Slide 22 / 113
SAMPLE PROBLEM #3Calculate the pH of the following solutions:
a) 0.85 M nitrous acid, HNO2b) 0.85 M HNO2 and 0.10 M potassium nitrite, KNO2
The Ka for nitrous acid is 4.5 x 10-4.
The Common-Ion Effect
Answers: a) b)
Slide 23 / 113
7 The ionization of HF will be decreased by the addition of:
A NaCl
B NaF
C HCl
D Both A and B
E Both B and C
Slide 24 / 113
8 The dissociation of Al(OH)3 will be decreased by the addition of:
A KOH
B AlCl3C Mg(OH)2
D Both A and B
E A, B and C
Slide 25 / 113
9 Which of the following substances will not decrease the ionization of H3PO4?
A K3PO4
B HCl
C Na3PO4
D None of them will decrease the ionization
E All of them will decrease the ionization
Slide 26 / 113
The Common-Ion Effect
The Common-ion effect can also be observed with weak bases.
Consider the ionization of ammonia, NH3, which is a weak base.
NH3 + H2O <---> NH4
+ + OH-
Slide 27 / 113
The Common-Ion Effect
Suppose a salt of the conjugate base is added to a solution of ammonia. What effect would this have on pH?
NH3 + H2O <---> NH4
+ + OH-
As with the previous Sample Problems, let us calculate the pH of the following:
a) a 0.45 M solution of NH3
b) a 0.45 M solution of NH3 that also contains 0.15 M NH4Cl, ammonium chloride -- a soluble salt that readily yields NH4
+ ions.
Slide 28 / 113
SAMPLE PROBLEM #4Calculate the pH of the following solutions:
a) a 0.45 M solution of NH3
b) a 0.45 M solution of NH3 that also contains 0.15 M NH4ClThe Kb for NH3 is 1.8 x 10-5.
The Common-Ion Effect
Slide 29 / 113
SAMPLE PROBLEM #4 - Answers
a) Calculate the pH of a 0.45 M solution of NH3. The Kb for NH3 is 1.8 x 10-5.
NH3 + H2O <---> NH4
+ + OH-
The Common-Ion Effect
[NH4+][OH- ]
[NH3]Kb = 1.8 x 10-5 =
x = [OH-] = _____________
pH = _________________
x2
(0.45)1.8 x 10-5 =
Slide 30 / 113
SAMPLE PROBLEM #4 - Answers (con't)
b) Calculate the pH of a 0.45 M solution of NH3 that also contains 0.15 M NH4Cl. The Kb for NH3 is 1.8 x 10-5.
NH3 + H2O <---> NH4
+ + OH-
The Common-Ion Effect
[NH4+][OH- ]
[NH3]Kb = 1.8 x 10-5 =
(0.15) (x)(0.45)
1.8 x 10-5 =
So, x = [OH-] = _____________
and pH = _________________
Slide 31 / 113
SAMPLE PROBLEM #5Calculate the pH of the following solutions:
a) 0.0750 M pyridine, C5H5Nb) 0.0750 M C5H5N and 0.0850 M pyridinium chloride, C5H5NHClNote that C5H5NHCl,( salt) dissociates into C5H5NH+ and Cl-.
The Kb for pyridine is 1.7 x 10-9.
The Common-Ion Effect
Answers: a) b)
Slide 32 / 113
Buffers
Buffers, or buffered solutions, are special mixtures that are resistant to large pH changes, even when small amounts of strong acid or strong base are added.
Slide 33 / 113
Buffers
· Buffers are able to resist large pH changes because they contain both an acidic component and a basic component.
· Buffers are prepared by mixing either:1) a weak acid and a salt containing its conjugate base OR2) a weak base and a salt containing its conjugate acid
Slide 34 / 113
Buffers
Consider a buffer solution composed of HF and NaF.
· The acidic component is HF. This component aids in the neutralization of any strong base that is added to the buffer.
· The basic component is the fluoride ion, F-. This component aids in the neutralization of any strong acid that is added to the buffer.
Slide 35 / 113
Buffers
If a small amount of KOH is added to this buffer, for example, the HF reacts with the OH− to make __________.
Consider a buffer solution composed of HF and NaF.
HF F-HF F-
F- HF
F- + H2O <-- HF + OH- H+ + F- --> HF
OH-H+
Buffer afteraddition of OH-
Buffer with equal conc. of weak acid and its conj. base
F-
HF
Slide 36 / 113
Buffers
Similarly, if a small amount of strong acid (H+) is added, the F− reacts with it to form ____________.
HF F-HF F-F- HF
F- + H2O <-- HF + OH- H+ + F- --> HF
OH- H+
Buffer afteraddition of OH-
Buffer with equal conc. of weak acid and its conj. base
Buffer after addition of H+
Slide 37 / 113
10 If a buffer is made of HA (weak acid) and NaA, which species will counteract the addition of a strong base?
A HAB Na+
C A-
D Both HA and Na+
E Both HA and A-
Slide 38 / 113
11 If a buffer is made of HNO2 and KNO2 , which species will counteract the addition of a strong acid?
A HNO2
B K+
C NO2-
D Both HNO2 and K+
E Both K+ and NO2-
Slide 39 / 113
Buffer capacity and pH
Two important characteristics of buffers:1- Its resistance to change in pH2- Its buffer capacity
Buffer capacity: The amount of acid or base the buffer can neutralize before the pH begins to change.
It depends on the amount of acid and base from which the buffer is made.
Slide 40 / 113
[ H3O+] = Ka
The above two combination of solution will have the same [H+].
The pH will depend on ka and the relative concentration of acid and base only.
The buffering capacity will be higher for the first buffer. It contains greater number of moles of CH3COOH and NaCH3COO-.It can neutralize more of the acid/base added.
The greater the amounts, greater resistance to pH change.
Buffer capacity and pH
Buffer A Buffer B
1.0 M CH3COOH and1.0 M CH3COONa
Total volume = 1.0 L
0.1 M CH3COOH and0.1 M CH3COONa
Total volume = 1.0 L
Ka = [H3O+][C2H3O2- ]
[HC2H3O2]
[HC2H3O2][C2H3O2
- ]= 1Since
Slide 41 / 113
12 Of the following solutions, which has the greatest buffering capacity?
A 0.1M NaF + 0.1MHF
B 0.5M NaF + 0.55M HF
C 0.8M NaF + 0.8M HF
D 0.2M NaF + 0.2M HF
E 1M NaF + 1M HF
Slide 42 / 113
Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
HA + H2O ↔ H3O+ + A-
Ka = [H3O+][A-]
[HA]
Slide 43 / 113
Buffer CalculationsRearranging slightly, this becomes
Ka = [H3O+] [A-][HA]
Taking the negative log of both sides, we get
-log Ka = -log [H3O+] +(-log )[A-][HA]
pKa = pH - log [base][acid]
Slide 44 / 113
Buffer Calculations
Since pKa = pH - log [base][acid]
Rearranging this:pH = pKa + log [base]
[acid]
This is the Henderson-Hasselbalch equation
Slide 45 / 113
Buffer calculations typically require you to calculate one or more of the following:
i) the pH of the buffer alone (Sample Prob #6)
ii) the pH of the buffer after a small amount of strong base has been added (and neutralized)(Sample Prob #7)
iii) the pH of the buffer after a small amount of strong acid has been added (and neutralized) (Sample Prob #8)
Because of the buffer system, the pH will not change drastically; it is usually less than a factor of 1.0 on the pH scale.
Buffer Calculations
Slide 46 / 113
SAMPLE PROBLEM #6
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 ´ 10−4.
Buffer Calculations
Method 1 - Traditional approachWrite the Ka expression. Solve for H+ and pH using ICE chart
HC3H5O3 (aq) + H2O(l) ↔ H3O+(aq) + C3H5O3-(aq)
0.12 0 0.10 -x +x +x 0.12-x x 0.10+x
1.4 ´ 10−4 = 0.10 x 0.12x = 0.000168PH =3.77
Slide 47 / 113
SAMPLE PROBLEM #6What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 ´ 10−4.
Buffer Calculations
Method 2 -Using the Henderson-Hasselbalch equation
Henderson–Hasselbalch Equation
pH = pKa + log [base][acid]
pH = -log (1.4 x 10-4) + log (0.10)(0.12)
pH = 3.85 + (-0.08)
pH = 3.77
Slide 48 / 113
13 The pH of a buffer solution that contains 0.818 M acetic acid (Ka = 1.76x10-5) and 0.172 M sodium acetate is __________.
A 4.077
B 5.434
C 8.571
D 8.370
E 9.922
Slide 49 / 113
14 The pH of a buffer solution containing 0.145 M HF and 0.183 M KF is _____. (Ka of HF is 3.5x10-4)
A 3.56 B 2.19 C 2.66 D 3.35
E 4.32
Slide 50 / 113
Recall that buffers resist drastic changes in pH, even when small amounts of either a strong acid or a strong base are added to it.
When this happens, we assume that all of the strong acid or strong base is consumed in the reaction. In other words, all of the strong acid or base gets completely neutralized by ONE of the components of the buffer system.
Addition of Strong Acid or Strong Base to a Buffer
Slide 51 / 113
X- + H3O+ ⇒ HX + H2O
HX + OH- ⇒ X- + H2O
RecalculateHX and X-
Use Ka, [HX] and X-
to calculate [H+] pH
Add strong acid
Add strong base
stoichiometric calculation equilibrium calculation
Remember that all of the strong acid or base is consumed in the reaction.
Addition of Strong Acid or Strong Base to a Buffer
Slide 52 / 113
15 If a buffer is made of NH3 and NH4Cl, which component of the buffer will neutralize a small amount of HCl that is added?
A NH3
B NH4+
C Cl-
D Both A and B
E Both B and C
Slide 53 / 113
16 If a buffer is made of HNO2 and KNO2 , which component of the buffer will neutralize a small amount of Ba(OH)2 that is added?
A HNO2
B K+
C NO2-
D Ba2+
E Both K+ and NO2-
Slide 54 / 113
17 If a buffer is made of NH3 and NH4Cl, which component will neutralize any KOH that is added?
A NH3
B NH4+
C Cl-
D Both HNO2 and K+
E Both K+ and NO2-
Slide 55 / 113
Addition of Strong Acid or Strong Base to a Buffer
· Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.· Use the Henderson–Hasselbalch equation to determine the new pH of the solution.
0.3M HC2H3O20.3M NaC2H3O2
add
0.02
M O
H-
add 0.02M H
+
pH =4.80
pH= 4.68
BufferpH= 4.74
Slide 56 / 113
Calculating pH Changes in Buffers
SAMPLE PROBLEM #7 A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Assume volume change is negligible.
Slide 57 / 113
Problem-solving strategy
Phase 1 - stoichiometric neutralization
· Identify the added substance. · Identify the component of the buffer that will neutralize the added substance.· Write down the equation for nutralization. · Solve for the new concentration of the buffer components.
Phase 2 - equilibrium · Write the relevant dissociation equation.· Create the ICE chart with the new [M] values(phase1) of acid and base components of the buffer.· Use Henderson-Hasselbalch equation to solve for pH.
Calculating pH Changes in Buffers
Slide 58 / 113
Phase 1 - Neutralization · strong acid or a strong base added ? · Which component of the buffer will neutralize the added substance?
SAMPLE PROBLEM #7 - AnswerA buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.
Calculating pH Changes in Buffers
HA(aq)acid
OH-(aq)added
substanceH2O(aq) A−(aq)
base
Before 0.3 0.02 0.3
Neutralization -0.02 -0.02 0.3+0.02
After 0.28 0.0 0.32
Slide 59 / 113
Phase 2 - equilibrium · Write the relevant dissociation equation.· Make a ICE chart using the amounts from Phase 1.
SAMPLE PROBLEM #7 - Answer (con't)A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.
Calculating pH Changes in Buffers
HA(aq) H2O(l) H3O+(aq) A−(aq) Initial 0.28 M 0 0.32 M
Change -x +x +x
At Equilibrium 0.28-x x 0.32+x
Slide 60 / 113
Calculating pH Changes in Buffers
[H3O+][A−][HA]Ka =
Use the quantities from the ICE chart to calculate pH. You will need to look up the Ka value.
(x) 0.32)(0.28)1.8 x 10-5 =
So, x = [H3O+] = _______________
and pH = ___________
Slide 61 / 113
Calculating pH Changes in BuffersAlternatively, you can use the Henderson-Hasselbalch equation to calculate the new pH:
pH = - log (1.8 x 10-5) + log (0.320)(0.280)
pH = 4.74 + 0.06
pH = 4.80
pH = pKa + log [base][acid]
Slide 62 / 113
Calculating pH Changes in BuffersSAMPLE PROBLEM #8 A buffer is made by adding 0.140 mol cyanic acid, HCNO, and 0.110 mol potassium cyanate, KCNO to enough water to make 1.00 L of solution. Assume volume changes are negligible. Ka for cyanic acid = 3.5 x 10-4
a) Calculate the pH of the buffer.b) Calculate the pH of the buffer after the addition of 0.015 mol KOH.c) Calculate the pH of the buffer after the addition of 0.018 mol HNO3.
a) 3.351b) 3.46c) 3.22
Answers
Slide 63 / 113
pH Range for Buffers
· The pH range is the range of pH values over which a buffer system works effectively.
· It is best to choose an acid with a pKa close to the desired pH.
· After studying titration graphs, you will see why buffers work best when pKa is close to the desired pH.
Slide 64 / 113
Calculating pH Changes in Buffers
SAMPLE PROBLEM #9 A buffer is made by adding 15.0 g ammonia, NH3, and 55.0 g ammonium chloride, NH4Cl. to enough water to make 1.00 L of solution. Kb for NH3 = 1.8 x 10-5. Assume volume changes are negligible.
a) Calculate the pH of the buffer.b) Calculate the pH of the buffer after the addition of 0.013 mol HClO4.c) Calculate the pH of the buffer after the addition of 0.015 mol KOH.
a) 9.193b) 9.181c) 9.206
Answers
Slide 65 / 113
Calculating pH Changes in BuffersSAMPLE PROBLEM #10 A buffer is made by adding 25.0 g ammonia, NH3, and 45.0 g ammonium chloride, NH4Cl. to enough water to make 1.75 L of solution. Kb for NH3 = 1.8 x 10-5. Assume volume changes are negligible.
a) Calculate the pH of the buffer.b) Calculate the pH of the buffer after the addition of 0.011 mol NaOH.c) Calculate the pH of the buffer after the addition of 0.016 mol HNO3.
a) 9.50b) 9.51c) 9.49
Answers
Slide 66 / 113
Titration
In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).
The concentration of the unknown will then be determined.
This is a quantitative analysis method.
Slide 67 / 113
Titration
Rinse the buret with distilled waterRinse with the titrantFill the buret with the titrantDrain a small portion of the titrant so that air bubbles near the tip of the burete is expelled and filled to the tip.
Important things to remember in getting ready for titration:
Slide 68 / 113
Rinse the pipette with distilled water
Rinse with the solution to be pipeted
Pipette the solution , adjust the level and dispense the fixed volume to a beaker or flask.
TitrationImportant things to remember in getting ready for titration:
Slide 69 / 113
TitrationA pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.
At the equivalence point,
# of moles of acid = # of moles of base
MaVa = MbVb
By using an indicator you could visually see the sudden change in color of the indicator at this point. This point is also known as the "end point" where you will stop adding the titrant.
The end point is the appearance of the first permanent color change of the indicator.
Note that this point will have a slight excess of the titrant in the solution
Slide 70 / 113
At the equivalence point, ( stoichiometric)
# of moles of acid = # of moles of base
MaVa = MbVb
By using an indicator you could visually see the sudden change in color of the indicator at this point. This point is also known as the "end point" where you will stop adding the titrant.
The end point is the appearance of the first permanent color change of the indicator.
Note that this point will have a slight excess of the titrant in the solution
This minute amount of the titrant is making the indicator color visible.
TitrationEquivalence point and End point
Slide 71 / 113
Slide 72 / 113
At the equivalence point,
# of moles of acid = # of moles of base
MaVa = MbVb
This equation is only applicable for reactions in which the ratio of
moles acid to moles base is 1:1.
Titration - Equivalence Point
Slide 73 / 113
HCl + NaOH --> NaCl + H2OHBr + KOH --> KBr + H2O
HNO3 + LiOH --> LiNO3 + H2OCH3COOH + NaOH --> CH3COONa + H2O
So for reactions such as these, use the equation above to calculate the molarity or volume of acid or base at the equivalence point.
MaVa = MbVb
Here are some examples of reactions in which the ratio of moles acid to moles base is 1:1.
Titration - Equivalence Point
Slide 74 / 113
Titration - Equivalence Point
2 HCl + Ca(OH)2 --> CaCl2 + 2 H2O2 HBr + Sr(OH)2 --> SrBr2 + 2 H2O
2 HNO3 + Ba(OH)2 --> Ba(NO3)2 + 2 H2O
Consider some examples of reactions in which the ratio of moles acid to moles base is 2:1.
moles acidmoles base
21= MaVa
MbVb=
Cross-multiplying, we obtain MaVa = 2 MbVb
Slide 75 / 113
H2SO4 + 2 KOH --> K2SO4 + 2 H2OH2SO3 + 2 LiOH --> Li2SO3 + 2 H2O
H2C2O4 + 2 NaOH --> Na2C2O4 + 2 H2O
Consider some examples of reactions in which the ratio of moles acid to moles base is 1:2.
moles acidmoles base
12=
Cross-multiplying, we obtain 2 MaVa = MbVb
MaVaMbVb
=
Titration - Equivalence Point
Slide 76 / 113
TitrationSummary of Equations for
Equivalence Point
Equation Acid Base
MaVa = MbVb HCl, HNO3, HBr, CH3COOH, HClO4
NaOH, KOH, LiOH
2 MaVa = MbVb H2SO4, H2SO3, H2C2O4
NaOH, KOH, LiOH
MaVa = 2 MbVb HCl, HNO3, HBr, CH3COOH, HClO4
Ca(OH)2, Sr(OH)2,Ba(OH)2
Slide 77 / 113
18 What is the concentration of hydrochloric acid if 20.0 mL of it is neutralized by 40mL of 0.10M sodium hydroxide?
A 0.025 M
B 0.050 M
C 0.10 M
D 0.20 MMaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
Slide 78 / 113
19 What is the concentration of KOH if 60 mL of it is neutralized by 20 mL 0.10M HCl?
A 0.005 MB 0.025 MC 0.033 MD 0.10 ME 0.30 M
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
Slide 79 / 113
20 What is the concentration of sulfuric acid if 50mL of it is neutralized by 10mL of 0.1M sodium hydroxide?
A 0.005M
B 0.01M
C 0.25M
D 0.5MMaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
Slide 80 / 113
21 How much 0.5 M HNO3 is necessary to titrate 25 mL of 0.05M Ca(OH)2 solution to the endpoint?
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
A 2.5 mL
B 5.0 mL
C 10 mL
D 20 mL
E 25 mL
Slide 81 / 113
22 How much 3.0 M NaOH is needed to exactly neutralize 20.0 mL of 2.5 mL H2SO3?
A 8.3 mL
B 17 mL
C 24 mL
D 33 mLE 48 mL
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
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23 How much 1.5 M NaOH is necessary to exactly neutralize 20.0 mL of 2.5 M H3PO4?
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
A 11 mLB 12 mLC 33 mLD 36 mLE 100 mL
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24 What is the molarity of a NaOH solution if 15 mL is exactly neutralized by 7.5 mL of a 0.02 M HC2H3O2 solution?
MaVa = MbVb
2 MaVa = MbVb
MaVa = 2 MbVb
A 0.005 M
B 0.010 M
C 0.020 M
D 0.040 M
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Typically, there are 4 "zones" in which you may be asked to calculate pH:
· Before any titrant is added from the buret
· After a small amount of titrant has been added
· At the equivalence point
· After the equivalence point
First, we will examine these zones on a titration graph. Then, we will review the pH calculation for each region.
The strategy for calculating pH is different for each zone.
Titration of a Strong Acid with a Strong Base
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Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly.
The low initial pH indicates that the substance being titrated is a strong acid.
(1) Before any titrant is
added from the buret
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Titration of a Strong Acid with a Strong Base
Just before (and after) the equivalence point, the pH increases rapidly.
(2) After a small amount of titrant has been added*
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Titration of a Strong Acid with a Strong Base
At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
(3) At the equivalence
point
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Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off.
(4) After the equivalence
point
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In summary, the four regions of any titration graph are:
· Before any titrant is added from the buret
· After a small amount of titrant has been added
· At the equivalence point
· After the equivalence point
The pH calculation differs for each "zone," so we will consider each one separately.
Titration of a Strong Acid with a Strong Base
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Solving Titration Problems: Strong Acid - Strong Base
1) Before any titrant is added from the buret
The pH depends up on the concentration of the acid or base.
Use the molarity of the acid or base to determine the pH.pH = - log [H3O+] OR pH = 14 - (-log [OH-]
Remember to check the acid or base is polyprotic or polyhydroxy.For example: · 20 ml 0.5M HCl is titrated with 0.25M NaOHThe original acid is 0.5M; The concentration of H+ = 0.5MMaVa = MbVb
· 20 ml 0.5M H2SO4 is titrated with 0.25M NaOHThe original acid is 0.5M; The concentration of H+ = 0.5x2 M2MaVa = MbVb
20 ml 0.5M HCl is titrated with 0.25M Ca(OH)2The original acid is 0.5M; The concentration of H+ = 0.5 MMaVa = 2MbVb2
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2) When some titrant is added from the buret, before the equivalence point is reachedExample: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH after 10 ml of base is added?
· It is recommended that you calculate the volume of the titrant needed to neutralize the acid at the beginning to see which segment of the titration we are in.
write down the neutralization reaction
HCl + NaOH --> NaCl + H2O0.01 mol 0.0025mol
-0.0025mol -0.0025mol
0.0075mol 0 mol left
[H+] after the 10 ml base had been neutralized by the acid is = 0.0075 mols /0.030L = 0.25MpH = 0.6
Solving Titration Problems: Strong Acid - Strong Base
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Solving Titration Problems: Strong Acid - Strong Base
3) At the equivalence point: Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH at the equivalence point ?
· You will know you have reached the equivalence point when MOLES of acid = MOLES of base
· Since there is no excess [H+] or [OH-], we must look to the salt that is formed to see if it will affect pH. · Since it is from a strong base and strong acid, will not undergo hydrolysis to change the pH.
· In cases such as these, when a strong acid and strong base are titrated, the pure water will have a pH = 7.0
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Solving Titration Problems: Strong Acid - Strong Base
4) Beyond the equivalence point, when excess titrant has been added from the buret
Example: 20 ml 0.5M HCl is titrated with 0.25M NaOH. What is the pH after 45 ml of base had been added?
0.5 x 20ml = 0.25 x V2mlV2 = 40 ml NaOH need to neutralize the 20 ml acid.
write down the neutralization reaction
HCl + NaOH --> NaCl + H2O0.01 mol 0.0125mol
-0.01mol -0.01mol
0 mol 0.0025mol excess
[OH-] after the 45 ml base added = = 0.0025 mols /0.065L = 0.0385MpOH = 1.415pH = 12.585
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Solving Titration Problems: Strong Acid - Strong Base
The pH at eqequivalence point will be =7Acid in the flask and titrant is the base.
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Solving Titration Problems: Strong Acid - Strong Base
Acid in the flask and base is the titrant
Base in the flask and acid is the titrant
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25 The moles of acid equals the moles of base at the equivalence point in a titration of _____ with _____.A strong acid, weak base
B strong base, weak acid
C strong acid, strong base
D weak acid, weak base
E All of the above
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26 What is the pH of a titration between a weak acid and a strong base at the equivalence point?
A Less than 7
B Equal to 7
C Greater than 7
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27 What is the pH of a titration between a weak base and a strong acid at the equivalence point?
A Less than 7
B Equal to 7
C Greater than 7
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Solving Titration Problems: Weak - Strong
1) Before any titrant is added from the buret
Consider the dissociation equation for the substance in the flask:
For a weak acid in the flask For a weak base in the flaskKa = x2 / [acid] Kb = x2 / [base]pH = - log x pH = 14 - (-log x)
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Solving Titration Problems: Weak - Strong
2) When some titrant is added from the buret, before the equivalence point is reachedexample# 11: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 10 ml NaOH is added: ka = 1.8 x 10-5
CH3COOH + NaOH --> Na+CH3COO- + H2O 0.01mol 0.0025mol-0.0025mol -0.0025mol + 0.0025mol0.0075mol 0.0 0.0025mol
0.25M (acid) 0.083M ( c.base)
Remember you have a buffer situation now because of the salt produced from the neutralizationof the weak acid.
Use the Henderson-Hasselbalch equation to solve for the pH of the solution at this point.
pH = pKa + log ( [base] / [acid] )
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Sample # 12calculate the pH of the solution after adding 10 ml of 0.05M KOH to 40 ml of 0.025M Benzoic acid. Ka for benzoic acid = 6.3 x 10-5
Solving Titration Problems: Weak - Strong
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Solving Titration Problems: Weak - Strong
sample #13Calculate the pH of the solution after adding 10 ml of 0.1M HCl to 20 ml of 0.1 M NH3Kb of NH3 = 1.8 x 10-5
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Solving Titration Problems: Weak - Strong
3) At the equivalence pointexample # 14: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 40 ml NaOH is added: ka = 1.8 x 10-5
CH3COOH + NaOH --> Na+ CH3COO- + H2O 0.01mol 0.01mol-0.01mol -0.01mol + 0.01mol0.0mol 0.0 mol 0.01mol
0.166M (base)The salt Na+ CH3COO- has a strong conjugate base CH3COO-
The anion, CH3COO- will undergo hyrolysis as follows:
CH3COO- + H2O --> CH3COOH + OH-
0.166M 0 0-X +x +x0.166-X x x
Kb = X2 / 0.166 ; Solve for x Remember Kb = Kw / KaCalculate pHWill be slightly basic !
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Solving Titration Problems: Weak - Strong
sample # 15Calculate the pH at equivalence point when 40 ml of 0.025M benzoic acid is titrated with 0.05M KOH. Ka of benzoic acid is 6.3 10-5
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Solving Titration Problems: Weak - Strong
Sample # 16calculate the pH at the equivalence point when 40 ml of 0.1M NH3 is titrated with 0.2 M HCl.Kb of NH3 is 1.8 10-5
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Titration of a Weak Acid with a Strong Base
The pH at the equivalence point is above 7
weak acid vs strong base
strong acid vs strong base
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Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
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Titration of a Weak Base with a Strong Acid
· The pH at the equivalence point in these titrations is < 7.· Methyl red is the indicator of choice.
strong base with strong acid , pH =7weak base NH3 with strong acid , pH = 5.5
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Solving Titration Problems: Weak - Strong
4) Beyond the equivalence point, when excess titrant has been added from the buretexample: Titration of 20 ml of 0.5M acetic acid with 0.25M NaOH; 50 ml NaOH is added:
This is exactly the same as for strong acid-strong base titrations.
The strong base added after the end point will increase the OH- in the solution making it strongly basic.
Calculate the [OH-] from the excess base in the new volume and determine the pH.
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Titrations of Polyprotic Acids
When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.
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