acid-base reactions: titration
DESCRIPTION
Acid-Base Reactions: TITRATION. Neutralization Reactions. You can neutralize a base with an acid. The result will form water molecules NaOH + HCl → NaCl + H 2 0. Acid-Base Reactions. Strong acid + strong base HCl + NaOH ----> SALT WATER Strong acid + weak base - PowerPoint PPT PresentationTRANSCRIPT
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Acid-Base Acid-Base
Reactions:Reactions: TITRATIONTITRATION
Neutralization ReactionsNeutralization Reactions
You can neutralize a base You can neutralize a base with an acid. with an acid.
The result will form water The result will form water moleculesmolecules
NaOH + HCl NaOH + HCl → NaCl + H→ NaCl + H2200
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Acid-Base ReactionsAcid-Base Reactions
• Strong acid + strong base HCl + NaOH ----> SALT WATER
• Strong acid + weak base HCl + NH3 ---> ACID
• Weak acid + strong base HOAc + NaOH ---> BASIC
• Weak acid + weak base HOAc + NH3 ---> Ka / Kb
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Let us first calculate the pH of a 0.25 M NH3 solution.
[NH3] [NH4+] [OH-]
initial 0.25 0 0
change -x +x +x
equilib 0.25 x x
Let us first calculate the pH of a 0.25 M NH3 solution.
[NH3] [NH4+] [OH-]
initial 0.25 0 0
change -x +x +x
equilib 0.25 x x
QUESTION: What is the effect on the pH of adding NH4Cl to 0.25 M NH3(aq)?
NH3(aq) + H2O NH4+(aq) + OH-(aq)
pH of Aqueous NHpH of Aqueous NH33
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Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x2
0.25 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x2
0.25 - x
pH of Aqueous NHpH of Aqueous NH33
Assuming x is << 0.25, we have
[OH-] = x = [Kb(0.25)]1/2 = 0.0021 M
This gives pOH = 2.67
and so pH = 14.00 - 2.67
= 11.33 for 0.25 M NH3
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We expect that the pH will decline on adding NH4Cl. Let’s test that!
[NH3] [NH4+] [OH-]
I 0.25 0.1 0
C -x +x +x
E 0.25 - x 0.10 + x x
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
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Kb 1.8 x 10-5 = [NH4
+][OH- ][NH3 ]
= x(0.10 + x)
0.25 - xKb 1.8 x 10-5 =
[NH4+][OH- ]
[NH3 ] =
x(0.10 + x)0.25 - x
pH of NHpH of NH33/NH/NH44++ Mixture Mixture
[OH-] = x = (0.25 / 0.10)Kb
= 4.5 x 10-5 M
This gives pOH = 4.35 and pH = 9.65
pH drops from 11.33 to
9.65 on adding a common
ion.
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TitrationsTitrationsTitrationsTitrations
pHpHpHpH
Titrant volume, mLTitrant volume, mLTitrant volume, mLTitrant volume, mL
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Strong acid titrated with a strong base
Figure 18.4Figure 18.4
10Acetic acid titrated with NaOHAcetic acid titrated with NaOH
Figure 18.5Figure 18.5
Weak acid titrated with a strong base
11Acid-Base TitrationsAcid-Base Titrations
Adding NaOH from the buret to acetic acid in
the flask, a weak acid. In the beginning
the pH increases very slowly.
Additional NaOH is added. pH increases and then levels off as NaOH is added beyond
the equivalence point.
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Acid-Base TitrationSection 18.3
You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?
HBz + NaOH ---> Na+ + Bz- + H2O
You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?
HBz + NaOH ---> Na+ + Bz- + H2O
CC66HH55COCO22H = HBzH = HBz Benzoate ion = Bz-
Kb = 1.6 x 10-10
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QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.0.100 M NaOH to the equivalence point.
pH at pH at equivalence equivalence point?point?
pH at pH at equivalence equivalence point?point?
pH of solution of pH of solution of benzoic acid, a benzoic acid, a weak acidweak acid
pH of solution of pH of solution of benzoic acid, a benzoic acid, a weak acidweak acid
Benzoic acid Benzoic acid + NaOH+ NaOHBenzoic acid Benzoic acid + NaOH+ NaOH
pH at pH at half-way half-way point?point?
pH at pH at half-way half-way point?point?
14Acid-Base ReactionsAcid-Base Reactions
Strategy — find the conc. of the conjugate base Bz- in the solution AFTER
the titration, then calculate pH.
This is a two-step problem
1.stoichiometry of acid-base reaction
2.equilibrium calculation
Strategy — find the conc. of the conjugate base Bz- in the solution AFTER
the titration, then calculate pH.
This is a two-step problem
1.stoichiometry of acid-base reaction
2.equilibrium calculation
QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final solution?
15Acid-Base ReactionsAcid-Base Reactions
STOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d
(0.100 L)(0.025 M) = 0.0025 mol HBz
This requires 0.0025 mol NaOH
2.Calc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH req’d
STOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d
(0.100 L)(0.025 M) = 0.0025 mol HBz
This requires 0.0025 mol NaOH
2.Calc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH req’d
16Acid-Base ReactionsAcid-Base Reactions
STOICHIOMETRY PORTION
25 mL of NaOH req’d
3. Moles of Bz- produced = moles HBz = 0.0025 mol
4. Calc. conc. of Bz-
There are 0.0025 mol of Bz- in a TOTAL SOLUTION VOLUME of
STOICHIOMETRY PORTION
25 mL of NaOH req’d
3. Moles of Bz- produced = moles HBz = 0.0025 mol
4. Calc. conc. of Bz-
There are 0.0025 mol of Bz- in a TOTAL SOLUTION VOLUME of
QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
125 mL125 mL
[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M
17Acid-Base ReactionsAcid-Base Reactions
Equivalence PointEquivalence Point
Most important species in solution is benzoate ion, Bz-, the weak conjugate base of benzoic acid, HBz.
Bz- + H2O HBz + OH- Kb = 1.6 x 10-10
[Bz-] [HBz] [OH-]
I 0.020 0 0
C - x +x +x
E 0.020 - x x x
Equivalence PointEquivalence Point
Most important species in solution is benzoate ion, Bz-, the weak conjugate base of benzoic acid, HBz.
Bz- + H2O HBz + OH- Kb = 1.6 x 10-10
[Bz-] [HBz] [OH-]
I 0.020 0 0
C - x +x +x
E 0.020 - x x x
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Acid-Base ReactionsAcid-Base Reactions
x = [OH-] = 1.8 x 10-6
pOH = 5.75 -----> pH = 8.25
Kb 1.6 x 10-10 =
x2
0.020 - x Kb 1.6 x 10-10 =
x2
0.020 - x
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QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?
QUESTION:QUESTION: You titrate 100. mL of a You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. 0.100 M NaOH to the equivalence point. What is the pH at half-way point?What is the pH at half-way point?
pH at pH at half-way half-way point?point?
pH at pH at half-way half-way point?point?
Equivalence Equivalence point point pH = 8.25pH = 8.25
Equivalence Equivalence point point pH = 8.25pH = 8.25
20Acid-Base ReactionsAcid-Base Reactions
You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH.
What is the pH at the half-way point?
[H3O ] = [HBz]
[Bz- ]• Ka[H3O ] =
[HBz]
[Bz- ]• Ka
At the half-way point, [HBz] = [Bz-]Therefore, [H3O+] = Ka = 6.3 x 10-5
pH = 4.20 = pKa of the acid
Both HBz and Both HBz and BzBz-- are present. are present.
This is a This is a BUFFER!BUFFER!
Both HBz and Both HBz and BzBz-- are present. are present.
This is a This is a BUFFER!BUFFER!
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Figure 18.6
Weak diprotic acid (H2C2O4) titrated with a
strong base (NaOH)
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Figure 18.7Figure 18.7
Weak base (NH3) titrated with a strong
acid (HCl)
IndicatorIndicatora compound that can a compound that can reversibly change color reversibly change color depending on the depending on the concentration of Hconcentration of H33OO++ ions ions..
Used to measure pHUsed to measure pH
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Indicators for Acid-Base Titrations
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