Acid-Base ReactionsAcid-Base Reactions
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
Strong Acid and Strong Base
HCl + NaOH
HCl + HHCl + H22O --->O --->HH33OO++ + Cl + Cl--
NaOH(aq) ---> NaOH(aq) ---> NaNa++(aq) + OH(aq) + OH--(aq)(aq)
HH33OO++ + Cl + Cl- - + + NaNa+++ + OHOH-- 2 H2 H22OO + + ClCl- - + + NaNa++
Net ionic Equation doesn’t show spectator ions!
Acid-Base ReactionsAcid-Base Reactions
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
H2O + H2O H3O+ + OH- K = 1 x 10-14
HH33OO++ + + OHOH-- 2 H2 H22OO
K = 1/ 1 x 10-14 = 1 x 1014
NEUTRALIZATION REACTION, pH = 7
Acid-Base ReactionsAcid-Base Reactions
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
Strong Acid and Weak Base
NHNH33(aq)(aq) + H + H22O(liq) O(liq) NHNH44++(aq)(aq) + + OHOH--(aq)(aq)
HCl + NH3
HH33OO++ + + OHOH-- 2 H 2 H22OO
HH33OO++ + + NHNH33 H H22O + NHO + NH44
++
Knet = Kb (1/Kw) = 1.8 x 109
Acid-Base ReactionsAcid-Base Reactions
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
STRONG Acid and Weak Base
Mixing equal molar quantities gives an acidic solution.
Weak Acid and STRONG Base
Mixing equal molar quantities gives a basic solution.
Acid-Base ReactionsAcid-Base Reactions
Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida
Weak Acid and Weak Base
pH depends on Kb and Ka of conjugate base and acid.
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Stomach Acidity &Stomach Acidity &Acid-Base ReactionsAcid-Base Reactions
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Example 18.1—Calculate pH
100 mL of 0.10 M HCl with 50 mL of 0.20 M NH3
HCl + HCl + NHNH33 Cl Cl-- + NH + NH44
++
0.100 L x 0.10 mol/L = 0.0100 mol HCl =0.0100 mol Cl-
0.050 L x 0.20 mol/L = 0.0100 mol NH3
=0.0100 mol NH4+
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Example 18.1—Calculate pH
100 mL of 0.10 M HCl with 50 mL of 0.20 M NH3
HCl + HCl + NHNH33 Cl Cl-- + NH + NH44
++
0.0100 mol Cl- / 0.150 L solution = 0.0667 M Cl-
0.0100 mol NH4+
/ 0.150 L solution = 0.0667 M NH4+
HH22O + NHO + NH44++ HH33OO++ + + NHNH33
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Example 18.1—Calculate pH
Equilibria of NH4+ contributes to pH
EQUIL Ka = x (x) / (0.0667 – x)
Ka = 5.6 x 10-10 and x = 6.1 x 10-6 M
HH22O + NHO + NH44++ HH33OO++ + + NHNH33
Ka = [Ka = [HH33OO++ ] [ ] [NHNH33] / ] / [[NHNH44++] ]
pH = 5.21
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Acid-Base ReactionsAcid-Base ReactionsSection 18.4Section 18.4
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution? solution?
HBz + NaOH ---> NaHBz + NaOH ---> Na++ + Bz + Bz-- + H + H22OO
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution? solution?
HBz + NaOH ---> NaHBz + NaOH ---> Na++ + Bz + Bz-- + H + H22OO
CC66HH55COCO22H = HBzH = HBzBenzoate ion = Bz-
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Acid-Base ReactionsAcid-Base ReactionsSection 18.4Section 18.4
The product of the titration of benzoic acid, The product of the titration of benzoic acid, the benzoate ion, Bzthe benzoate ion, Bz--, is the conjugate base , is the conjugate base of a weak acid. of a weak acid. The final solution is basic.The final solution is basic.
BzBz- - + H+ H22O HBz + OHO HBz + OH--
The product of the titration of benzoic acid, The product of the titration of benzoic acid, the benzoate ion, Bzthe benzoate ion, Bz--, is the conjugate base , is the conjugate base of a weak acid. of a weak acid. The final solution is basic.The final solution is basic.
BzBz- - + H+ H22O HBz + OHO HBz + OH--
++
KKbb = 1.6 x 10 = 1.6 x 10-10-10
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
Strategy —Strategy — find the conc. of the find the conc. of the conjugate base Bconjugate base B-- in the solution AFTER in the solution AFTER the titration, then calculate pH.the titration, then calculate pH.
This is a two-step problemThis is a two-step problem
1.1. stoichiometrystoichiometry of acid-base reaction of acid-base reaction
2.2. equilibrium calculationequilibrium calculation
Strategy —Strategy — find the conc. of the find the conc. of the conjugate base Bconjugate base B-- in the solution AFTER in the solution AFTER the titration, then calculate pH.the titration, then calculate pH.
This is a two-step problemThis is a two-step problem
1.1. stoichiometrystoichiometry of acid-base reaction of acid-base reaction
2.2. equilibrium calculationequilibrium calculation
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d1. Calc. moles of NaOH req’d
(0.100 L HBz)(0.025 M) = (0.100 L HBz)(0.025 M) = 0.0025 mol HBz0.0025 mol HBz
This requires This requires 0.0025 mol NaOH0.0025 mol NaOH
2.2. Calc. volume of NaOH req’dCalc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.0025 mol (1 L / 0.100 mol) = 0.025 L0.025 L
25 mL of NaOH req’d 25 mL of NaOH req’d
STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d1. Calc. moles of NaOH req’d
(0.100 L HBz)(0.025 M) = (0.100 L HBz)(0.025 M) = 0.0025 mol HBz0.0025 mol HBz
This requires This requires 0.0025 mol NaOH0.0025 mol NaOH
2.2. Calc. volume of NaOH req’dCalc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.0025 mol (1 L / 0.100 mol) = 0.025 L0.025 L
25 mL of NaOH req’d 25 mL of NaOH req’d
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION
25 mL of NaOH req’d 25 mL of NaOH req’d
3. Moles of Bz3. Moles of Bz-- produced = moles HBz = produced = moles HBz = 0.0025 mol0.0025 mol
4. Calc. conc. of Bz4. Calc. conc. of Bz--
There are 0.0025 mol of BzThere are 0.0025 mol of Bz-- in a in a TOTAL SOLUTION TOTAL SOLUTION VOLUMEVOLUME of of 125 mL125 mL
[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M
STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION
25 mL of NaOH req’d 25 mL of NaOH req’d
3. Moles of Bz3. Moles of Bz-- produced = moles HBz = produced = moles HBz = 0.0025 mol0.0025 mol
4. Calc. conc. of Bz4. Calc. conc. of Bz--
There are 0.0025 mol of BzThere are 0.0025 mol of Bz-- in a in a TOTAL SOLUTION TOTAL SOLUTION VOLUMEVOLUME of of 125 mL125 mL
[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
EQUILIBRIUM PORTIONEQUILIBRIUM PORTION
BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]
initialinitial changechange
equilibequilib
EQUILIBRIUM PORTIONEQUILIBRIUM PORTION
BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]
initialinitial changechange
equilibequilib
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
EQUILIBRIUM PORTIONEQUILIBRIUM PORTION
BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]
initialinitial 0.0200.020 00 00
changechange -x-x +x+x +x+x
equilibequilib 0.020 - x0.020 - x xx xx
EQUILIBRIUM PORTIONEQUILIBRIUM PORTION
BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]
initialinitial 0.0200.020 00 00
changechange -x-x +x+x +x+x
equilibequilib 0.020 - x0.020 - x xx xx
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
EQUILIBRIUM PORTIONEQUILIBRIUM PORTION
BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]
equilibequilib 0.020 - x0.020 - x xx xx
Solving in the usual way, we findSolving in the usual way, we find
x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6, pOH = 5.75, and , pOH = 5.75, and pH = 8.25pH = 8.25
EQUILIBRIUM PORTIONEQUILIBRIUM PORTION
BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10
[Bz[Bz--]] [HBz][HBz] [OH[OH--]]
equilibequilib 0.020 - x0.020 - x xx xx
Solving in the usual way, we findSolving in the usual way, we find
x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6, pOH = 5.75, and , pOH = 5.75, and pH = 8.25pH = 8.25
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?
Kb 1.6 x 10-10 = x2
0.020 - xKb 1.6 x 10-10 =
x2
0.020 - x
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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions
HBzHBz + H+ H22O O HH33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5
At the half-way point, [HBz] = [BzAt the half-way point, [HBz] = [Bz--], so], so
[H[H33OO++] = 1 (K] = 1 (Kaa ) = 6.3 x 10 ) = 6.3 x 10-5-5
pH = 4.20pH = 4.20
HBzHBz + H+ H22O O HH33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5
At the half-way point, [HBz] = [BzAt the half-way point, [HBz] = [Bz--], so], so
[H[H33OO++] = 1 (K] = 1 (Kaa ) = 6.3 x 10 ) = 6.3 x 10-5-5
pH = 4.20pH = 4.20
QUESTION:QUESTION: You titrate 100. mL of a 0.025 M You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point?What is the pH at the half-way point?
][
]][[ 3
HA
AOHKa
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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved
Henderson-Hasselbalch Henderson-Hasselbalch EquationEquation
Take the negative log of both sides of this Take the negative log of both sides of this equationequation
[H3O] [Acid]
[Conj. base]•Ka[H3O]
[Acid][Conj. base]
•Ka][
]][[ 3
HA
AOHKa
pH pKa - log[Acid]
[Conj. base]
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Henderson-Hasselbalch Henderson-Hasselbalch EquationEquation
oror
This is called the Henderson-Hasselbalch equation.This is called the Henderson-Hasselbalch equation.
pH pKa + log[Conj. base]
[Acid]pH pKa + log
[Conj. base][Acid]
pH pKa - log[Acid]
[Conj. base]
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Henderson-Hasselbalch Henderson-Hasselbalch EquationEquation
This shows that the pH is determined largely This shows that the pH is determined largely by the pKby the pKaa of the acid and then adjusted by of the acid and then adjusted by the ratio of acid and conjugate base.the ratio of acid and conjugate base.
pH pKa + log[Conj. base]
[Acid]pH pKa + log
[Conj. base][Acid]
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