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28-3-2011

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Diprotic and Polyprotic Acids

Have more than one acidic proton.

If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

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H2CO3H+HCO3-

Initial0.1000

Change-xX+X+

Equil.0.1-xxx

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H3PO4H+H2PO4-

Initial0.1000

Change-xX+X+

Equil.0.1-xxx

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Species concentrations of diprotic acids

Evaluate concentrations of species in a 0.10 M H2SO4 solution.

Solution:H2SO4 H+ + HSO4

– completely ionized

(0.1–0.1) 0.10+y 0.10-y

HSO4–

H+ + SO42– Ka2 = 0.012

0.10–y 0.10+y y Assume y = [SO42–]

(0.10+y) y————— = 0.012(0.10-y)

[SO42–] = y = 0.01M

[H+] = 0.10 + 0.01 = 0.11 M; [HSO4

–] = 0.10-0.01 = 0.09 M

Y2 + 0.112 y – 0.0012 = 0

- 0.112+0.1122 + + 4*0.0012y = —————————————— = 0.0098

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Another approach: Two steps method

Solution:H2SO4 H+ + HSO4

– completely ionized

(0.1–0.1) 0.10 0.10

In the 2nd equilibrium, substitute in [H+] = 0.1 M and [HSO4-] = 0.1 M

HSO4–

H+ + SO42– Ka2 = 0.012

0.10–x 0.10+x x

(0.10+x) x————— = 0.012, assume 0.1>>x, RE = {0.012/0.1}*100 = 12%

(0.10-x) Assumption is not OK

[SO42–] = y = 0.01M

[H+] = 0.10 + 0.01 = 0.11 M; [HSO4

–] = 0.10-0.01 = 0.09 M

Y2 + 0.112 y – 0.0012 = 0

- 0.112+0.1122 + + 4*0.0012y = —————————————— = 0.0098

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Species concentrations of weak diprotic acids

Evaluate concentrations of species in a 0.10 M H2S solution.

Solution:H2S H+ + HS– Ka1 = 1.02*10-7

(0.10–x) x+y x-y Assume x = [HS–]

HS– H+ + S2– Ka2 = 1.0*10-13

x–y x+y y Assume y = [S2–]

(x+y) (x-y) (x+y) y————— = 1.02*10-7 ———— = 1.0*10-13

(0.10-x) (x-y)

[H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = x y = 1.0*10–4 M; [S2–] = y = 1.0*10-13 M

0.1>> x >> y: x+ y = x-y = x

x = 0.1*1.02*10-7 = 1.00*10-4

y = 1*10-13

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Another approach: Two steps methodSolution:

H2S H+ + HS– Ka1 = 1.02*10-7 HS–

H+ + S2– Ka2 = 1.0*10-13

Ka1/Ka2 >102 , therefore the 2nd equilibrium can be neglected for the moment. From the 1st equilibrium:

H2S H+ + HS– Ka1 = 1.02*10-7

0.1 – x x x 1.02*10-7 = x2/(0.1-x), x = 10-4 and RE = {10-4/0.1}*100 = 0.1% (OK)

Therefore, [H+] = [HS-] = 10-4M

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From the second equilibrium, we have:

HS– H+ + S2– Ka2 = 1.0*10-13

10-4 -x 10-4 +x x

1.0*10-13 = (10-4 +x)x/(10-4 -x)

Assume 10-4 >>x

X = 1.0*10-13 which is very small compared to 10-4

Therefore, the concentrations of the different species are:

[H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = 10-4 - x = 10–4 M; [S2–] = x = 1.0*10-13 M

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NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH3]

Kb is the base ionization constant

Kb

weak basestrength

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

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Ionization Constants of Conjugate Acid-Base Pairs

HA (aq) H+ (aq) + A- (aq)

A- (aq) + H2O (l) OH- (aq) + HA (aq)

Ka

Kb

H2O (l) H+ (aq) + OH- (aq) Kw

KaKb = Kw

Weak Acid and Its Conjugate Base

Ka = Kw

Kb

Kb = Kw

Ka

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Ka Kb and Kw

H+ + Base Conjugate_acid of Base+

Acid H+ + Conjugate_base of Acid-

For example:

NH3 + H2O NH4+ + OH-

Ka for NH4+ = Kw / Kb for NH3

HA H+ + A- Kb for A- = Kw / Ka for HA

Thus, Ka Kb = Kw for conjugate pairs.

A- + H2O HA + OH-

[HA] [OH-] [H+]Kb = ————— ———

[A-] [H+]

[HA] = ———— [OH-] [H+] [A-] [H+]

1 = —— Kw Ka

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Ka Kb and Kw – another way

HA + H2O (l) H3O+ (aq) + A– (aq) Ka of HA +) A– + H2O (l) OH–(aq) + HA (aq) Kb of A–

2 H2O (l) H3O+ (aq) + OH– (aq) Kw = Ka Kb

When you add two equations to get a third, what are the relationship between the K’s?

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Weak Bases

Bases react with water to produce hydroxide ion.

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pH of Basic Solutions

What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH−]

[NH3]Kb = = 1.8 10−5

NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)

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NH3(aq) + H2O(l) NH4+ (aq) + OH−(aq)

Initial 0.15 0.00 0.00Change -x +x +x

Equilibrium 0.15 – x x x

1.8*10-5 = x2/(0.15 – x)Kb seems very small, therefore assume that 0.15>>x

X = 1.6*10-3

Relative error = {1.6*10-3/0.15}*100% = 1.1%, which is acceptable.[OH-] = 1.6*10-3 M, pOH = 2.79, therefore pH = 14-2.79 = 11.2

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Aniline is an organic chemical used in dye synthesis and is a weak base, with a Kb of 3.8 x 10-10. What is the pH of a 1.5 M ArNH2 solution?

ArNH2 (aq) + H2O(l) ArNH3+

(aq) + OH-(aq)

Kb = [ArNH2]

Concentration (M) ArNH2 H2O ArNH3+ OH-

Initial 1.5 ---- 0 0Change -x ---- +x +xEquilibrium 1.5 - x ---- x x

making the assumption: since Kb is small: 1.5 M - x = 1.5 M

[ArNH3+] [OH-]

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Substituting into the Kb expression and solving for x:

Kb = = = 3.8 x 10-10[ArNH3

+] [OH-]

[ArNH2]

(x)(x)

1.5 - x

x = 2.4 x 10-5 = [OH-] = [ArNH3+]

Calculating pH:

[H3O+] = = = 4.2 x 10-10Kw

[OH-]

1.0 x 10-14

2.4 x 10-5

pH = -log[H3O+] = - log (4.2 x 10-10) = 9.4Or calculate pOH = 4.6 pH = 14 – 4.6 = 9.4

x = 2.4 x 10-5 , RE = {2.4*10-5/1.5}*100% = 1.6*10-3%