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Ionic Equilibria in Aqueous Systems Section 19.2

Acid-Base Titration Curves

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Acid-Base Indicators

An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In–).

The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

The ratio [HIn]/[In–] is governed by the [H3O+] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction.

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Figure 19.6 Colors and approximate pH range of some common acid-base indicators.

pH

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Figure 19.7 The color change of the indicator bromthymol blue.

pH < 6.0 pH > 7.5 pH = 6.0-7.5

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Acid-Base Titrations

In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration.

The equivalence point of the reaction occurs when the number of moles of OH– added equals the number of moles of H3O+ originally present, or vice versa.

The end point occurs when the indicator changes color. - The indicator should be selected so that its color change occurs at a

pH close to that of the equivalence point.

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Figure 19.8 Curve for a strong acid–strong base titration.

The pH increases gradually when excess base has been added.

The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00.

The initial pH is low.

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Strong Base-Strong Acid Titration Curve

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Calculating the pH during a strong acid–strong base titration

Initial pH

[H3O+] = [HA]init

pH = –log[H3O+]

pH before equivalence point initial mol H3O+ = Vacid x Macid

mol OH– added = Vbase x Mbase

mol H3O+remaining = (mol H3O+

init) – (mol OH–added)

[H3O+] = pH = –log[H3O+] mol H3O+remaining

Vacid + Vbase

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pH at the equivalence point pH = 7.00 for a strong acid-strong base titration.

Calculating the pH during a strong acid–strong base titration

pH beyond the equivalence point initial mol H3O+ = Vacid x Macid

mol OH– added = Vbase x Mbase

mol OH–excess = (mol OH–

added) – (mol H3O+init)

[OH–] =

pOH = –log[OH–] and pH = 14.00 - pOH

mol OH–excess

Vacid + Vbase

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Example:

40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH.

The initial pH is simply the pH of the HCl solution:

[H3O+] = [HCl]init = 0.1000 M and pH = –log(0.1000) = 1.00

To calculate the pH after 20.00 mL of NaOH solution has been added:

OH– added = 0.02000 L NaOH x 0.1000 mol 1 L

= 2.000x10–3 mol OH–

Initial mol of H3O+ = 0.04000 L HCl x 0.1000 mol 1 L

= 4.000x10–3 mol H3O+

H3O+ remaining = 4.000x10–3 – 2.000x10–3 = 2.000x10–3 mol H3O+

The OH– ions react with an equal amount of H3O+ ions, so

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[H3O+] = 2.000x10–3 mol 0.04000 L + 0.02000 L

= 0.03333 M

pH = –log(0.03333) = 1.48

To calculate the pH after 50.00 mL of NaOH solution has been added:

The equivalence point occurs when mol of OH– added = initial mol of HCl, so when 40.00 mL of NaOH has been added.

OH– added = 0.05000 L NaOH x 0.1000 mol 1 L

= 5.000x10–3 mol OH–

OH– in excess = 5.000x10–3 – 4.000x10–3 = 1.000x10–3 mol OH–

[OH–] = 1.000x10–3 mol 0.04000 L + 0.05000 L

= 0.01111 M

pOH = –log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05

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Figure 19.9 Curve for a weak acid–strong base titration.

The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution.

The pH increases slowly beyond the equivalence point.

The pH at the equivalence point is > 7.00 due to the reaction of the conjugate base with H2O.

The initial pH is higher than for the strong acid solution.

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Calculating the pH during a weak acid–strong base titration

Initial pH

[H3O+] = √Ka x [HA]init

pH = –log[H3O+]

[H3O+][A–] [HA]

Ka =

pH before equivalence point

[H3O+] = Ka x or

pH = pKa + log [base] [acid]

[HA] [A–]

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Calculating the pH during a weak acid–strong base titration

A–(aq) + H2O(l) HA(aq) + OH–(aq)

pH at the equivalence point

[OH–] =

where [A-] = and Kw Ka

Kb = mol HAinit

Vacid + Vbase

[H3O+] ≈ and pH = -log[H3O+] Kw

pH beyond the equivalence point

[OH–] =

pH = –log[H3O+]

mol OH–excess

Vacid + Vbase

[H3O+] = Kw

[OH–]

√Kb x [A–]

√Kb x [A–]

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Acid/Base Titration Problems • “What do I have in solution?”

• Two parts of calculations

• Stoichiometric part

• Equilibrium part

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Sample Problem 19.4 Finding the pH During a Weak Acid–Strong Base Titration

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10–5) after adding the following volumes of 0.1000 M NaOH:

PLAN: The initial pH must be calculated using the Ka value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH– added. Once we know the volume of base required to reach the equivalence point we can calculate the pH based on the species present in solution.

(a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL

SOLUTION: (a) [H3O+] = √Ka x [HPr]init = √(1.3 x 10–5) (0.100) = 1.1x10–3 M

pH = –log(1.1x10–3) = 2.96

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Sample Problem 19.4

(b) 30.00 mL of 0.1000 M NaOH has been added.

Initial amount of HPr = 0.04000 L x 0.1000 M = 4.000x10–3 mol HPr

Amount of NaOH added = 0.03000 L x 0.1000 M = 3.000x10–3 mol OH–

Each mol of OH– reacts to form 1 mol of Pr–, so HPr(aq) + OH–(aq) → Pr-(aq) + H2O(l) Concentration (M)

Initial 0.004000 0.003000 0 - Change −0.003000 −0.003000 +0.003000 -

Equilibrium 0.001000 0 - 0.003000

[H3O+] = Ka x [HPr] [Pr-]

pH = −log(4.3x10–6) = 5.37

0.001000 0.003000

= (1.3x10–5) x = 4.3x10–6 M

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Sample Problem 19.4

(c) 40.00 mL of 0.1000 M NaOH has been added. This is the equivalence point because mol of OH− added = 0.004000 = mol of HAinit. All the OH− added has reacted with HA to form 0.004000 mol of Pr−.

0.004000 mol 0.04000 L + 0.04000 L

[Pr−] = = 0.05000 M

Pr− is a weak base, so we calculate Kb = Kw

Ka

= 1.0x10−14 1.3x10−5

= 7.7x10−10

1.0x10-14 = 1.6x10-9 M

pH = –log(1.6x10-9) = 8.80

[H3O+] ≈ Kw = √Kb x [Pr−] √(7.7x10-10)(0.05000)

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Sample Problem 19.4

(d) 50.00 mL of 0.1000 M NaOH has been added.

Amount of OH– added = 0.05000 L x 0.1000 M = 0.005000 mol

Excess OH– = OH–added – HAinit = 0.005000 – 0.004000 = 0.001000 mol

[OH–] = mol OH–

excess

total volume

= 0.001000 mol

0.09000 L = 0.01111 M

[H3O+] = Kw

[OH–] =

1x10–14

0.01111 = 9.0x10–13 M

pH = –log(9.0x10–13) = 12.05

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Figure 19.10 Curve for a weak base–strong acid titration.

The pH at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H2O.

The pH decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution.

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Figure 19.11 Curve for the titration of a weak polyprotic acid.