Principles of Neutralization Titrations: Determining Acids, Bases and the Ph of Buffer

Chapter 12

Principles of Neutralization Titrations: Determining Acids, Bases and the pH of Buffer Solutions

Class Notes:

Types of Titrations:

-  strong acid titrated with a strong base

-  strong base titrated with a strong acid

-  weak acid or weak base titrated with a strong acid or strong base

-  weak acid or base titrated with weak base or acid

*Can’t titrate a strong acid or base to a weak acid or base

Acids:

Monoprotic Acid – HA

Diprotic Acid – H2A

Triprotic Acid – H3A

Bases: for the titrations of which one equivalent H+, 2 or 3 equivalents of acidity

A3-

H2A2-

HA-

HA requires 1 H3O+ to titrate

H2A requires 2H3O+ to titrate

H3A requires 3H3O+ to titrate

Clinical/Medical Application for Titrations and pH:

Tears

Perspiration

Circulatory

Extracellular All have distinct pH values

Intracellular

Urine

Excrement

Tissue Based pH:

Kidneys pH > 8-9

Lower GI Tract pH à 2-3

Example:

Titrate 100ml of 0.1M HCL with 0.1000M NaOH

Phenylpthalein Indicator

NaOH + HCl à NaCl + Water

OH- + H+ à H2O

Before the addition of any base:

pH = -logAH+

[HCl] = 0.1M à [H+] = [H3O+] = 0.1M

pH = -log(0.1) = 1.0

So, pH of HCl before any titration with base = 1.0

After adding 25ml of 0.1000M Base NaOH

(25ml)(0.1000mmol/ml) = 2.5 mmol NaOH added

total mmol HCl = (100ml)(0.1mmol/ml) = 10 mmol HCl

mmol HCl left = 10 mmol – 2.5 mmol = 7.5 mmol HCl

[H3O+] = [HCl] = 7.5 mmol divided by new volume in titration 100ml(HCl) + 25ml(NaOH) = 125 ml

7.5 mmol = 0.06M

125 ml

pH = -log0.06 = 1.22 Pre-equivalence Point

50 ml 0.1M Base = 5 mmol Base added from buret initially had 100ml (0.1M) = 10 mmol HCl – 5 mmol Base = 5 mmol acid left

[H3O+] = 5 mmol / 150ml = 0.0333M

pH = -log0.0333 = 1.48 Equivalence Point

(99.9ml)(0.1M OH) = 9.99 mmol Base

acid left = 10 – 9.99 = 0.01 mmol HCl

0.01  mmol / 199.9ml = 0.00005 M

pH = -log0.00005 M = 4.3 Post-equivalence Point

A strong acid titrated with strong base has an end point at pH 7.0.

A weak acid titrated with weak base has its end point at pH 7.0.

pH 7 should be at the middle of the inflection on the reaction’s titration curve.

Post-equivalence Point for Base:

(125ml)(0.1M OH) = 12.5 mmol OH

mmol Base in excess = 12.5 mmol – 10 mmol = 2.5 mmol excess NaOH

[OH-] excess = 2.5 mmol – 225 ml = 0.00111M

pOH = -log0.00111

pOH = 1.88 – 14 = 12.11

Example:

Titrate

50ml of 0.1M HOAc (weak acid) with a 0.1M KOH (strong base)

At 0ml Base added:

HOAc + H2O ↔ OAc- + H3O+

Ka = [H3O+][OAc-] = 1.78 x 10-5

[HOAc]

*If K(100) is still smaller than [HA] than the x in the denominator is negligible.

1.78x10-5 = (X)(X) = X2 =

.1 .1

X= √1.78x10-5 = 1.34x10-3M

pH = -log 1.34x10-3

pH = 2.87

After adding 15ml of Base:

15ml (0.1M KOH) = 1.5 mmol

mmol HOAc = 5 mmol KOH – 1.5 mmol = 3.5 mmol

Ka = [H3O+][OAc-] = [H3O+](1.5mmol)

[HOAc] ______65ml______= 4.14x10-5

(3.5mmol)

65ml

pH = -log 4.14x10-5

pH= 4.48

50% equivalence point:

(25ml)(0.1M) = 2.5 mmol = [OAc-] = 2.5 mmol / 75ml = 0.0333M

5 mmol – 2.5 mmol = 2.5 mmol

[HOAc] = 2.5 mmol / 75ml = 0.0333M

1.78x10-5 = [H3O+][0.0333M] = [H3O+]

0.03333M

pH = -log 1.78x10-5

pH = 4.75 at 25ml

[HOAc] = [OAc]

pKa = pH

[OAc] = 5 mmol / 100ml = 0.05M

OAc- + H2O ↔ HOAc + OH- (hydrolysis reaction)

Kb = Kw/Ka = 1.0x10-14 / 1.78x10-5 = [HOAc][OH-] = (X)(X)

[OAc-] 0.05M

5.6x10-10 = (X)(X) = 5.29x10-6

0.05M

pOH = -log 5.29x10-6

pOH = 8.77 at 100ml

concentration of all bases = [OH-]1 + [OH-]2 + [OH-]3

[OH]1 = OAc- hydrolysis

[OH]2 = excess OH (effects pH)

[OH]3 = autoionization of water

Text Notes:

Neutralization titrations are dependent on a chemical reaction between an analyte and a standard reagent. The equivalence point of the titration is usually marked by a chemical indicator (color change) or instrumental measurement.

I.  Standard Solutions

-  Strong acids or strong bases because they react more completely with an analyte than weak acids and bases leading to a more clear end points.

-  The larger the Ka of a particular solute the sharper the end point it will have.

Problem 12-12

Solute with sharper end point in a titration with 0.10M NaOH?

(a)  0.10M nitrous acid or 0.10M iodic acid?

The iodic acid has a higher Ka value so it will have the sharper end point.

(b)  0.10M anilinium hydrochloride or0.10M benzoic acid?

Benzoic acid has larger Ka à better end point.

(c)  0.10M hypochlorous acid or 0.10M pyruvic acid?

Pyruvic Acid

(d)  0.10M salicylic acid or 0.10M acetic acid?

Salicylic Acid

-  Dilutions of hydrochloric, perchloric or sulfuric are typically used as acidic standard solutions and NaOH and KOH are the most commonly used strong bases.

II.  Acid / Base Indicators

-  Indicators display a color that is dependent on the pH of the solution in which they are dissolved so the solution can be identified as acidic or basic.

-  The acid / base indicator is usually a weak organic acid or weak organ base whose undissociated form differs in color from its conjugate form.

-  An acid indicator will change color when an acidic H+ attaches to it and show its base color when the H+ comes off:

HIn + H2O ↔ In- + H3O+

HIn shows acid color

In- shows base color

Pure acid color will be seen when [HIn] / [In-] ≥ 10 / 1

Pure base color will be seen when [HIn] / [In-] ≤ 1/ 10

For full acid color the hydronium concentration need o change it to that color will be:

[H3O+] = 10Ka

And for base it would be:

[H3O+] = 0.1Ka

The pH range of these expressions is found by the negative logarithm:

pH(acid color) = -log(10Ka) = pKa + 1

pH(basic color) = -log(0.1Ka) = pKa – 1

-  The pH interval over which a given indicator exhibits a color change is affected by temperature, ionic strength of the medium and by the presence of organic solvents and colloidal particles. These factors can cause a pH shift of1 to 2 pH units.

A.  Common Acid / Base Indicators

-  Thymol blue, methyl yellow, bromocresol green, phenol red and phenolphthalein are just a few that all indicate different pH increments.

III.  Calculating pH in Titration of Strong Acids and Strong Bases

-  In solutions of a strong acid that are more concentrated than 1x10-6M, it can be assumed that the equilibrium concentration of the hydronium ion is equal to the analytical concentration of the acid. The same idea can be assumed for the hydroxide ion in solutions of strong bases.

A.  Titrating a Strong Acid with a Strong Base

Problem 12-23

20ml HCl ( 0.2000M HCl) = 4.0 mmol HCl

a)  [HCl] = 4 mmol HCl / 20ml + 25ml = 0.0889M

pH = -log 0.0889 = 1.05

b)  1.05

c)  [HCl] = 4 mmol HCl – (25ml)(0.132M NaOH) / 45ml = -log = 1.81

d)  1.81

e)  [NaOH] = 25ml(0.232M NaOH) – 4 mmol HCl / 45ml = -log –14 = 12.60

1.  Titration Curve- curve produced as the acid progresses through preequivalence, equivalence and postequivalence points.

a)  preequivalence point – concentration of the acid is determined from its starting concentration and the amount of base that has been added.

b)  Equivalence point – point at which hydronium and hydroxide ion concentrations are equal (pH 7).

c)  Postequivalence point – analytical concentration of the base is calculated and that is assumed to be equal or a multiple of the hydroxide concentration.

Problem 12-18

14g HCl x 1.054g soln x 1 mmol HCl = 4.047M

100g ml soln 0.03646g/mmol

[HCl] = 4.047M

pH = -log 4.047 = -0.607

Problem 12-19

9.00g HCl x 1.098g soln x 1mmol NaOH = 2.471M = [OH-]

100g ml 0.04g NaOH

-log 2.471 = -0.393

pOH = 14 – (-0.393) = 14.393

d)  The larger the concentration of the titrant the larger the change in pH in the equivalence point region. A lesser concentration will still have an equivalence point but it will be markedly less than the titrant of greater concentration.

IV.  Titrating a Strong Base with a Strong Acid

-  A titration curve for strong bases are constructed in the same fashion as for a strong acid. Except for the equivalence point the solution is very basic. The hydroxide ion concentration is numerically related to the molarity of the base. The solution is neutral at the equivalence point and able to become basic is at the post equivalence point as the hydronium concentration becomes equal to the analytical concentration of excess strong acid.

Problem 12-28

NH3 + H2O ↔ NH4+ + OH- Kb = 1.00x10-14

5.70x10-10

a) 1.00x10-1 [NH4+][OH-] = [OH] = [OH]2 = 1.75x10-5 =

[NH3] 0.100-[OH] 0.100

√1.75x10-5 (0.100) = 1.323x10-3

pH = 14 – (-log 1.323x10-3) = 11.12

b) use the same process just substitute 0.100 with 0.0100M and pH = 10.62

c) proceed in same fashion just substitute 0.000100M for 0.0100M and pH= 9.62

V.  Buffer Solutions

-A buffer is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid that resists changes in pH of a solution

-A buffer solution consisting of a conjugate acid/base pair formed when a weak acid is titrated with a strong base or a weak base with a strong acid.

-Buffers are used to maintain a pH of a solution at a constant and predetermined level.

Problem 12-45

Volume of 0.200M HCl that needs to be added to 250ml of 0.300M sodium mandelate to make a buffer solution with a pH of 3.37.

V = milliliters of 0.200M HCl that must be added

HCl added = V (0.200 mmol)

NaA remaining = original amount of NaA – V added =

250ml (0.300 mmol/ml) – V (0.200 mmol/ml)

total volume of solution = 250ml + V

[HCl] = 0.200 V

250 + V

[NaA] = 75 – 0.200

250 + V

[NaA] = (75 – 0.200 V) / (250 + V) = 0.9377

[HCl] (0.200 V) / (250 + V)

75 – 0.200 V = 0.200 V (0.9377) = 0.1875 V

V = 75 / (0.200 + 0.1875) = 194ml HCl added

A.  Buffer Properties

1.  the pH of a buffer remains basically independent of dilution

2.  resist pH change after addition of small amounts of strong acids or bases

3.  the buffer capacity of a buffer is the number of moles of strong acid or strong base that causes one liter of the buffer to change pH by one unit

VI.  Calculating pH in Weak Acid or Weak Base Titrations

1.  At the beginning the solution contains only a weak acid or a weak base and the pH can be calculated from the concentration of that solute and its dissociation constant

2.  After various increments of titrant have been added, the solution consists of a series of buffer. The pH of each buffer can be calculated from the analytical concentrations of the conjugate base or acid and the residual concentrations of the weak acid or base.

3.  At the equivalence point, the solution contains only the conjugate of the weak acid or base being titrated and the pH is calculated from the concentration of this product.

4.  Beyond the equivalence point, the excess of the strong acid or base titrant represses the acidic or basic character of the reaction product to such an extent that the pH is governed largely by the concentration of the excess titrant.

VII.  How Buffer Solutions Change as a Function of pH

- Plot the relative equilibrium concentration of the weak acid and conjugate base as a function of the pH of the solution. The relative concentrations are known as alpha values. CT is the sum of the analytical concentrations of the acid and base at any point on the plot. This all leads to the development of the formula:

α = [OAc] = ____Ka_____

CT [H3O+] + Ka