AP CHEMISTRY TOPIC 13B: TITRATIONS

Beep, beep

, beep, beep….

Goes with chapter 18 – Silberberg: Principles of General Chemistry

Mrs. Laura Peck, 2013

1

2

TITRATION A titration can be used to determine the [] of an

unknown solution. The calculations involve stoichiometric

principles.

3

Titrations Before equivalence point

At equivalence point After equivalence point

Strong acid titrated w/strong base

Species in sol’t which affect pH

Use excess [H+] ion to calculate pH. pH < 7

H+

pH = 7

H2O

Use excess [OH-] ion to caclulate pH. pH > 7

OH-

Weak acid titrated w/strong base

Species in sol’t which affect pH

Weak acid + c.base (a buffer problem) pH < 7

HA, A-

Salt hydrolysis problem

pH > 7

A-

Use excess OH- ion to calculate pHpH > 7

OH-

Weak base titrated with a strong acid

Species in sol’t which affect the pH.

Weak base + c.acid (a buffer problem) pH > 7

B, HB+

Salt hydrolysis

pH < 7

HB+

Excess H+ ion

pH < 7

H+

-At any point during the titration of a strong acid with a strong Base (or vice-versa), the pH can be calculated from the molarity Of the ion present in excess after the complete reaction.

- At the equivalence point of any type of titration,- Moles H+ = mole OH-

-For a strong acid/strong base titration, the pH equals 7

EXAMPLE #16 Consider the titration of 40.0 ml of 0.200M HBr by 0.100M

KOH. Calculate the pH of the resulting solution when the following volumes of KOH have been added.

A) 10.00ml B) 80.0ml C) 100.0ml

4

1st, using the volume and molarity of the HBr, calculate the moles of H+ to Titrated. The resulting moles of H+ appears in the first column of your table.

2nd, for the titration in question, calculate the moles of OH- added from the Volume and [] of KOH. The resulting moles of OH- used in each titration Appear in the 2nd column of the table.

3rd, calculate the moles of ion in excess, H+ or OH-. The results for each Titration appear in column three in the table.

4th, using the total volume, in column 4 of the table, and the mols of ion inExcess, determine the molarity of the ion in excess. The molarity of the ionIn excess for each of the three titrations is in column 5.

It is the molarity of the ion in excess which determines the pH. In part b ofThis example, neither ion will be in excess. The pH equals 7 because theTitration is at the equivalence point. Be careful in part c, the excess is the OH-Ion. You must determine pOH first, then calculate pH.

Problem

Mol H+ Mol OH- Mol ion in excess

Total volume

Molarity ion in excess

pH

A 8.00x10-3 1.00x10-3 (H+) 7.00x10-3

0.0500L (H+)0.140M

0.854

B 8.00x10-3 8.00x10-3 NoneEquivalence point

Not needed

None 7.00

C 8.00x10-3 10.0x10-3 (OH-)2.0x10-3

0.140L (OH-) 0.014M

12.15

TITRATION OF A WEAK ACID WITH A STRONG BASE

There are three characteristic types of calculations for the titration of a weak acid with a strong base: Before the equivalence point, at the equivalence point, and after the equivalence point.

5

Example #17: a 25.0mL sample of 0.100M HC3H5O2 is titrated with0.100M NaOH. Calculate the pH when the following volumes of 0.100M NaOH are added: Ka = 1.3x10-5 for HC3H5O2

a) 8.0ml b) 12.5ml c) 25.0ml d) 30.0ml

-Part A of the example involves the titration before the equivalence point. -perform the stoich and record the results in a table under the reaction.-the first three lines in the table are filled with:

- Calculate the initial moles of the weak acid present in sample- 0.0250L x 0.100mol/L = 2.50x10-3 mol HC3H5O2

- Calculate the moles of strong base, OH-, added- 0.0080L x 0.100mol/L = 8.0x10-4 mol

-Calculate the new initial moles of weak acid and its c.base present afterThe OH- has reacted. -this problem is now identical to a ‘buffer’ problem-make new line under the initial moles called ‘mol OH- added’-calculate the new initial moles of weak acid and its c.base, after the Weak acid has reacted with the strong base, by subtracting or adding the Number of moles OH- added.

HC3H5O2 + HOH H3O+ + C3H5O2-

Initial mole 2.50x10-3 0 0

Add mol OH- -8.0x10-4 0 +8.0x10-4

New initial Mol 1.70x10-3 0 8.0x10-4mol

I 1.70x10-3/0.0330L 8.0x10-

4/0.0330L

=0.0515M = 0.0242M

C -x +x +x

E 0.0515-x x 0.0242+x

Ka = [H+][C3H5O2-] 1.3x10-5 = [H+](0.0242+x) ~ [H+]0.0242

[HC3H5O2] 0.0515-x 0.0515

2.77x10-5M = [H+] pH = -log[H+] = 4.56

**alternatively, you could’ve used the Henderson-Hasselbalch.

HALFWAY TO THE EQUIVALENCE POINT In part B of example #17, a 25.00mL sample

of 0.100M HC3H5O2 is titrated with 12.5 of 0.100M NaOH. Ka is 1.3x10-5 for HC3H5O2.

Half of the acid being titrated is neutralized. This point is halfway to the equivalence

point. Since half the acid is neutralized and half

remains, [HA]=[A-]

Ka = [H+][A-]

[HA] pKa = pH

pKa = -log(1.3x10-5) = 4.89

6

AT THE EQUIVALENCE POINT In part C of example #17, a 25.00mL sample of 0.100M HC3H5O2 is

titrated with 25.0ml of 0.100M NaOH. Ka is 1.3x10-5 for HC3H5O2

Calculate the initial moles of the weak acid in the sample. Calculate the moles of strong base, OH-, added.

7

0.0250L x 0.100mol/L = 2.50x10-3mol OH- = mol C3H5O2-

There is no weak acid present once the strong base added completely reacts. All that is present is A- (C3H5O2

-), the c.base of weak acid, HA (HC3H5O2)

HA + NaOH NaA + HOH *the problem from this point on is a saltHydrolysis problem.

C3H5O2- + HOH OH- + HC3H5O2

I 2.50x10-3/0.0500L 0 0 = 0.0500MC -x +x +x E 0.0500-x x x

Kb = Kw = 1.0x10-14 = 7.7x10-10 Kb = [OH-][HC3H5O2] = x2 ~ x2 Ka 1.3x10-5 [C3H5O2-] 0.0500-x 0.0500

X = [(7.7x10-10)(0.0500)]1/2 = 6.2x10-6M OH-

pOH = 5.21; pH = 14.00-5.21 = 8.79

AFTER THE EQUIVALENCE POINT The calculations involved here are identical

to those in a strong acid-strong base titration.

In part D of example #17, a 25.00mL sample of 0.100M HC3H5O2 is titrated with 12.5 of 0.100M NaOH. Ka is 1.3x10-5 for HC3H5O2.

0.0300L x 0.100mol/L = 0.00300mol NaOH 0.02500L x 0.100mol/L = 0.002500mol

HC3H5O2

Excess OH- = 0.00300mol – 0.002500mol/0.0500L = 0.010M

OH-

pOH = 2.00; pH = 14 – pOH = 12.00

8

TITRATION OF A WEAK BASE WITH A STRONG ACID

The calculations for a weak base-strong acid titration are very similar to the weak acid-strong base titration.

The differences are these: The pH before the equivalence point is greater

than 7 At the equivalence point, the pH is less than 7

since the salt formed is acidic After the equivalence point, the pH is determined

by the molarity of the excess H+

9

CHARACTERISTICS OF TITRATION CURVES.

Example #18: The questions below refer to the following titration curves. All solutions are equimolar.

10

a) Which titration curve represents a Strong acid titrated by a strong base?

b) Which titration curve represents aWeak acid titrated by a strong base?

c) Which titration curve represents aWeak base titrated by a strong acid?

d) Which titration curve represents aWeak acid titrated by a weak base?

INTERPRETING GRAPHS A = Weak acid, HA

B-C= Weak acid, HA and its c.base, A-

D=Basic ion, A-

E=Strong base, OH-

11

INTERPRETING GRAPHS CONT…..

12

INTERPRETING GRAPHS CONT……

13

THE END…… FOR NOW….

14