8.7 Acid-Base Titration

Learning Goals …… determine the pH of the solution formed in a neutralization reaction

• Titrations are used to determine the concentration of an acid or a base

• The equivalence point is the point in the titration when the acid and the base completely react with each other

• If you know the volumes of both solutions at the equivalence point and the concentration of one of them,

you can calculate the unknown concentration.

Titrations Involving Strong Acids and Strong Bases

Calculate the pH of HNO3 before any base is added.pH = - log [H+] = - log (0.200) = 0.699

Since HNO3 is a strong acid, [H+] = [HNO3] = 0.200 M

Calculate the pH of the solution after 50.0 mL of NaOH is added.

pH = - log [H+] = - log (0.05) = 1.30

HNO3 + NaOH H2O + NaNO3

nCV

0.005 mol0.010 mol0.200 M0.050 L

0.100 M0.050 L

LR

initialusedXS

0.010 mol

0.005 mol0.005 mol

[H+] = [HNO3] = 0.005 mol 0.100 L = 0.05 M

XS

What volume of NaOH will have been added at the equivalence point? What is the pH at the equivalence point?

HNO3 + NaOH H2O + NaNO3

nCV

0.010 mol0.010 mol0.200 M0.050 L

0.100 M0.100 L

At the equivalence point all the acid and base have been neutralized. Since this is a titration between a strong acid and a strong base, the salt produced (NaNO3) will have no effect on pH.

The pH at the equivalence point will be 7.0 (neutral)

• The equivalence point is the middle of the steep rise that occurs in a titration curve. The endpoint of a titration occurs when the indicator changes colour, which

happens over a range of about 2 pH units. The pH changes rapidly near the equivalence point.

• Chemists use indicators to observe the endpoint of the titration. Some common indicators are the following:

Phenolphthalein (colourless to pink)pH 8.2 – 10.0

Methyl Red (red to yellow)pH 4.3 – 6.2

Titration Curve for a Strong Acid with a Strong Base:

• These titrations have a pH of 7 at equivalence (neutral)

Titration of a Weak Acid with a Strong Base:Recall: The anion from the salt produced will react with water

Ex 1) 100 mL of 0.1M cyanic acid (HCN) is titrated with 100 mL of 0.1M KOH. Calculate the pH at equivalence.

0.1 M 0.1 M

HCN + KOH H2O + KCNnCV 0.100 L 0.100 L

0.01 mol 0.01 mol 0.01 mol

0.20 L0.05 M

CN- + H2O HCN + OH-i

ce

0.05 M 0 0- x + x+ x

0.05 - x xx

Kb for CN- = Kw = 1x10-14 = 1.6x10-5

Ka 6.2x10-10

Ka for HCN = 6.2x10-10

Can we simplify? 0.05/Kb > 100 YES

x2 = 1.6x10-5

(0.05)

x2 = 8.0x10-7

x = 8.94x10-4

[OH-] = x = 8.94x10-4

pOH = - log [OH-] = 3.05

pH = 10.95

x2 = 1.6x10-5

(0.05-x)

• These titrations have pH values that are greater than 7 at equivalence.

• The conjugate bases of weak acids will react with water • At the equivalence point, the weak acid is neutralized by

the base but the conjugate base is present, therefore increasing the pH

0.2 M 0.2 M

NH3 + HCl NH4+

+ Cl-

nCV 0.020 L 0.020 L

0.004 mol 0.004 mol 0.004 mol

0.040 L0.1 M

Titration for a Weak Base with a Strong AcidRecall: The cation from the salt produced will react with water

Ex 1) 20 mL of 0.2M NH3 is titrated against 0.2M HCl. Calculate the pH at equivalence.

NH4+ + H2O NH3 +

H3O+ice

0.1 M 0 0- x + x+ x

0.1 - x xx

Ka for NH4+ = Kw = 1x10-14 = 5.56x10-10

Kb 1.8x10-5

Kb for NH3 = 1.8x10-5

Can we simplify? 0.1/Ka > 100 YES

x2 = 5.56x10-10

0.1 x = 7.45x10-6

[H3O+] = x = 7.45x10-6

pH = - log [H3O+]

pH = 5.13

x2 = 5.56x10-10

0.1-x

• These titrations have pH values that are less than 7 at equivalence.• The conjugate acids of weak bases will react with water• At the equivalence point, the weak base is neutralized by

the acid but the conjugate acid is present, therefore decreasing the pH

HOMEWORKp547 #1,2p554 #1,2p557 #1-8

Self CheckHow prepared am I to start my homework? Can I …… determine the pH of the solution formed in a neutralization reaction