CHM 152 Titration Handout

CHM 152 Titration Handout

Calculating pH values

Strong Acid (known) with Strong Base (titrant)

Example: 20.00 mL of 0.20 M HCl is titrated with 0.20 M NaOH.

Determine the volume of base needed to reach the equivalence point.

Write a balanced equation: HCl(aq) + NaOH(aq) à H2O(l) + NaCl(aq)

Moles HCl = moles NaOH (because 1:1 stoichiometric ratio)

(0.20 M)(20.00 mL) = (0.20 M)(x mL) x = 20.00 mL à 20.00 mL of base to reach equivalence point.

1) Before any titrant is added (0.00 mL NaOH).

All acid; strong acid: 0.20 M HCl = 0.20 M H3O+ pH = -log (0.20 M) = 0.70

2) Before equivalence point is reached (10.00 mL NaOH).

Find mmol of excess acid: mmol acid – mmol base

(0.20 M)(20.00 mL) – (0.20 M(10.00 mL) = 2.00 mmol acid in excess

Find concentration of acid: mmol excess acid / total volume of solution

2.00 mmol / 30.00 mL = 0.06667 M HCl = 0.06667 M H3O+ pH = -log(0.06667 M) = 1.18

3) At equivalence point (20.00 mL NaOH).

Since this is a strong acid-strong base system, only a neutral salt (NaCl) and water are in the container. The equivalence point is reached at pH = 7 (for strong acid-strong base systems ONLY).

4) After equivalence point (30.00 mL NaOH).

All base; Find mmol of excess base: mmol base – mmol acid

(0.20 M)(30.00 mL) – (0.20 M)(20.00 mL) = 2.00 mmol base in excess

Find concentration of base: mmol base / total volume

2.00 mmol / 50.00 mL = 0.040 M NaOH = 0.040 M OH-

pOH = -log(0.040 M) = 1.40 OR [H3O+] = 1 x 10-14 / 0.0400 = 2.5 x 10-13 M

pH = 14 – 1.40 = 12.60 pH = -log(2.5 x 10-13 M) = 12.60

Weak Acid (known) with Strong Base (titrant)

Example: 20.00 mL of 0.20 M HCN is titrated with 0.20 M NaOH.

Determine the volume of base needed to reach the equivalence point.

Write a balanced equation: HCN(aq) + NaOH(aq) à H2O(l) + NaCN(aq)

Moles HCN = moles NaOH (because a 1:1 stoichiometric ratio)

(0.20 M)(20.00 mL) = (0.20 M)(x mL) x = 20.00 mL à Volume of base needed to reach equivalence point.

Ka of HCN = 4.9 x 10-10

1) Before any titrant is added (0.00 mL NaOH).

All acid (no base has been added); weak acid: ICE table

Ka = x2 / (0.20 – x) Assume x is small. Ka = x2 / 0.20 x = 9.899 x 10-6 M

[H3O+] = 9.899 x 10-6 M pH = -log(9.899 x 10-6 M) = 5.00

2) Before equivalence point is reached (10.00 mL NaOH). BUFFER ZONE!

Base is the limiting reactant; all base reacts to form conjugate base of acid (CN- in this case)

Find concentration (or moles) of excess acid AND concentration (or moles) of conjugate base

Excess acid: mmol acid - mmol base

[(0.20 M)(20.00 mL) – (0.20 M)(10.00 mL)] = 2.00 mmol excess HCN

Amount of CN-: all strong base (limiting reactant) is converted to conjugate base

(0.20 M)(10.00 mL) = 2.00 mmol CN-

Can use ICE table using concentrations (L) or Henderson-Hasselbach equation using mmoles (J)

pH = pKa + log (base / acid) = - log (4.9 x 10-10) + log (2.00 mmol / 2.00 mmol)

pH = -log(4.9 x 10-10) = 9.30

This point is the half-equivalence point (half of base needed for equivalence point has been added).

At this point, pH = pKa. Why?

3) At equivalence point (20.00 mL NaOH).

We have Na+, CN-, and H2O in solution. H3O+ is no longer produced.

Equilibrium is established by the reaction of conjugate base (CN-) with water (hydrolysis).

[CN-] is calculated by concentration of base: mmol base / total volume

(0.20 M)(20.00 mL) / 40.00 mL = 0.10 M base

Since we have a base produced, we have to use Kb, not Ka. Kb = Kw / Ka

Kb = 1.0 x 10-14 / 4.9 x 10-10 = 2.04 x 10-5

Kb = x2 / (0.10 – x) Assume x is small. Kb = x2 / 0.100

x = 1.428 x 10-3 M = [OH-] [H3O+] = 1 x 10-14 / 1.428 x 10-3 M = 7.003 x 10-12 M

pH = -log(7.003 x 10-12 M) = 11.15

4) After equivalence point (30.00 mL NaOH).

Solution consists of excess base. pH is calculated by excess OH-.

[OH-] = (mmol base – mmol acid) / total volume

[(0.20 M)(30.00 mL) – (0.20 M)(20.00 mL)] / 50.00 mL = 0.0400 M OH-

pOH = -log(0.040 M) = 1.40

pH = 14 – 1.40 = 12.60 (Does this value look familiar? Check the stong acid-strong base titration.)

Why do both titrations yield the same pH when there is excess base?