Titrations are AWESOME!!

We looked at titrations today. Remember, titrations are just stoichiometry problems applied to a specific system/type of problem, they're not completely new information, approach them the same way you would approach any other stoichiometry problem:

1. Write a balanced chemical equation

2. Convert whatever you know the most about to moles

3. Using the mole ratio from the balanced chemical equation, convert moles of what you know to moles of what you're looking for

4. Convert moles of what you're looking for into whatever you want to know about it (grams, volume, concentration, etc.)

5. Check that your answer is reasonable (if possible)

On to today's problems...

30.00mL of 0.713M HNO₂(aq) is titrated to the equivalence point with 28.43mL of NaOH(aq) of an unknown concentration. What is the concentration of the NaOH(aq) stock solution? What was the pH of the HNO₂(aq) solution before the titration begins? What is the pH at the equivalence point? {K_a(HNO₂) = 4.0x10^-4}

HNO₂(aq) + NaOH(aq) ⇄ H₂O(l) + NaNO₂(aq)

(0.03000L HNO₂(aq)) (0.713M HNO₂(aq)) = 0.02139mols HNO₂

(0.02139mols HNO₂) (1mol NaOH / 1mol HNO₂) = 0.02139mols NaOH

(0.02139mols NaOH) / (0.02843L NaOH(aq)) = 0.752M NaOH(aq)

NaOH(aq) should be slightly more concentrated than HNO₂(aq), so this answer is reasonable

Before the titration begins, this is an aqueous solution of a weak acid, so we can calculate the pH using a K_a-type approach. Setting up a table...

	HNO2(aq) +	H2O(l) ⇄	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.713M	XXXX	0	0
Δ[ ]	- x	XXXX	+ x	+ x
[ ]equilibrium	(0.713 – x) M	XXXX	x M	x M

Assuming that “x” is much less than 0.713, the K_a expression simplifies to:

K_a = (x)(x) / (0.713) = 4.0x10^-4

x = 0.01689 = [H₃O⁺]

pH = -log[H₃O⁺] = -log(0.01689) = 1.77

At the equivalence point, all of the HNO₂(aq) that was originally in the reaction has reacted with OH^-1(aq) to form nitrite ions, NO₂^-1(aq). The titration started with:

(0.03000L HNO₂(aq)) (0.713M HNO₂(aq)) = 0.02139mols HNO₂

So at the equivalence point we have a solution that contains 0.02139mols of NO₂^-1(aq) in (30.00mL + 28.43mL = 58.43mL) of solution. The concentration of NO₂^-1(aq) at the equivalence point is:

(0.02139mols of NO₂^-1(aq)) / (0.05843L) = 0.3659M NO₂^-1(aq)

This can now be plugged in to a K_b-type equilibrium to solve...

	NO2-1(aq) +	H2O(l) ⇄	OH-1(aq) +	HNO2(aq)
[ ]initial	0.3659M	XXXX	0	0
Δ[ ]	- x	XXXX	+ x	+ x
[ ]equilibrium	(0.3659 – x) M	XXXX	x M	x M

Assuming that “x” is much less than 0.3659, the K_b expression simplifies to:

K_b = (x)(x) / (0.3659) = 2.5x10^-11

x = 3.02x10^-6 = [OH^-1]

pOH = -log[OH^-1] = -log(3.02x10^-6) = 5.519

pH = 14 – 5.519 = 8.48

15.00mL of sulfurous acid of unknown concentration is titrated to the second equivalence point with 23.18mL of 0.332M NaOH(aq). What is the concentration of the sulfurous acid stock solution?

H₂SO₃(aq) + 2 NaOH(aq) ⇄ H₂O(l) + Na₂SO₃(aq)

(0.02318L NaOH(aq)) (0.332M NaOH(aq)) = 7.696x10^-3mols NaOH

(7.696x10^-3mols NaOH) (1mol H₂SO₃ / 2mol NaOH) = 3.848x10^-3mols H₂SO₃

(3.848x10^-3mols H₂SO₃) / (0.01500L H₂SO₃(aq)) = 0.257M H₂SO₃(aq)

NaOH(aq) should be slightly more concentrated than H₂SO₃(aq), so this answer is reasonable

I'll be in tomorrow morning, let me know if you have any questions.