2012-07-12

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? {Ka(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 Ka-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 Ka expression simplifies to:
Ka = (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 Kb-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 Kb expression simplifies to:
Kb = (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.

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