Volumetric Acid Determination. Week 1 - Standardize NaOH w/ pure KHP Determine Unknown %KHP Week 2 – Continue Unknown % KHP pH meter titration of an unknown acid. From last week…why Boil Water for Preparation of NaOH Solution?.
Week 1 - Standardize NaOH w/ pure KHP Determine Unknown %KHP
Week 2 – Continue Unknown % KHP
pH meter titration of an unknown acid
KHP K+(aq) + HP–(aq)
KHP is potassium hydrogen phthalate
Ka = 3.91 x 10-6
MW = 204.2236
HP–(aq) + OH– (aq)→ P2–(aq) + H2O()
pH? Kb = [HP-][OH-]
─────── = 2.56 x 10-9
P2- + H2O → HP- + OH-
[P2-]-x x x at equilibrium
[P2-] = n(P2-)/Vol
n(P2-) = 0.5 g/204.2236 g/mol = .00245 mol
V(OH-) = .00245 mol/(.08 mol/L) = 30 mL = .030 L
Vtotal = 50 + 30 mL = 0.080 L
[P2-] = 0.00245 mol/ (.080 L) = .030 M
───── = 2.56 x 10-9 ; x = 8.76 x 10-6 = [OH-]
.030 – x
pOH = 5.063 ; pH = 14 - 5.063 = 8.937
HP- + OH- → P2- + H2O complete at endpoint
P2- + H2O → HP- + OH- K = ? = Kb = Kw/Ka
Let’s derive it …
H+ + P2- → HP- 1/Ka
H2O → H+ + OH- Kw
P2- + H2O → HP- + OH- Kb = Kw/Ka = (1.00 x 10-14)/(3.91 x 10-6) = 2.56 x 10-9
The color change in phenolphthalein is due to a change in structure of the molecule.
In acid, the molecule is in its H2In form containing a central 5-membered ring, which is somewhat strained.
In base the In-2 structure opens up and becomes flatter.
Quick Check of Precision
Use to calculate
To determine % KHP
vol NaOH moles NaOH moles KHP mass KHP mass% KHP
(concentration of NaOH)(stoichiometry)(molar mass)mass KHP x 100%
Start unknown %KHP
Then go back to pure KHP if necessary
pH meter- half the class starts first, then teach the second half by 3:15 PM
Continue titrations with time left
Clean up after yourselves
HA & A-
Volume of NaOH
Use the equivalence pt & halfway equivalence point for calculations...how?
End point needs to be past the equivalence point
Then clean up after yourselves! Wash vials and leave on rack near storage dessicators and ovens. Otherwise, points will be deducted!!!