Question

In: Chemistry

A big part of my grade in quantitative analysis is preforming a procedure. If I could...

A big part of my grade in quantitative analysis is preforming a procedure. If I could get some help on how to carry out this procedure (pretty sure titration is involved) and the calculations after, I would greatly appreciate it!

Problem: You must determine the molarity of an unknown solution of phosphoric acid. The concentration of the phosphoric acid is between 0.09 and 0.15 M. You have access to all the glassware in your laboratory drawers in addition to anything in the stockroom.

Chemicals and reagents you will have available to you are:

Approximately 2 grams of dried potassium hydrogen phthalate (KHP), MM 204.22 grams
~250 mL of NaOH, approximate molarity 0.1 M
~250 mL of the phosphoric acid sample
A solution of ~1.8M (20 %w/w) CaCl2

You have access to the following indicators:

1) Methyl Orange, pKa = 3.75
     HIn: Red              In-: Yellow

2) Congo Red, pKa = 4.0,
     HIn: Blue/violet          In-: Red

3) Bromocresol Green, pKa = 4.6,
     HIn: Yellow          In-: Blue

4) Calmagite, pKa2 = 8.1, pKa3 = 12.4,
    H2In-: Red,          HIn-2: Blue,             In-3: Orange

5) Phenolphthalein, pKa = 8.8,
    HIn: Colorless          In-: Pink

6) Alizarin Yellow, pKa = 11.9
    HIn: Yellow      In-: Orange

The following information will be useful:

Phosphoric acid: pKa1 = 2.148, pKa2 = 7.198, pKa3 = 12.375
Ksp of Ca3(PO4)2: 3.45 x 10-30

Solutions

Expert Solution

1)

First exact concentration of NaOH can be determined by using KHP.

NaOH(aq) + KHC8H4O4(aq) = KNaC8H4O4(aq) + H2O(l)

Moles of NaOH = Moles of KHP

Moles of KHP = Mass of KHP (2 gram) / 204.22 g/mol = 0.0098

[NaOH] = Moles of NaOH / Volume of NaOH used

2)

pH at first equivalence point:

= (pKa1 + pKa2)/2 = (2.15 + 7.2)/2

= 4.7

pH at second equivalence point:

= (pKa2 + pKa3)/2 = (7.2 + 12.4)/2

= 9.8

pH at third equivalence point:

pKb = 14 – pKa3

= 1.62

Kb = 10-1.62 = 0.024

~[PO43-] = 0.1 M

[OH-] = (Kb * [PO43-])1/2

= 0.05

pOH = - log(0.05) = 1.3

pOH = 14 – 1.3 = 12.7

H3PO4 + NaOH → NaH2PO4 + H2O

H3PO4 can be titrated as a monoprotic acid with standard NaOH, against indicator changing color around pH 4.7 (methyl orange).

3)

Phosphoric acid (H3PO4) can be titrated as triprotic PO43- anion. It is precipitated first using metal ions of Ca2+ from CaCl2.

2H3PO4 + 3CaCl2 → Ca3(PO4)2 + 6HCl

After precipitation, HCl produced can be titrated against phenolphthalein.


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