Question

Experiment 26: The Solubility Product of Ba(IO3)2 : Processing the Data
Experiment 26 Data and Calculations: The Solubility Product of Ba(IO3)2 The calculations here are similar to those in Experiment 23, and are outlined in the Experimental Procedure section. To use a spreadsheet, set up a table like the one below. Test Tube Number mL 0.0350 M KIO mL 0.0200 M Ba(NO3)2 1.00 2.00 3.00 4.00 5.00 5.00 5.00 5.00 5.00 5.00 12.00 12.00 12.00 12.00 12.00 0.382 0.396 0.804 0.494 0.616 2.02540-3 1.063103 1.0854/00 H 1.6538103 Total volume in mL Absorbance of soln. [10; } in soln. 1.326x10-3 Processing the Data Initial moles Ba? Initial moles 10,- TIL | Moles 10,- in equil. soln. (12 mL) Moles 10,- precipitated (continued on following page)

Answer 1

Initial moles Ba2+

Moles Ba2+ = (volume of Ba(NO3)2 in L)*(molarity of Ba(NO3)2)

Take test tube 1 as an example.

Moles Ba2+ = (5.00 mL)*(1 L)/(1000 mL)*(0.0200 M)

= 1.00*10-4 mol

Fill in the table.

Test Tube Number	1	2	3	4	5
Initial mols Ba2+	1.00*10-4	1.00*10-4	1.00*10-4	1.00*10-4	1.00*10-4

Initial moles IO3-

Moles IO3- = (volume of KIO3 in L)*(molarity of KIO3)

Take test tube 1 as an example.

Moles IO3- = (1.00 mL)*(1 L)/(1000 mL)*(0.0350 M)

= 3.50*10-5 mol.

Fill in the table.

Test Tube Number	1	2	3	4	5
Initial mols IO3-	3.50*10-5	7.00*10-5	1.05*10-4	1.40*10-4	1.75*10-4

Moles IO3- in solution

Moles IO3- = (volume of solution in L)*(calculated molarity of IO3- in solution)

Take test tube 1 as an example.

Molarity IO3- in solution = 1.025*10-3 M.

Moles IO3- in solution in test tube 1 = (12.00 mL)*(1 L)/(1000 mL)*(1.025*10-3 M)

= 1.23*10-5 mol

Fill in the table.

Test Tube Number	1	2	3	4	5
Mols IO3- in solution	1.23*10-5	1.2756*10-5 ≈ 1.27*10-5	1.3008*10-5 ≈ 1.30*10-5	1.5912*10-5 ≈ 1.59*10-5	1.9836*10-5 ≈ 1.98*10-5

Moles IO3- precipitated

Moles IO3- precipitated = (initial mols IO3-) – (mols IO3- in solution)

Take test tube 1 as an example.

Moles IO3- precipitated = (3.50*10-5) – (1.23*10-5)

= 2.27*10-5

Fill in the table.

Test Tube Number	1	2	3	4	5
Mols IO3- precipitated	2.27*10-5	5.73*10-5	9.20*10-5	1.241*10-4 ≈ 1.24*10-4	1.552*10-4 ≈ 1.55*10-4

Moles Ba2+ precipitated

Consider the ionization of Ba(IO3)2.

Ba(IO3)2 (s) <=======> Ba2+ (aq) + 2 IO3- (aq)

As per the stoichiometric equation above,

1 mol Ba2+ = 2 mols IO3-

Therefore,

Mols Ba2+ precipitated = ½*(mols IO3- precipitated)

Take test tube 1 as an example.

Mols Ba2+ precipitated = ½*(2.27*10-5 mol)

= 1.135*10-5

Fill in the table.

Test Tube Number	1	2	3	4	5
Mols Ba2+ precipitated	1.135*10-5	2.865*10-5	4.60*10-5	6.20*10-5	7.75*10-5

Moles Ba2+ in equilibrium solution

Moles Ba2+ in equilibrium solution = (initial mols Ba2+) – (mols Ba2+ precipitated)

Take test tube as an example.

Mols Ba2+ in equilibrium solution = (1.00*10-4) – (1.135*10-5)

= 8.865*10-5

Fill in the table.

Test Tube Number	1	2	3	4	5
Mols Ba2+ in equilibrium solution	8.865*10-5	7.135*10-5	5.40*10-5	3.80*10-5	2.25*10-5

[Ba2+] in equilibrium solution

Volume of the equilibrium solution = 12.00 mL = (12.00 mL)*(1 L)/(1000 mL)

= 0.012 L

[Ba2+] = (mols Ba2+ in the equilibrium solution)/(0.012 L)

= 7.3875*10-3 M

≈ 7.39*10-3 M

Fill in the table below.

Test Tube Number	1	2	3	4	5
[Ba2+] in equilibrium solution, M	7.39*10-3	5.94*10-3	4.50*10-3	3.16*10-3	1.875*10-3 ≈ 1.87*10-3

Ksp for Ba(IO3)2

Ksp = [Ba2+][IO3-]2

Take test tube 1 as an example.

Ksp = (7.39*10-3)*(1.025*10-3)2

= 7.76*10-9 (ignore units)

Fill in the table below.

Test Tube Number	1	2	3	4	5
Ksp	7.76*10-9	6.71*10-9	5.29*10-9	5.56*10-9	5.11*10-9

Mean value of Ksp

Mean value = 1/5*(7.76 + 6.71 + 5.29 + 5.56 + 5.11)*10-9

= 6.086*10-9

≈ 6.09*10-9

Standard deviation of Ksp = √Σ(x - x̅)2/(n – 1)

where x̅ is the mean of the Ksp values and n is the number of measurements.

Standard Ksp = 1.12*10-9 (ignore units)