This is the procedure: http://www3.chem21labs.com/labfiles/UofC_GL22_Lab.pdf?rf=

Question

This is the procedure: http://www3.chem21labs.com/labfiles/UofC_GL22_Lab.pdf?rf=7495

State and Explain how each of these errors would affect (high, low, no change) your calculated values of solubility and Ksp.

• The saturated solution of Cu(IO3)2 is not filtered to remove undissolved Cu(IO3)2.

State and Explain how each of these errors would affect (high, low, no change) your calculated values of solubility and Ksp.

• The saturated solution of Cu(IO3)2 is not filtered to remove undissolved Cu(IO3)2.

Explanation / Answer

Solubility Equilibrium.

The solubility product of a compound is the product of molar concentrations from constituent ions, each one elevated to the potency of its stoichiometry coefficient in the equilibrium equation.

Cu(IO3)2 (s) <==> Cu2+ (aq) + 2 IO3-

So,

Kps = [Cu2+][IO3-]2

It indicates the solubility of a ionic compound, that is, the lower the Kps the lower the solubility of the compound. When the solution is saturated, the ionic product (initial concentrations) will be equal to the Kps.

The solubility of a slightly soluble salt that contains basic anions increases with the increment of H+ ions concentrations. The presence of a common ion in this kind of compounds decreases the solubility.

If Cu(IO3)2 is not completely filtered, there will be quantitative errors in the calculation of Kps and solubility.

Assuming Q = Kps:

Kps of copper iodate = 5.1x10-12 ; m.wt = 413.3513 g/mol.

5.1x10-12 = s(2s)2

s3 = 5.1x10-12/4

s = 1.08x10-4 M

--

solubility = 1.08x10-4 mol / 1L Sln x 413.3513 g/ 1 mol

= 4.48x10-2 g/L

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