The Solubility Constant

Purpose

To measure the molar solubility of calcium iodate in pure water as well as in a solution containing added potassium iodate, in order to determine and compare the solubility product constant.

Introduction

Calcium iodate has been shown to be completely dissociated in water into calcium ions Ca2+, and iodate ions, IO3-2, as expressed by the equation

Ca(IO3)2 (s) ↔ Ca2+ (aq) + 2IO3- (aq)

The concentration of a dissolved salt, in this case calcium iodate, is the molar solubility. Molar solubility is expressed in units of moles per liter. In calcium iodate, the calcium ion concentration is equal to the molar solubility, and the iodate ion concentration is equal to 2 times the molar solubility. This is true because each mole of calcium iodate that dissolves gives one mole of calcium ions and two moles of iodate ions. The concentrations of Ca2+ and IO3- will be related to molar solubility, s, by

[Ca2+] = s

[IO3-] = 2s

The equilibrium product constant, Ksp, can then be determined:

Ksp = [Ca2+] [IO3-] 2 = 2(2s)2 = 4s3

This would be the equilibrium constant product of calcium iodate in water.

The solubility of calcium iodate would be lower in a solution of potassium iodate, however, according to Le Chatelier’s principle. Since potassium iodate is a strong electrolyte that completely dissociates in water, it could be predicted that its iodate would add to the iodate ion from Ca(IO3)2 (situation referred to as the common ion effect), thus shifting the equation at equilibrium expressed above towards the left.

Procedure

A 50 mL buret was rinsed with standardized 15 mL of sodium thiosulfate, recording the initial reading. An Erlenmeyer flask was filled up to 50 mL of water, adding then 2 g of KI and 10 mL of saturated calcium iodate solution (for which temperature was calculated first). HCl was added to the flask until the solution turned brown, and was immediately titrated with sodium thiosulfate until it changed to yellow. At that point, 5...