Some ionic compounds are highly soluble in water while others are not. Add salt to a solution of silver nitrate and a solid precipitate of silver chloride will form--because silver chloride, unlike silver nitrate or sodium chloride, is only slightly soluble in water. Using the solubility product or Ksp for a given salt, you can predict how much precipitate will form following a simple reaction in solution. The solubility products for various salts are experimentally determined and are available in the table under the Resources section at the end of this article.
Write the reaction equation and net ionic equation for the reaction you want to use. Use your solubility rules to predict which compounds, if any, will precipitate (use the solubility rules link from the Resources below). The net ionic equation excludes spectator ions--ions that don't play a direct role in the reaction--so it will only include the ions that form the precipitate. If you were to add salt to a solution of silver nitrate, for example, the overall equation would be Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) <=> Na+(aq) + NO3(aq) + AgCl(s), while the net ionic equation would be Ag+(aq) + Cl-(aq) <=> AgCl(s), since sodium and nitrate are spectator ions.
Calculate the concentration of the ions in solution for each of the reactants in the net ionic reaction. The concentration is equal to the number of moles of a specific ion divided by the total volume of the solution. If we were to add 1 mole of NaCl to 1 litre of water, for example, the concentration of NaCl would be 1 mole per litre. The concentration of sodium or chloride ions in the solution would also be 1 mole per litre, since NaCl will dissociate in water.
Set up the solubility product equation for the net ionic equation. When an ionic compound dissolves, it dissociates into ions in solution. The product of the concentration of these ions is equal to the solubility product constant or Ksp (again, Ksp values for many common salts are available in the table under the Resources section below). Note also that if any ion in the equation has a coefficient greater than 1, the concentration of that ion will have an exponent equal to its coefficient from the equation. If we were looking at calcium fluoride dissolving in water, for example, the solubility product equation would be Ksp = [Ca+2][F-][F-]; the concentration of the fluoride ion is squared in this equation because each formula unit of calcium fluoride releases two fluoride ions when it dissolves.
Solve the solubility product equation to find the molar solubility or concentration of each ion when the reaction is at equilibrium. An easy way to do this is to assume the salt is in solid form initially and treat the change in the number of moles of solid as x, then use these assumptions to find x for each ion released when the solid dissolves and plug x into the equation. When calcium fluoride dissolves in water, for example, each mole of calcium fluoride that dissolves produces 1 mole of calcium ions and 2 moles of fluoride ions, so the solubility product equation becomes Ksp = [x][2x] to the 2. Since we know Ksp, we can use it to find x and hence the molar solubility.
Use the difference between the molar solubility of the salt and the initial concentration of reactant ions to determine how much precipitate will form. The salt will continue to precipitate until the product of the ion concentration is equal to Ksp. If we had two moles each of silver and chloride ions in one litre of water to start with, for example, but the molar solubility of silver chloride is only 1.3 x 10 to the -5 moles per litre, the amount of precipitate would be 2 moles minus 1.3 x 10 to the -5 moles.
The above calculations are oversimplified in that they don't take into account a number of factors that will be important in many real-life problems. For example, if there were unequal concentrations of silver and chloride ions in solution, you would need to calculate the amount of precipitate formed using the solubility product equation. Since Ksp is equal to the product of the concentrations, you could use the concentration of one ion to determine the maximum amount of the other that could remain in solution.
Tips and warnings
- The above calculations are oversimplified in that they don't take into account a number of factors that will be important in many real-life problems. For example, if there were unequal concentrations of silver and chloride ions in solution, you would need to calculate the amount of precipitate formed using the solubility product equation. Since Ksp is equal to the product of the concentrations, you could use the concentration of one ion to determine the maximum amount of the other that could remain in solution.