Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later. Consider, for example, the effect of adding a soluble salt, such as CaCl 2 , to a saturated solution of calcium phosphate [Ca 3 PO 4 2 ].
The common ion effect usually decreases the solubility of a sparingly soluble salt. Calculate the solubility of calcium phosphate [Ca 3 PO 4 2 ] in 0.
A The balanced equilibrium equation is given in the following table. We can insert these values into the ICE table. Thus 0. This value is the solubility of Ca 3 PO 4 2 in 0. Calculate the solubility of silver carbonate in a 0. The solubility of silver carbonate in pure water is 8. The solubility product K sp is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product Q describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product K sp , is a measure of the solubility of a compound.
Whereas solubility is usually expressed in terms of mass of solute per mL of solvent, K sp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product Q describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and K sp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed.
The solubility of the salt is almost always decreased by the presence of a common ion. Learning Objectives To calculate the solubility of an ionic compound from its K sp. The Solubility Product When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions.
Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate K sp. Solution A We need to write the solubility product expression in terms of the concentrations of the component ions. A crystal of calcite CaCO 3 , illustrating the phenomenon of double refraction.
When a transparent crystal of calcite is placed over a page, we see two images of the letters. Image used with permisison from Wikipedia. Calculate the mass of solute in mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero.
We can insert these values into the table. Calculate the following: the molarity of a saturated solution the mass of silver carbonate that will dissolve in mL of water at this temperature Answer 1. The Ion Product The ion product Q of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. The solution is unsaturated, and more of the ionic solid, if available, will dissolve. The solution is saturated and at equilibrium.
The solution is supersaturated, and ionic solid will precipitate. Given: K sp and volumes and concentrations of reactants Asked for: whether precipitate will form Strategy: Write the balanced equilibrium equation for the precipitation reaction and the expression for K sp. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product Q.
Compare the values of Q and K sp to decide whether a precipitate will form. Reply to this topic Start new topic. Recommended Posts. Posted July 21, Link to comment Share on other sites More sharing options Vanamonde Posted July 21, Not sure what this is about. Could you clarify your question, please? Godot Posted July 21, Seems like he means the solubility product constant That acronym really needs clarification, especially on a KSP forum.
Incidentally, I'm of absolutely no help. Sorry to have wasted your time. Temeter Posted July 21, KSP doesn't NaCl.
If you get the right answer when you substitute this concrete example into your equation, it must be written correctly. The value of K a for an acid is proportional to the strength of the acid.
If we find the following K a values in a table, we can immediately conclude that formic acid is a stronger acid than acetic acid. The following base-ionization equilibrium constants imply that methylamine is a stronger base than ammonia. Unfortunately, there is no simple way to predict the relative solubilities of salts from their K sp 's if the salts produce different numbers of positive and negative ions when they dissolve in water.
Click here to check your answer to Practice Problem 5. Imagine what happens when a few crystals of solid AgNO 3 are added to this saturated solution of AgCl in water. According to the solubility rules, silver nitrate is a soluble salt. In more formal terms, we can argue that the ion product Q sp for the solution is larger than the solubility product K sp for AgCl. The ion product is literally the product of the concentrations of the ions at any moment in time.
When it is equal to the solubility product for the salt, the system is at equilibrium. The reaction eventually comes back to equilibrium after the excess ions precipitate from solution as solid AgCl. Now imagine what happens when a few crystals of NaCl are added to a saturated solution of AgCl in water.
There are two sources of the chloride ion in this solution. Point C describes a solution at equilibrium that was prepared by dissolving two sources of the Cl - ion in water, such as NaCl and AgCl. Any point that is not along the solid line in the above figure represents a solution that is not at equilibrium.
Any point below the solid line such as Point D represents a solution for which the ion product is smaller than the solubility product. Points above the solid line such as Point E represent solutions for which the ion product is larger than the solubility product.
The solution described by Point E will eventually come to equilibrium after enough solid AgCl has precipitated. Practice Problem 2: Several compounds were studied as possible sources of the fluoride ion for use in toothpaste.
Practice Problem 3: Use the K sp for calcium fluoride to calculate its solubility in grams per liter. Practice Problem 4: Calculate the solubility in grams per liter of silver sulfide in order to decide whether it is accurately labeled when described as an insoluble salt.
Saturated solution of AgCl in water:. Saturated solution of AgCl to which NaCl has been added:.
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