Uses of Solubility Product Constants

When the solubility product for a compound is known, the solubility of the compound in H2O can be calculated, as the following example illustrates. 

Calculate the molar solubilities, concentrations of the constituent ions, and solubilities in grams per liter for (a) silver chloride, AgCl (Asp = 1.8 X 10 ), and (b) zinc hydroxide, Zn(OH)2 (Asp = 4.5 X 10  

We are given the value for each solubihty product constant. In each case we write the appropriate balanced equation, represent the equihbrium concentrations, and then substitute 

Each formula unit of AgCl that dissolves produces one Ag+ and one Cl . We let X = mol/L of AgCl that dissolves, that is, the molar solubility. 

Substitution into the solubihty product expression gives 

---

Uses of Solubility Product Constants 20-4 Fractional Precipitation Qsp >... 

The general approach illustrated by Example 18.7 is widely used to determine equilibrium constants for solution reactions. The pH meter in particular can be used to determine acid or base equilibrium constants by measuring the pH of solutions containing known concentrations of weak acids or bases. Specific ion electrodes are readily adapted to the determination of solubility product constants. For example, a chloride ion electrode can be used to find [Cl-] in equilibrium with AgCl(s) and a known [Ag+]. From that information, Ksp of AgCl can be calculated. 

B We first use the solubility product constant expression for Pbl2 to determine the [i" j needed in solution to just form a precipitate when [Pb2+] = 0.010 M. We assume that the volume of solution added is small and that [Pb2+] remains at 0.010 M throughout. 

In Chapter 9, as in most of Unit 4, you learned about equilibrium reactions. In this section, you analyzed precipitation reactions. You mainly examined double-displacement reactions—reactions in which two soluble ionic compounds react to form a precipitate. You used the solubility product constant, Ksp, to predict whether or not a precipitate would form for given concentrations of ions. In Unit 5, you will learn about a class of reactions that will probably be new to you. You will see how these reactions interconvert chemical and electrical energy. 

In complex solutions such as estuarine waters, where the proportion of ions in solution is commonly not the same as that in the solids from which they are derived, it is necessary to use the solubility product constant defined as KSp = [A ] [B-]. To determine the degree to which a solution is supersaturated or undersaturated, the ion activity product (IAP) can be compared to the Ksp. 

Using the solubility product constants for calcite and strontianite and assuming a calcium activity of 1.6 mmol/L, a distribution coefficient of 0.8 for strontium and 0.98 for calcite, and a ratio of 50 1 (=0.02) in the solid-solution mineral, the following equation gives the activity of strontium ... 

Again, as in previous chapters, the best place to investigate for determining the modes of removal is the table of solubility products constants as shown in Table 14.1. A precipitation product that has the lowest K p means that the substance is the most insoluble. As shown in the table, the phosphate ion can be precipitated using a calcium precipitant producing either Ca5(P04)3(0H)(j) or Ca3(P04)2. Of these two precipitates, Ca5(P04)3(0H)(j) has the smaller K p of 10" thus, it will be used as the criterion for the precipitation of phosphates. Ca5(P03)3(OH)(j) is also called calcium hydroxy apatite. 

You have been given the solubility product constant for CuCOj. The copper and carbonate Ion concentrations are in a one-to-one relationship with the molar solubility of CUCO3. Use the solubility product constant expression to solve for the solubility. Because K p is of the order of 10 °, you can predict that the solubility will be the square root of K p, or about 10 . ... 

How can you use the solubility product constant to calculate the solubility of a sparingly soluble ionic compound ... 

You can calculate the molar solubility of an ionic compound using the solubility product constant expression. 

A table of solubility-product constants for numerous inorganic salts is found in Appendix 2. The examples that follow demonstrate some typical uses of solubility-product expressions. Further applications are considered in later chapters. 

CAN YOU ANSWER THIS Is the water in your community hard or soft Use the solubility-product constants from Table 15.2 to calculate the molar solubility ofCaCO and MgCOs-How many moles of CaCO3 are in 5 L of water that is saturated... 

Even very small solubility products can be measured electrically, and these values are listed in chemical tables. As the above exercise illustrates, the solubility of a substance can be derived from its solubility product. Solubility products are generally listed only for slightly or sparingly soluble substances. If is very small, the substance is often termed insoluble (in water). In the case of moderately and highly soluble substances (such as NaCl or NaOH), the use of solubility products is not very useful. This is because instead of defining an equilibrium constant in terms of concentrations we would have to... 

Lead(II) sulfate is often used as a test for lead(II) ion in qualitative analysis. Using the solubility product constant (Table 18.1), calculate the molar solubility of lead(II) sulfate in water. 

Mercury(I) chloride, Hg2Cl2, is an unusual salt in that it dissolves to form Hg2 " and 2CP. Use the solubility product constant (Table 18.1) to calculate the following (a) the molar solubility of Hg2Cl2 (b) the solubility of Hg2Cl2 in grams per liter. 

Solubility of a slightly soluble salt is found using the solubility product constant. 

To describe the ionization of a weak acid, we use the ionization constant K. For a solubility equilibrium, we use the solubility product constant K p. The equilibrium constant that is used to deal with a complex-ion equilibrium is called the formation constant. The formation constant, Kf, of a complex ion is the equilibrium constant describing the formation of a complex ion from a central ion and its attached groups. For reaction (18.7) this equilibrium constant expression is... 

Note that the brackets, [ ], refer to the concentration of the species. K,p is the solubility product constant hence [Cu " ] and [OH] are equal to the molar concentrations of copper and hydroxyl ions, respectively. The K p is commonly used in determining suitable precipitation reactions for removal of ionic species from solution. In the same example, the pH for removal of copper to any specified concentration can be determined by substituting the molar concentration into the following equation ... 

The common-ion effect is an application of Le Chatelicr s principle to equilibrium systems of slightly soluble salts. A buffer is a solution that resists a change in pH if we add an acid or base. We can calculate the pH of a buffer using the Henderson-Hasselbalch equation. We use titrations to determine the concentration of an acid or base solution. We can represent solubility equilibria by the solubility product constant expression, Ksp. We can use the concepts associated with weak acids and bases to calculate the pH at any point during a titration. 

Once the composition of the aqueous solution phase has been determined, the activity of an electrolyte having the same chemical formula as the assumed precipitate can be calculated (11,12). This calculation may utilize either mean ionic activity coefficients and total concentrations of the ions in the electrolyte, or single-ion activity coefficients and free-species concentrations of the ions in the electrolyte (11). If the latter approach is used, the computed electrolyte activity is termed an ion-activity product (12). Regardless of which approach is adopted, the calculated electrolyte activity is compared to the solubility product constant of the assumed precipitate as a test for the existence of the solid phase. If the calculated ion-activity product is smaller than the candidate solubility product constant, the corresponding solid phase is concluded not to have formed in the time period of the solubility measurements. Ihis judgment must be tempered, of course, in light of the precision with which both electrolyte activities and solubility product constants can be determined (12). 

Le Chatelier s principle is a powerful tool for explaining how a reaction at equilibrium shifts when a stress is placed on the system. In this experiment, you can use Le Chatelier s principle to evaluate the relative solubilities of two precipitates. By observing the formation of two precipitates in the same system, you can infer the relationship between the solubilities of the two ionic compounds and the numerical values of their solubility product constants (K ). You will be able to verify your own experimental results by calculating the molar solubilities of the two compounds using the Ksp for each compound. 

In this section, you determined the solubility product constant, Kgp, based on solubility data. You obtained your own solubility data and used these data to calculate a value for Kgp. You determined the molar solubility of ionic solutions in pure water and in solutions of common ions, based on their Ksp values. In section 9.3, you will further explore the implications of Le Chatelier s principle. You will use a reaction quotient, Qsp, to predict whether a precipitate forms. As well, you will learn how selective precipitation can be used to identify ions in solution. 

When equilibrium is reached, solubility product constants are used to describe saturated solutions of ionic compounds of relatively low solubility. When the ion concentration in solution reaches saturation, equilibrium between the solid and dissolved ions is established. 

Write the equation for the slight dissolution of the insoluble compound and calculate the AG°(reaction) using Equation (6) from Chapter 2. Once the AG°(reaction) value is obtained, the K value is calculated using Equation (8) also from Chapter 2. Keeping in mind that the concentration (activity) of the insoluble compound is defined as 1, it is recognized that the K value for the dissolution of the insoluble compound is the solubility product constant, Ksp. Alternatively, the Ksp value may be available from compilations of such values as presented in Lange s Handbook of Chemistry or the CRC Handbook of Chemistry and Physics. ... 

---