Ionic compounds molar solubility

The solubility product depends on the temperature, and at a given temperature it is constant for a particular ionic compound. Molar solubility is defined as the number of moles of solute dissolved in one liter of its saturated solution. Using the molar solubility of a compound, the solubility product of that compound can be determined or vice versa. 

The reverse of Example 16.4 involves finding Rq, of a compound given its solubility. The solubilities of many ionic compounds are determined experimentally and tabulated in chemical handbooks. Most solubility values are given in grams of solute dissolved in 100 grams of water. To obtain the molar solubility in moles/L, we have to assume that the density of the solution is equal to that of water. Then the number of grams of solute per 100 g water is equal to the number of grams of solute per 100 mL of solution. This assumption is valid because the mass of the compound in solution is small. To solve for IQp, find the molar solubility of the solute and determine the concentration of its component ions. Substitute into the IQp expression. 

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. 

Calculate the molar solubilities of the two ionic compounds from their Ksp values. 

In this section the solubilities of NaCl, CaCl2, Na2C03 and CaC03 are compared to investigate the effects of having doubly charged ions in a compound. The solubilities of the compounds are given in Table 3.17 in terms of mass % and molar concentration. Sodium and calcium cations are chosen for this comparison because they have almost identical ionic radii Na h 102 pm, Ca2+ 100 pm. 

The Kelvin equation may also be applied to the equilibrium solubility of a solid in a liquid. In this case the ratio p/p0 in Equation (40) is replaced by the ratio a/a0, where a0 is the activity of dissolved solute in equilibrium with a flat surface, and a is the analogous quantity for a spherical surface. For an ionic compound having the general formula MmXn, the activity of a dilute solution is related to the molar solubility S as follows ... 

In dilute aqueous solutions, it has been demonstrated experimentally for poorly soluble ionic salts (solubilities less than 0.01 molL ) that the mathematical product of the total molar concentrations of the component ions is a constant at constant temperature. This product, is called the solubility product. Thus for a saturated solution of a simple ionic compound AB in water, we have the dynamic equilibrium ... 

You have learned that the solubility product constant can be used to determine the molar solubility of an ionic compound. You can apply this information as you do the CHEMLAB at the end of this chapter. also can be used to find the concentrations of the ions in a saturated solution. 

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

For part (a), we write the appropriate chemical equations and solubility product expression, designate the equilibrium concentrations, and then substitute into the solubility product expression. For part (b), we recognize that NaF is a soluble ionic compound that is completely dissociated into its ions. MgF2 is a slightly soluble compound. Both compounds produce F ions so this is a common ion effect problem. We write the appropriate chemical equations and solubility product expression, represent the equilibrium concentrations, and substitute into the solubility product expression. For part (c), we compare the molar solubilities by calculating their ratio. 

These are the compounds in Example 20-5. We recognize that Na2S04 is a soluble ionic compound and that the molarity of 804 is equal to the molarity of the Na2S04 solution. We are given for BaS04, so we solve for [Ba +]. 

As the above examples show, solubility and solubility product are related. If we know one, we can calculate the other, but each quantity provides different information. Table 16.3 shows the relationship between molar solubility and solubility product for a number of ionic compounds. 

Determining fCjp from Solubility The solubilities of ionic compounds are determined experimentally, and several chemical handbooks tabulate them. Most solubility values are given in units of grams of solute dissolved in 100 grams of H2O. Because the mass of compound in solution is small, a negligible error is introduced if we assume that TOO g of water is equal to 100 mL of solution. We then convert the solubility from grams of solute per 100 mL of solution to molar solubility, the amount (mol) of solute dissolved per liter of solution (that is, the molarity of the solute). Next, we use the equation for the dissolution of the solute to find the molarity of each ion and substitute into the ion-product expression to find the value of K p. 

The solubility of an ionic compound M2X3 (molar mass = 288g)is3.6 X 10 g/L. What is for the compound ... 

Since the pure solid A( ) has a constant activity, it follows that the activity of the dissolved material in equilibrium with it is also constant, at a given temperature. In general, when ionic strength and solubility are both not much above 0.1 M, activities of uncharged species can be taken equal to their molarities for accuracy within 2 %. The solubility of such compounds as benzene, ethyl acetate, etc., in water is then the equilibrium constant for this reaction. 

---