Ion-product expression

The ion product expression for magnesium sulfate is written as follows ... 

You can use the relationship between the ion product expression and the solubility product expression to predict whether a precipitate will form in a given system. One common system involves mixing solutions of two soluble ionic compounds, which react to form an ionic compound with a very low solubility. If Qsp > Kp. based on the initial concentrations of the ions in solution, the sparingly soluble compound will form a precipitate. 

Step 2 Substitute the ion concentrations into the ion product expression for silver chloride to determine Qsp. 

Before we can use these quantities in the ion product expression, we will have to convert them to molarities. We need to make sure we are using the combined volume of the two solutions to determine the molarity in the final mixture ... 

Write the equation for the dissolving of PbCl2 and the ion product expression, Osp-... 

As in previous cases, we incorporate the constant concentration of the solid, [PbS04], into the value of Q, which gives the ion-product expression, Q p. 

At saturation, the concentration terms have their equilibrium values, so we can write the ion-product expression directly with the symbol K p. For example, the equation and ion-product expression that describe a saturated solution of Cu(OH)2 are... 

Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds Problem Write the ion-product expression for each of the following compounds ... 

Plan We write an equation that describes a saturated solution and then write the ion-product expression, K p, according to Equation 19.2, noting the sulfide in part (d). Solution (a) Magnesium carbonate ... 

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. 

Plan We are given the solubilities in various units and must find K p. Por each compound, we write an equation for its dissolution to see the number of moles of each ion, and then write the ion-product expression. We convert the solubility to molar solubility, find the molarity of each ion, and substitute into the ion-product expression to calculate K p. Solution (a) Por PbS04. Writing the equation and ion-product (K p) expression ... 

Comment 1. In part (b), the formula Pbp2 means that [P ] is twice [Pb " ]. Then we square this value of [P ]. Always follow the ion-product expression explicitly. 

Plan We write the dissolution equation and the ion-product expression. We know K p (6.5X10 ) to find molar solubility (S), we set up a reaction table that expresses [Ca ] and [OH ] in terms of 5, substitute into the ion-product expression, and solve for S. Solution Writing the equation and ion-product expression ... 

Comment 1. Note that we did not double and rfien square [OH ]. 25 is the [OH ], so we just squared it, as the ion-product expression required. 

Plan Addition of Ca ", the common ion, should lower the solubility. We write the equation and ion-product expression and set up a reaction table, with [Ca"" ] ni, coming from Ca(N03)2 and S equal to [Ca" ]fromCa(OH)2- To simplify the math, we assume that, because /Csp is low, S is so small relative to [Ca ]ini, that it can be neglected. Then we solve for S and check the assumption. 

Plan First, we must decide which slightly soluble salt could form and look up its value in Appendix C. To see whether mixing these solutions will form the precipitate, we find the ion concentrations by calculating the amount (mol) of each ion from its concentration and volume, and then dividing by the total volume because one solution dilutes the other. Finally, we write the ion-product expression, calculate Q p, and compare Q p with K p. Solution The ions present are Ca ", Na", F, and N03. All sodium and all nitrate salts are soluble (Table 4.1), so the only possibility is CaF2 (K p = 3.2X10 ... 

Write the ion-product expressions for (a) iron(III) hydroxide (b) barium phosphate (c) tin(II) sulfide. 

SAMPLE PROBLEM 19.4 Writing Ion-Product Expressions (or Slightly... 

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