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Solubility of Inorganic Compounds: Calculating Equilibrium Constants and Solubilities, Summaries of Chemistry

University of North Florida (UNF)Chemistry

Guidance on calculating the solubility of inorganic compounds using the solubility product constant (ksp). It includes examples of solubility calculations for lead(ii) sulfate, lead(ii) iodide, and the interaction between calcium sulfate and lead(ii) sulfate in solution. Students will learn how to determine the amount of a compound that dissolves in a given volume of water and how to predict which metal ion will precipitate first when multiple ions are present.

What you will learn

  • What is the solubility product constant (Ksp) and how is it used to calculate the solubility of a compound?
  • If two metal ions are present in a solution, how can you determine which one will precipitate first?

Typology: Summaries

2021/2022

Uploaded on 09/27/2022

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Page IV-15-1 / Solubility Guide
CH 223 Guide to Solubility Calculations
The extent to which an insoluble salt dissolves can be expressed in terms of the equilibrium
constant. For lead(II) iodide, the expression can be written as:
PbI2(s) <=> Pb+2(aq) + 2 I-(aq)
Ksp = [Pb+2][I-]2
The equilibrium constant, Ksp, is called the solubility product constant or simply the solubility
constant. The concentration of the solid, PbI2(s), is omitted from the equilibrium expression
because it is a constant - all solids and liquids are removed from equilibrium constants.
The solubility of any compound can be observed graphically using the following diagram:
As solid is first introduced to the system, it is immediately dissolved until a certain threshold
value is obtained - here referred to as "Ksp". Past the threshold, the solid is insoluble in the
solution - no more dissolves. Each solvent has a certain capacity to dissolve solute, and the
threshold value signifies the limit of the solvent to dissolve the solid solute.
Examples of typical solubility calculations follow.
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CH 223 Guide to Solubility Calculations

The extent to which an insoluble salt dissolves can be expressed in terms of the equilibrium constant. For lead(II) iodide, the expression can be written as: PbI2(s) <=> Pb+2(aq) + 2 I-(aq) Ksp = [Pb+2][I-]^2 The equilibrium constant, Ksp, is called the solubility product constant or simply the solubility constant. The concentration of the solid, PbI2(s), is omitted from the equilibrium expression because it is a constant - all solids and liquids are removed from equilibrium constants. The solubility of any compound can be observed graphically using the following diagram: As solid is first introduced to the system, it is immediately dissolved until a certain threshold value is obtained - here referred to as "Ksp". Past the threshold, the solid is insoluble in the solution - no more dissolves. Each solvent has a certain capacity to dissolve solute, and the threshold value signifies the limit of the solvent to dissolve the solid solute. Examples of typical solubility calculations follow.

1. If 55 mg of lead(II) sulfate are placed in 250. mL of pure water, what mass of the lead compound remains undissolved? In a lead(II) sulfate solution, Ksp = 1.8 * 10-^8. The amount that dissolves, x (which is also referred to as the solubility of PbSO 4 ), can be expressed using Ksp: Ksp = [Pb2+][SO 42 - ] = xx = x^2 , and x = (Ksp)0.5^ = (1.810-^8 )0.5^ = 1.310-^4 M This is the amount that dissolves in solution; i.e. the concentration of both lead(II) ions and sulfate ions will equal 1.310-^4 M at equilibrium. This concentration can be converted to grams: (1.310-^4 mol ions / L) * 0.250 L * (1 mol PbSO 4 / 1 mol ions) * (303 g PbSO 4 / mol PbSO 4 ) = 0.010 g * (10^3 mg / g) = 10. mg PbSO 4 This answer means that 10. mg of lead(II) sulfate will dissolve in 250. mL of water. If 55 mg of lead(II) sulfate are placed in 250. mL of water, only 10. mg will dissolve. Therefore, at equilibrium 55 mg - 10. mg = 45 mg undissolved PbSO 4 Note that if only 7 mg of PbSO 4 were placed in 250. mL of water, all of it would dissolve. 10. mg is the threshold for the PbSO 4 system; if less than 10. mg is added, all of it dissolves, and if more than 10. mg is added, only 10. mg of it will dissolve and the remainder will exist as solid precipitate. 2. A solution contains Ca2+^ and Pb2+^ ions, both at a concentration of 0.010 M. If sodium sulfate is added, which metal ion precipitates first? What is the concentration of the first ion to precipitate as the second more soluble ion begins to precipitate? The first step is to determine the salts that will be precipitating; these are CaSO 4 (Ksp = 2.410-^5 ) and PbSO 4 (Ksp = 1.810-^8 ). To find which one will precipitate first, we need to calculate the concentration of the sulfate ion necessary to initiate precipitation: For PbSO 4 : [SO 42 - ] = Ksp / [Pb2+] = 1.810-^8 / 0.010 = 1.810-^6 M For CaSO 4 : [SO 42 - ] = Ksp / [Ca2+] = 2.410-^5 / 0.010 = 2.410-*^3 M The concentration of sulfate required to precipitate PbSO 4 is much less than the concentration of sulfate required to precipitate CaSO 4 ; hence, PbSO 4 will begin to precipitate out of solution before CaSO 4 begins to precipitate.