K eq calculations

Here the value of K eq, which has no units, is a constant for any particular reaction, and its value does not change unless the temperature of the system is changed. It does not depend on the initial concentrations used to reach the point of equilibrium. H 2 (g) + I 2 (g) 2 HI (g) K eq = [HI] 2 [I 2 ] [H 2 ]

2H 2 S (g) 2H 2(g) + S 2(g) Calculate the value of k eq given that the concentration for S 2 and H 2 S are and mol/L respectively

The value for K eq is the same despite differences in equilibrium concentrations for the individual participants. K eq relates the concentrations of products to reactants at equilibrium. For aqueous solutions, concentration is often measured as Mol · L -1. For gases, concentration is often measured as partial pressure.

The concentrations of both aqueous solutions and gases change during the progress of a reaction. For reactions involving a solid or a liquid, while the amounts of the solid or liquid will change during a reaction, their concentrations (much like their densities) will not change during the reaction.

Instead, their values will remain constant. Because they are constant, their values are not included in the equilibrium constant expression. For example, consider the reaction showing the formation of solid calcium carbonate from solid calcium oxide and carbon dioxide gas: CaO (s) + CO 2 (g) CaCO 3 (s)

K eq =

Ksp - background The equilibrium between solids and ions is different from the equilibrium between gases The equilibrium between solids and ions is a “phase” equilibrium (e.g. NaCl(aq)) NaCl(s) Na + (aq) + Cl - (aq) Ksp deals with a phase equilibrium: (s)  (aq)

Kc - ignoring solids and liquids Concentrations of liquids and solids do not change. Concentrations of ions and gases do.

Ksp - molar solubility A solid always dissolves until no more can dissolve - called molar solubility (mol/L or M) An equilibrium is established when the amount dissolving equals the amount precipitating This can only be true if there is some solid (thus, we can usually see if there is an equilibrium) A solution with solid remaining (I.e. in equilibrium) is called “saturated” Note: adding more solid will not affect equilibrium

Solubility and the Solution Process II Saturated solution: maximum amount of solute is dissolved in solvent. Trying to dissolve more results in undissolved solute in container. Solubility: Amount of solute that dissolves in a solvent to produce a saturated solution. (Solubility often expressed in g/100 mL.) E.g g of I 2 dissolved in 1000 g of H 2 O. Unsaturated solution: less than max. amount of solute is dissolved in solvent. E.g g of I 2 dissolved in 1000 g of H 2 O. Supersaturation = more solute in solution than normally allowed; we call this a supersaturated solution.

Finding the solubility product Ksp First we need an equation To find the equation we need to know the dissolving equation Let’s use the example of CaCl 2 The equation would be CaCl 2(s)  Ca +2 (aq) + 2Cl - (aq)

The equation would be CaCl 2(s)  Ca +2 (aq) + 2Cl - (aq) All equilibrium constants have the equation Constant = [Products] [Reactants] And unlike the rate equation equilibrium constants are raised to the power of the coefficient in the balanced equation The equilibrium balance is not effected by how much solid is left Ksp = [Ca +2 ].[Cl - ] 2

Find the Ksp equation for a saturated solution of Fe(OH) 3 Ksp = [Fe +3 ].[OH - ] 3

Steps for finding Ksp ( Fe(OH) 3 ) Step 1 write out ion equation Step 2 write out Ksp equation Don’t forget to raise each ion to the power of it’s coefficient Step 3 substitute in concentrations Step 4 solve

Find the equation for K 3 PO 4 What is step 1 What ions would be formed What would be the dissolving equation How many K+ ions would be formed

Calculate the solubility product constant for PbCl 2, if a saturated solution was found to contain 0.1 M of Pb +2 and 0.2 M of Cl - PbCl 2  Pb Cl - Ksp = [Pb +2 ]. [Cl-] 2 Ksp = [0.1 M][0.2 M ] 2 Ksp = 0.1 M x 0.04 M 2 Ksp = M 3