Lecture 18

Lecture 18 - 2/24/10 What is the equilibrium constant...

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2/24/10 1 What is the equilibrium constant (solubility product constant) for the dissolution of silver carbonate in water? 111 A) B) C) D) Ag 2 CO 3 (s) ! 2Ag + (aq) + CO 3 2- (aq) K sp = Ag + [ ] 2 CO 3 2 ! [ ] K sp = Ag + [ ] CO 3 2 ! [ ] K sp = 2 Ag + [ ] CO 3 2 ! [ ] K sp = Ag + [ ] 2 CO 3 2 ! [ ] Ag 2 CO 3 [ ] We are starting with the handout from Monday. This question does not appear explicitly on the handout – refers to equilibrium on bottom of page 38. Solubility : the amount of solid that must dissolve to produce a saturated solution (mol/L or g/L) 9.1. Relationship between solubility and K sp Saturated solution : a solution with the maximum concentration of dissolved ions, with the solution in contact with extra undissolved solid 9.1.1. Determining K sp from solubility At 25 o C, only 0.00349 g of silver carbonate (Ag 2 CO 3 ) dissolves in 100 mL of water, any excess solid remaining undissolved. Calculate the K sp . Step 1: Convert solubility to mol/L (molar solubility) s = (3.49 ! 10 -3 g/ 0.100 L)/(275.75g/mol) 112 = 1.27 ! 10 -4 mol/L Dimensionless (like all equilibrium constants) When determining K sp , MUST use units of mol/L for solubilities. 113 I C E solid 0.00 M 0.00 M +2s solid + s solid +2s + s Ag 2 CO 3 (s) ! 2Ag + (aq) + CO 3 2- (aq) = 4(1.27 ! 10 -4 mol/L) 3 K sp = Ag + [ ] 2 CO 3 2 ! [ ] = 8.19 ! 10 -12 = (2s) 2 s = 4s 3 9.1.2 Determining solubility from K sp Determine the solubility (in mol/L) of MgF 2 in water at 25 o C, given that K sp (MgF
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Lecture 18 - 2/24/10 What is the equilibrium constant...

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