Home > Chemistry > Chemistry 1A Part Deux: Lecture 19 Notes

Chemistry 1A Part Deux: Lecture 19 Notes

Overview

Today is a 51 minute lecture on the solubility product.  They have a test coming up.

Details

Solubility product equilibrium constants can be used to predict the solubility of ionic compounds.

CAT scans are computed axial tomography.  X rays in 3 dimensions.  Absorption or non-absorption will lead to images.  BaSO4, a salt, is used as a contrast agent.  We’d like for barium sulfate to not be absorbed.  It has a low solubility.

NaCl Solubility

NaCl(s) = Na+(aq) + Cl-(aq)

NOT NaCl s = NaCl aq

If ion ion attraction in the solid are similar to the ion dipole attraction in solution, then the substance will dissolve.  This is “the rule of spreading out” or entropy.

Molecular View of Solution Formation found online.  Google if desired.

If you keep adding salt to water, you’ll reach saturation and pass it eventually.  Precipitate will form.

Saturated solution is described by solubility s in moles/Liter.

NaCls = Na+ aq + Cl-

s = Na+ = Cl- = 6.1 M

Equilibrium Constant

K sp = [Na+][ Cl-] = 37.3

K sp = s^2

s = +6.1 M

K sp =  solubility product

PbI2 s  = Pb +2 + 2I+

K sp = [Pb+2 ] [2I+] / [ PbI2]= 8.7 x 10^ -9

K sp = (x) (2x^2) = 8.7 x 10^ -9

x= 4×3

x = 1.3 x 10^-3 = [Pb2+]

[I-] = 2x

The Common Ion Effect

PbI2 = Pb+2 + 2I-

Ksp = Pb+2 I-T^2  = 8.7x 10+9

solubility in pure water= s = 1.3 x 10^-3

Reaction Quotient

Q = Pb2+ I- = 0.001 .0020 + 1

Q>Ksp

Therefore, we cannot add any more PbI2

How many moles of PbI2 can dissolve in 1L of aqueous 1M NaI solution?

Ksp = (x) (2x)/  (12) = 8.7 x 10^-9

x = 8.7 x 10^-9

This is smaller than in pure water.  Solubility in pure water is 1.3 x 10^-3.

Review

PV = nRT

Ideal Gas Law: high temp, low pressure, random motion, elastic collisions, point sources

G = H – TS

K = Products/Reactants

van der Waal gas: (P + n^2/a^2)(V – nb) = nRT

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