Solubility Unit III Lesson 1. Solubility is a measure of the maximum amount of solid that will...

SolubilityUnit IIILesson 1

Solubility is a measure of the maximum amount of solid that will dissolve in a volume of water.

Units: g/L mol/L g/100mL

Or even: mL/L for CO2(g) in water

The unit must have an amount on the top and volume on the bottom!

In order to determine the solubility you must completely fill or saturate the solution!

To saturate the solution you must add weighed portions of the solid to a given volume of water and stir until no more will dissolve. A small amount of excess solid will be present in the saturated solution.

NaCl 10.0 g

Determining The Solubility of NaCl

Take 100 mL of water

Add measured portions of NaCl and stir to dissolve

NaCl


NaCl


The solution is stirred and it all dissolves

Na+ Cl-

Amount NaCl Dissolved

10.0 g


The solution is stirred and it all dissolvesRate of dissolving > Rate of crystallization

Na+ Cl-

NaCl 10.0 g Amount NaCl Dissolved

10.0 g


Add more since the solution is not full

Na+ Cl-

NaCl


10.0 g


Na+ Cl-NaCl


10.0 g



Na+ Cl-


10.0 g10.0 g



Na+ Cl-


10.0 g10.0 g

NaCl 10.0 g



Na+ Cl-


10.0 g10.0 g

NaCl


Na+ Cl-


10.0 g10.0 g10.0 g



Na+ Cl-


10.0 g10.0 g10.0 g

NaCl 3.0 g



Na+ Cl-


10.0 g10.0 g10.0 g

NaCl


Na+ Cl-


10.0 g10.0 g10.0 g 3.0 g



Na+ Cl-


10.0 g10.0 g10.0 g 3.0 g

NaCl 1.0 g



Na+ Cl-


10.0 g10.0 g10.0 g 3.0 g


NaCl

Na+ Cl-


10.0 g10.0 g10.0 g 3.0 g

33.0 g/100 mL


NaCl(s)

Rate of dissolving = Rate of crystallization

A bit of solid (1 g) sits on the bottom

The solution is stirred and it does not dissolve!The solution is full or saturated.


10.0 g10.0 g10.0 g 3.0 g

33.0 g/100 mL

The solution is stirred and it does not dissolve!The solution is full or saturated.


Na+ Cl-

NaCl(s)

Rate of dissolving = Rate of crystallization

A bit of solid sits on the bottomThis tells you that the solution is saturated and that it is in equilibrium

Calculate the solubility in units of g/L and mole/L.

Solubility = 33.0 g0.100 L

= 330. g/L

s = molar solubility = the molarity to saturate

= 33.0 g x 1 mole 58.5 g

0.100 L

= 5.60 M

33.0 g/100 mL

Na+ Cl-

NaCl(s)

A bit of solid sits on the bottomThis tells you that the solution is saturated and that it is in equilibrium

Since this is an equilibrium lets write the Equilibrium

Equation: NaCl(s) ⇌ Na+ + Cl-

Expression: Keq = [Na+][Cl-] do not count a solid!

Because this Keq is special and based on the solubility of a solid it is called a solubility product or Ksp.Ksp = [Na+][Cl-]

This unit is all about the Ksp and the solubility of saturated solutions.

There are three types of solutions. Unsaturated Contains less than the maximum amount of dissolved solid

Not at equilibrium

Add more solid and it will dissolve

The rate of dissolving > the rate of crystallizing

There are three types of solutions.

Saturated

Contains the maximum amount of dissolved solid

At equilibrium

Add more solid and it does not dissolve

The rate of dissolving is equal to the rate of crystallization

There are three types of solutions.

Supersaturated

Contains more than the usual maximum amount of dissolved solid

Not at equilibrium

Add more solid and more solid precipitates

The rate of dissolving is less than the rate of crystallization

What we just calculated was exact solubility. The data book page 4 gives us relative solubility.

High Solubility means > .1 MLow Solubility means .1M Classify as high or low solubility. Na3PO4


High Solubility means > .1 MLow Solubility means .1M Classify as high or low solubility. Na3PO4 High

H2SO4



H2SO4 High

Ca(NO3)2



H2SO4 High

Ca(NO3)2 High

CaSO4



H2SO4 High

Ca(NO3)2 High

CaSO4 Low

FeSO4



H2SO4 High

Ca(NO3)2 High

CaSO4 Low

FeSO4 High

Classify as high or low solubility. Ag2SO4

Classify as high or low solubility. Ag2SO4 low

CuSO4


CuSO4 high

CuCl2


CuSO4 high

CuCl2 high

CuCl


CuSO4 high

CuCl2 high

CuCl low BaS


CuSO4 high

CuCl2 high

CuCl low BaS high K2CO3


CuSO4 high

CuCl2 high

CuCl low BaS High K3CO3 high

Ionic Solutions Molecular Solutions NaCl(aq) C6H12O6(aq) metal

Ca(OH)2(aq) C12H22O11(aq) nonmetal

(NH4)3PO4(aq) CH3OH(aq)

Ca(CH3COO)2(aq) O2(aq)

H2SO4(aq) N2H4(aq)

Conduct electricity Do not conduct electricity

Write dissociation (ionic) or dissolving (molecular) equations for the first six compounds to show how each dissolves in water.The first five are solids and the sixth one is a liquid.

NaCl(s) Na+ + Cl-

C6H12O6(s) C6H12O6(aq)

Ca(OH)2(s) Ca2+ + 2OH-

C12H22O11(s) C12H22O11(aq)

(NH4)3PO4(s) 3NH4+ + PO4

3-

CH3OH(l) CH3OH(aq)

Equilibrium Expressions for Saturated Solutions

Write the equation for equilibrium present in a saturated solution of Al2(SO4)3 solution.

Al2(SO4)3(s) ⇌ 2Al3+ + 3SO42-

Write the equilibrium expression or solubility product

Solubility Product = Ksp = [Al3+]2[SO42-]3

Equilibrium Expressions for Saturated Solutions

Write the equation for equilibrium present in a saturated solution of Al2(SO4)3 solution.

Ca3(PO4)2(s) ⇌ 3Ca2+ + 2PO43-

Write the equilibrium expression or solubility product

Solubility Product = Ksp = [Ca2+]3[PO43-]2

Solubility Unit III Lesson 1. Solubility is a measure of the maximum amount of solid that will...

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