Chapter 7 – SOLUBILITY & REACTIONS

Saturated Solution – a solution in which no more solute will dissolve

the solubility of a substance is the concentration of a saturated solution of that substance

solubility is usually measured in grams of solute per 100mL and is very dependant on temperature e.g. the solubility of sodium sulphate in water at 0ºC is 4.76g/100mL

this means that 4.76g of sodium sulphate can be dissolved in 100mL of water at 0ºC…..and if you add more it won’t dissolve!

Solubility Curves

Graphs of Solubility(Maximum Concentration) vs. Temperature allow a fast and easy reference (See Figure 2 on page 316)

Solubility of Gases

Gases can & do dissolve in liquid

Examples; chlorine in swimming pools

oxygen in rivers & streams

carbon dioxide in cans of pop

Does temperature effect the solubility of a liquid as it does in solids?

DO LAB EXERCISE 7.1.1 on Page 318.

SOLUBILITY IN WATER – GENERALIZATIONS & EXAMPLES

SOLUTE / LOWER
TEMP. / HIGHER
TEMP. / DOESN’T
DISSOLVE / IN
WATER / E.G. (as
xg/100mL)
180g@0ºC
487g@100ºC
SOLIDS /
GASES /
LIQUIDS / polar / non-polar* / small
polar **
ELEMENTS / Low solubility
in water
incl.
halogens & oxygen

* = immiscible with water (forms a layer)

** = miscible (dissolves completely)

Crystallization

When the solute comes out of solution as hard particles this is called crystallization

as solvent evaporates there is not enough solvent to keep the solute in solution

this process can be speeded up by heating a solution to evaporate off the solvent-- used industrially to produce table salt & table sugar

Solubility Categories

solubilities can range from high to extremely low

we use the term insoluble to mean negligible solubility

we can categorize the solubilities as;

HIGH SOLUBILITY / Maximum concentration at SATP
of greater than or equal to 0.01mol/L
LOW SOLUBILITY / Maximum concentration at SATP of less than 0.1mol/L
INSOLUBLE / Substance that has a negligible solubility at SATP

SOLUBILITY OF COMMON CATIONS & ANIONS AT SATP

COMMON ANIONS

Cl-1,Br-1,I-1 / S2- / OH-1 / SO4-2 / CO32-,PO4-4 / C2H3O2- / NO3-1
High
Solubility
> or =
0.1mol/L
(at SATP) / Most / Group
I & II,
NH4+1 / Group
I,
NH4+,
Sr2+,
Ba2+,
Tl+ / Most / Group I
&
NH4+1 / Most / All
Low
Solubility
< or =
0.1mol/L
(at SATP) / Ag+,Pb2+,
Tl+,Hg2+,
Hg+,Cu+ / Most / Most / Ag+,
Pb2+,
Ca2+,
Ba2+,
Sr2+,
Ra2+ / Most / Ag+1 / Most

Hard Water Treatment

hard water is water containing a higher concentration than normal of calcium and magnesium ions

usually results in floating scum, when using soap in a bath, soap not working up a good lather, etc.

to counteract this water softenerscan be used

water can be softenedby removing the excess calcium & magnesium ions

our municipalities use the soda-lime process but home water-softening involves an ion exchange process where sodium ions replace the calcium & magnesium

Answer questions 1-4 on Page 329.

7.3 Reactions in Solution

The solubility table on Page 324 can be useful in predicting whether the products of a reaction will be soluble or insoluble.

Net Ionic Equations

if we mix lead (II) nitrate with potassium iodide, we will produce a yellow precipitate.

If we look at our solubility table using the products we expect to be formed: lead iodide + potassium nitrate we can hypothesize;

  1. lead iodide will have a low solubility
  2. potassium nitrate will have a high solubility

(1) Pb(NO3)2(aq) + 2KI(aq) -- PbI2(s) + 2KNO3(aq)

(2) Pb(NO3)2(aq) + NaI(aq) -- PbI2(s) + 2NaNO3(aq)

(3) Pb(C2H3O2)2(aq) + MgI2(aq) -- PbI2(s) + Mg(C2H3O2)2(aq)

using Arrhenius’ Theory of Dissociation we can rewrite (1)

Pb(NO3)2(aq) + 2KI(aq) -- PbI2(s) + 2KNO3(aq)

as:

Pb2+(aq) + 2NO3-1(aq) + 2K+1(aq) + 2I-1(aq)

----PbI2(s) + 2K+1(aq) + 2NO3-1(aq)

This is called a TOTAL IONIC EQUATION

Any ion (atom, molecule, etc.) present in a reaction system that does not change or become involved during a chemical reaction is called a spectator.

If we rewrite an equation for the reaction, showing only the entities that change we have what’s called a net ionic equation.

In other words;

Pb2+(aq) + 2NO3-1(aq) + 2K+1(aq) + 2I-1(aq)

----PbI2(s) + 2K+1(aq) + 2NO3-1(aq)

would be re-written as:

Pb2+(aq) + 2I-1(aq) ----PbI2(s) ……..our net ionic equation

Writing Net Ionic Equations

  1. Write the balanced chemical equation with full chemical formulas for all the reactants and products.
  2. Using the solubility table (see p. 324), rewrite the formulas for all the high solubility ionic compounds as dissociated ions, to show the total ionic equation.
  3. Cancel identical amounts of identical entities appearing on both sides of the equation.
  4. Write the net ionic equation reducing the coefficients if necessary.

Sample Problem 1

Q: Write the net ionic equation for the reaction for the reaction of

aqueous barium chloride and aqueous sodium sulphate.

A: step 1: BaCl2(aq) + Na2SO4(aq) --- BaSO4(s)+ 2NaCl(aq)

step 2 & 3: Ba2+(aq) + 2Cl-1(aq) + 2Na+1(aq)+ SO4-2(aq)

---- BaSO4(s) + 2Na+1(aq) + 2Cl-1(aq)

step 4: Ba2+(aq) + SO4-2(aq) ---- BaSO4(s) …..net ionic equation

Sample problem 2

Q: Write the net ionic equation for the reaction of zinc metal and

aqueous copper (II) sulphate and then write a statement to

communicate the meaning of the net ionic equation..

A: step 1: Zn(s) + CuSO4(aq) --- Cu(s) + ZnSO4(aq)

steps 2 & 3: Zn(s) + Cu2+(aq) + SO42-(aq)

--- Cu(s) + Zn2+(aq) + SO42-(aq)

step 4: Zn(s) + Cu2+(aq) ---Cu(s) + Zn2-(aq) ….net ionic equation

Solid zinc in an aqueous solution containing copper (II) ions will

produce solid copper and aqueous zinc ions.

7.4 Waste Water Treatment

7.5 Qualitative Chemical Analysis

  1. Qualitative Analysis – identification of the specific substances present.
  2. Quantitative Analysis – measurement of the quantity of a substance present.

1-(a) Qualitative analysis – can be done by colour

ION / SOLUTION COLOUR
Group 1,2,17 / colourless
Cr2+(aq) / blue
Cr3+(aq) / green
Co2+(aq) / pink
Cu+(aq) / green
Cu2+(aq) / blue
Fe2+(aq) / pale green
Fe3+(aq) / yellow-brown
Mn2+(aq) / pale pink
Ni2+(aq) / green
CrO4-(aq) / yellow
Cr2O72-(aq) / orange
MnO4-1(aq) / purple

In conjunction with colour solution tests, we can also use flame tests to determine qualitatively what substances are present.

(see Fig. 2 on p. 342)

1-(b)-Sequential Qualitative Chemical Analysis

this is a type of test where we set up a double displacement reaction using one unknown solution and one known solution

in this test, we would predict that if a precipitate forms then a certain ion must have been present in the unknown solution

Book example: if you were given a solution that contains lead (II) ions or strontium ions (or neither or both) what would we do?

To Complete A Sequential Analysis

1. Locate the possible cations on the solubility table.

2. Determine which anions would precipitate the possible cations cations.

3. Plan a sequence of precipitation reactions that would use use anions to precipitate a single cation at a time.

4. Use filtration between the steps to remove cation

precipitates that might interfere with subsequent

additions of anions.

5. Draw a flow chart to assist testing & communication.

Filter

White precipitate no precipitate

Quantitative Analysis

often quantitative analysis will follow a qualitative analysis.

e.g. a police officer sees erratic driving, stops the car, smells

alcohol on the breath of the driver, and then does a breathalyzer

to get an actual readingof the diver’s blood alcohol.

light is passed through a solution detects colour change (if any) and light that passes through is converted by a photocell into an electric current which indicates blood alcohol content

a blood test may be done but it takes longer

in Ontario, the legal limit is 0.080g/100mL (800ppm)

[0.08]

Solution Stoichiometry

most solutions take place in aqueous conditions

we need to know how to find the concentration of reactants in solutions

Stoichiometry Calculations

  1. Write a balanced equation for the reaction, to obtain the mole ratios.
  2. Convert the given value to an amount in moles using the appropriate conversion factor.
  3. Convert the given amount in moles to the required amount in moles, using the mole ratio from the balanced equation.
  4. Convert the required amount in moles to the required value again using the appropriate conversion factor.

Molar Concentration Problems

many well known fertilizers include ammonium hydrogen phosphate

made commercially by reacting concentrated aqueous solutions of ammonia and phosphoric acid

Q: What volume of 14.8mol/L NH3(aq) would be needed to react

completely with each 1.00kL (1.00m3) of 12.9mol/L H3PO4(aq) to

produce fertilizer in a commercial operation?

A: step 1:

2NH3(aq) + H3PO4(aq) --- (NH4)2HPO4(aq)

v=? v=1.00kL

C=14.8mol/L C=12.9mol/L

step 2:

to find the number of moles of H3PO4, we use the formula C=n/v

so, n=v x C--- nH3PO4=1.00kL x 12.9mol/L =12.9kmol

step 3:

using the mole ratio of NH3 : H3PO4

2 : 1

nNH3 = 12.9kmol x 2/1 = 25.8kmol

step 4: again using C=n/v, to find the volume of NH3

v=n/C

vNH3=25.8kmol

14.8mol/L

v=1.74kL

Therefore the volume of NH3 needed would be 1.75kL.

Sample Problem 1

Q: A 10.00mL sample of sulfuric acid reacts completely with 15.9mL

of a 0.150mol/L potassium hydroxide solution. Calculate the molar

concentration of the sulfuric acid.

A:

step 1: H2SO4(aq) + 2KOH----2H2O(l) + K2SO4(aq)

v=10.00mL v=15.9mL

C=? C=0.150mol/L

step 2: CKOH =nKOH

vKOH

therefore nKOH =vKOH x CKOH

=15.9mL x 0.150mol/L

=2.39mmol

step 3: nH2SO4 =2.39mmol x 1

2

nH2SO4 =1.19mmol

step 4: to find C of H2SO4

CH2SO4 = n/v = 1.19mmol

10.00mL

= 0.119mol/L

Therefore the molar concentration of the sulfuric acid is 0.119mol/L.