Solubility Guidelines for Aqueous

Solutions

Agenda

• Day 68 – Solubility Rules & Precipitation Reactions and Net Ionic Equations

• Lesson: PPT,

• Handouts: 1. Solubility/ Net Ionic/ Stoichiometry Handout 2.

• Text: 1. P. 424-427 -Memorize Rule 1 and 2

• HW: 1. Worksheets, P. 428 # 1-9, P. 455- 459- Review questions

•

How is a precipitate different from insoluble?

• Solute ions that don’t dissociate… ~ insoluble

•Already dissolved ions get together to form a new compound ~precipitate

What does the formation of a precipitate indicate?

Precipitates indicate a chemical reaction occurred

How can we know if an ionic compound will dissolve?

Two ways:

1. Mix different solutions together and see what happens, or…

2. Learn from what others have already done.

• When two aqueous ionic compounds are mixed together, the compounds will either dissociate , or precipitate

• If no reaction occurs, write NR.

Evidence of a double displacement reaction:

• Precipitate

• Gas

• Water

Precipitate Reactions

• Based on the solubility guidelines

• Example:

Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(S) + 2NaNO3(aq)

Reactions Forming a Gas

Reactions that produce a compound that decomposes into a gas and water

Example:

K2SO3(aq) + 2 HNO3(aq) 2 KNO3(aq) + SO2(s) + H2O(l)

CaCO3(aq) +2 HCl(aq) CaCl2(aq) + CO2 (s) + H2O(l)

Reactions Forming Water

• Neutralization reaction between an acid and a base to produce water and a salt

Example:

H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2 H2O(l)

Solubility •Precipitation refers to the formation of a solid from ions. A precipitate is “insoluble” •Soluble and insoluble are general terms to describe how much of a solid dissolves. •Solubility can be predicted from rules ( Table 1 on pg.424) •You will have to memorize some of these rules, and you will need to know how to use the rules to predict solubility.

Slightly Soluble?

Consider it insoluble

Almost everything dissolves at least a little bit.

General Solubility Rules • Salts are generally more soluble in HOT water (Gases are

more soluble in COLD water)

• Alkali Metal salts are very soluble in water.

NaCl, KOH, Li3PO4, Na2SO4 etc...

• Ammonium salts are very soluble in water.

NH4Br, (NH4)2CO3 etc…

• Salts containing the nitrate ion, NO3-, are very soluble in

water.

• Most salts of Cl-, Br- and I- are very soluble in water - exceptions are salts containing Ag+ and Pb2+.

soluble salts: FeCl2, AlBr3, MgI2 etc...

“insoluble” salts: AgCl, PbBr2 etc...

Soluble or Insoluble? a) Ca(NO3)2 - Soluble rule (salts containing NO3

- are soluble) b) FeCl2 - Soluble rule (all chlorides are soluble) c) Ni(OH)2 - Insoluble rule (all hydroxides are insoluble) d) AgNO3 - Soluble rule (salts containing NO3

- are soluble) e) BaSO4 - Insoluble rule (Sulfates are soluble, except … Ba2+) f) CuCO3 - Insoluble rule (containing CO3

2- are insoluble)

Net ionic equations

Na+

Al3+

S2–

2Ca2+ PO43–

3Cl–

Review: forming ions • Ionic (i.e. salt) refers to +ve ion plus -ve ion • Usually this is a metal + non-metal or metal + polyatomic

ion (e.g. NaCl, NaClO3, Li2CO3) • Polyatomic ions are listed in the nomenclature package • (aq) means aqueous (dissolved in water) • For salts (aq) means the salt exists as ions • NaCl(aq) is the same as: Na+(aq) + Cl–(aq) • Acids form ions: HCl(aq) is H+(aq) + Cl–(aq), Bases form ions: NaOH(aq) is Na+ + OH– Q - how is charge determined (+1, -1, +2, etc.)? A - via valences (periodic table or see the nomenclature

package ) • F, Cl gain one electron, thus forming F–, Cl– • Ca loses two electrons, thus forming Ca2+

Background: valences and formulas • Charge can also be found via the compound • E.g. in NaNO3(aq) if you know Na forms Na+, then NO3

must be NO3– (NaNO3 is neutral)

• By knowing the valence of one element you can often determine the other valences

Q - Write the ions that form from Al2(SO4)3(aq)? Step 1 - look at the formula: Al2(SO4)3(aq) Step 2 - determine valences: Al3 (SO4)2

(Al is 3+ according to the periodic table)

Step 3 - write ions: 2Al3+(aq) + 3SO42–(aq)

• Note that there are 2 aluminums because Al has a subscript of 2 in the original formula

Practice with writing ions Q - Write ions for Na2CO3(aq)

A - 2Na+(aq) + CO32–(aq) (from the PT Na is 1+. There are 2,

thus we have 2Na+. There is only one CO3. It must have a 2- charge)

• Notice that when ions form from molecules, charge can be separated, but the total charge (and number of each atom) stays constant.

Q - Write ions for Ca3(PO4)2(aq) & Cd(NO3)2(aq)

A - 3Ca2+(aq) + 2PO43–(aq)

A - Cd2+(aq) + 2NO3–(aq)

Q - Write ions for Na2S(aq) and Mg3(BO3)2(aq)

A - 2Na+(aq) + S2–(aq), 3Mg2+(aq)+ 2BO33–(aq)

Demo:

• When a colourless solution of lead(II) nitrate is added to a colourless solution of potassium iodide, a bright yellow precipitate of lead(II) iodide is obtained. Although, the lead(II) iodide is ionic, it is insoluble in water - hence the formation of the precipitate. This reaction can be represented by the following equation:

• Pb(NO3)2(aq) + 2KI(aq) KNO3(aq) + Pb(I)2(s)

Types of chemical equations Equations can be divided into 3 types 1) Molecular, 2) Ionic, 3) Net ionic • Here is a typical molecular equation: Cd(NO3)2(aq) + Na2S(aq) CdS(s) + 2NaNO3(aq)

• We can write this as an ionic equation (all compounds that are (aq) are written as ions):

Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq)

CdS(s) + 2Na+(aq) + 2NO3–(aq)

• To get the NET ionic equation we cancel out all terms that appear on both sides:

Net: Cd2+(aq) + S2–(aq) CdS(s)

Ionic Equations Ionic:

Ag+(aq) + NO3-(aq) + NH4

+(aq) + Cl-(aq)

Note: combine, in your head, the positive and

negative ions. If together a pair is insoluble, they

will form a precipitate (s).

In this case AgCl is insoluble

Ag+(aq) + NO3-(aq) + NH4

+(aq) + Cl-(aq)

AgCl(s) + NO3-(aq) + NH4

+(aq)

Net ionic: Ag+(aq) + Cl-(aq) AgCl(s)

If no solid is formed then write N.R.

Equations must be balanced • There are two conditions for molecular, ionic, and

net ionic equations

Materials balance

Both sides of an equation should have the same number of each type of atom

Electrical balance

Both sides of a reaction should have the same net charge

Q- When NaOH(aq) and MgCl2(aq) are mixed, _______ (s) and NaCl(aq) are produced. Write balanced molecular, ionic & net ionic equations

Mg(OH)2

Determining net ionic reactions

Step 1: Write formula for reactants and products using valences. Products are determined by switching +ve and –ve.

Step 2: Determine if any products are insoluble (use solubility rules). Note: all reactants must be soluble (i.e. aq) in order to mix.

If all products are aqueous: “no reaction”

Step 3: Balance the equation

Step 4: Write the ionic equation

Step 5: Write the net ionic equation

Step 4: Write the ionic equation

If you look at what we call the IONIC EQUATION, you will note that some of the ions are common to both the reactant and product side of the equation.

Such ions are called SPECTATOR IONS, as they play no part in the observed reaction. In fact, these spectator ions may be cancelled from each side of the equation, leaving only the

NET IONIC EQUATION.

Step 5: Write the net ionic equation

NaOH(aq) + MgCl2(aq) Mg(OH)2(s) + NaCl(aq)

Next, balance the equation

First write the skeleton equation

2 2

Ionic equation:

2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq) Mg(OH)2(s) + 2Na+(aq) + 2Cl-(aq)

Net ionic equation:

2OH-(aq) + Mg2+(aq) Mg(OH)2(s)

Write balanced ionic and net ionic equations: CuSO4(aq) + BaCl2(aq) CuCl2(aq) + BaSO4(s)

Fe(NO3)3(aq) + LiOH(aq) ______(aq) + Fe(OH)3(s)

Na3PO4(aq) + CaCl2(aq) _________(s) + NaCl(aq)

Na2S(aq) + AgC2H3O2(aq) ________(aq) + Ag2S(s)

LiNO3

Ca3(PO4)2

NaC2H3O2

Net Ionic Equation for Single Displacement Reaction

Write the net ionic equation for the reaction that occurs when aluminum metal is placed in a solution of copper (II) chloride.

• Go over Sample Problem 2 on P. 427

Cu2+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) Cu2+(aq)

+ 2Cl–(aq) + BaSO4(s)

Net: SO42–(aq) + Ba2+(aq) BaSO4(s)

Fe3+(aq) + 3NO3–(aq) + 3Li+(aq) + 3OH–(aq)

3Li+(aq) + 3NO3–(aq) + Fe(OH)3(s)

Net: Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

2Na3PO4(aq) + 3CaCl2(aq) Ca3(PO4)2(s)+ 6NaCl(aq)

6Na+(aq) + 2PO43–(aq) + 3Ca2+(aq) + 6Cl–(aq)

Ca3(PO4)2(s)+ 6Na+(aq) + 6Cl–(aq)

Net: 2PO43–(aq) + 3Ca2+(aq) Ca3(PO4)2(s)

2Na+(aq) + S2–(aq) + 2Ag+(aq) + 2C2H3O2–(aq)

2Na+(aq) + 2C2H3O2–(aq) + Ag2S(s)

Net: S2–(aq) + 2Ag+(aq) Ag2S(s)

• Day 69 – Solutions Stoichiometry

• Lesson: PPT,

• Handouts: 1. Solubility/ Net Ionic/ StoichiometryHandout

• Text: 1. P. 444-449

• HW: 1. Worksheets, P. 449 # 1-9

Stoichiometry overview • Recall that in stoichiometry the mole ratio

provides a necessary conversion factor:

grams (x) moles (x) moles (y) grams (y)

molar mass of x molar mass of y

mole ratio from balanced equation

• We can do something similar with solutions:

volume (x) moles(x) moles (y) volume(y)

mol/L of x mol/L of y

mole ratio from balanced equation

Question 1 Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used?

1 mol H2SO4

2 mol NH3 x

# mol H2SO4=

0.0244 L NH3 0.02684 mol

H2SO4

= 2.20 mol NH3

L NH3 x

H2SO4(aq) + 2NH3(aq) (NH4)2SO4(aq)

Calculate mol H2SO4, then c= n/V = n/0.0500 L

C = mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L

Question 2 Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. Calculate the volume of 0.0250 mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of 0.125 mol/L aluminum sulfate solution.

3 mol Ca(OH)2

1 mol Al2(SO4)3 x

# L Ca(OH)2=

0.0250

L Al2(SO4)3

= 0.375 L Ca(OH)2

0.125 mol Al2(SO4)3

L Al2(SO4)3 x

Al2(SO4)3(aq) + 3Ca(OH)2(aq) 2Al(OH)3(s) + 3CaSO4(s) Calculate mol Al3SO4, then mol of Ca(OH)2 then v = n/C

L Ca(OH)2

0.0250 mol

Ca(OH)2

x

Question 3 A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with an excess of 0.250 mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed?

3 mol Na2CO3

2 mol FeCl3

x

# L Na2CO3=

0.0750 L FeCl3

= 0.0900 L Na2CO3 = 90.0 mL Na2CO3

0.200 mol FeCl3

L FeCl3

x

2FeCl3(aq) + 3Na2CO3(aq) Fe2(CO3)3(s) + 6NaCl(aq)

L Na2CO3

0.250 mol

Na2CO3

x

Assignment

1. H2SO4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H2SO4 will be required to react completely with 75 mL of 0.50 mol/L NaOH?

2. How many moles of Fe(OH)3 are produced when 85.0 L of iron(III) sulfate at a concentration of 0.600 mol/L reacts with excess NaOH?

3. What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution.

Assignment

4. a) What volume of 0.20 mol/L AgNO3 will be needed to react completely with 25.0 mL of 0.50 mol/L potassium phosphate?

b) What mass of precipitate is produced from the above reaction?

5. What mass of precipitate should result when 0.550 L of 0.500 mol/L aluminum nitrate solution is mixed with 0.240 L of 1.50 mol/L sodium hydroxide solution?

Answers 1. H2SO4(aq) + 2NaOH(aq) 2H2O + Na2SO4(aq)

2 mol Fe(OH)3

1 mol Fe2(SO4)3 x

# mol Fe(OH)3=

85 L Fe2(SO4)3 0.600 mol Fe2(SO4)3

L Fe2(SO4)3 x

= 102 mol

1 mol H2SO4

2 mol NaOH x

# L H2SO4=

0.075 L NaOH

= 0.009375 L = 9.4 mL

0.50 mol NaOH

L NaOH x

2. Fe2(SO4)3(aq) + 6NaOH(aq) 2Fe(OH)3(s) + 3Na2SO4(aq)

L H2SO4

2.0 mol H2SO4

x

3. 2NaOH(aq) + ZnCl2(aq) Zn(OH)2(s) + 2NaCl(aq)

1 mol Zn(OH)2

2 mol NaOH x

# g Zn(OH)2=

0.0500

L NaOH

= 6.21 g

2.50 mol NaOH

L NaOH x

4a. 3AgNO3(aq) + K3PO4(aq) Ag3PO4(s) + 3KNO3(aq)

99.40 g Zn(OH)2

1 mol Zn(OH)2

x

3 mol AgNO3

1 mol K3PO4 x

# L AgNO3 =

0.025

L K3PO4

= 0.1875 L = 0.19 L

0.50 mol K3PO4

L K3PO4

x L AgNO3

0.20 mol AgNO3

x

1 mol Ag3PO4

1 mol K3PO4

x

# g Ag3PO4=

0.025

L K3PO4

= 5.2 g

0.50 mol K3PO4

L K3PO4

x 418.58 g Ag3PO4

1 mol Ag3PO4

x

4b. 3AgNO3(aq) + K3PO4(aq) Ag3PO4(s) + 3KNO3(aq)

1 mol Al(OH)3

1 mol Al(NO3)3

x

# g Al(OH)3=

0.550

L Al(NO3)3

21.4 g Al(OH)3 =

77.98 g Al(OH)3

1 mol Al(OH)3

x 0.500 mol Al(NO3)3

L Al(NO3)3

x

5. Al(NO3)3(aq) + 3NaOH(aq) Al(OH)3(s) + 3NaNO3(aq)

1 mol Al(OH)3

3 mol NaOH

x

# g Al(OH)3=

0.240

L NaOH

9.36 g Al(OH)3 =

77.98 g Al(OH)3

1 mol Al(OH)3

x 1.50 mol NaOH

L NaOH

x