section 6.4—solubility & precipitation how can we make sure everything that’s added to the...

Section 6.4—Solubility & Precipitation

How can we make sure everything that’s added to the sports drink will dissolve?

A Review of Double-Replacement Reactions

NaCl + AgNO3 AgCl + NaNO3

Double Replacement Reactions

The cations from two compounds replace each other.

ClCl

NaNa

AgAg

OO OO

NN

OO

ClCl

AgAg

NaNa

OO OONN

OO

Two ionic compounds switch ions

Double Replacement Reactions

A X B Z A XBZAA XX BB ZZ AA XXBBZZ

General format of a double replacement reaction:

Combine the cation of the first reactant with the anion of the second reactant

CaCl2 + AgNO3

1

Products of a Double Replacement

Combine the cation of the second reactant with the anion of the first reactant

CaCl2 + AgNO3

2

Products of a Double Replacement

& balance charges with subscripts when writing formulasRemember to write cations first …

AgCl

CaCl2 + AgNO3

3

Ca(NO3)2 +CaCl2 AgNO3+

Products of a Double Replacement

Only leave subscripts that are in the original compound there if they are a part of a polyatomic ion!

Precipitation Reactions

Precipitation Reactions

A precipitation reaction is when 2 soluble substances are mixed together and they form an insoluble substance. This is called a precipitate.

Reactants

2 soluble

chemicals:

NaOH and Cu(NO3)2

NaOHCu(NO3)2

Precipitation Reactions(DR Rxns)

Products: 1 soluble chemical: NaNO3 1 insoluble chemical (the precipitate): Cu(OH)2

Na+1

OH-1

Cu+2

NO3 -1

Na+1

NO3 -1

Cu(OH)2(S)

Solubility Rules

Solubility Rules Table

Use the table on the reference sheet!

Insoluble = Precipitate

Let’s Practice #1

Example:Decide whether each is soluble

or not

NaNO3

Fe(C2H3O2)2

CaBr2

Ba(OH)2

Cu(OH)2

Let’s Practice #1

Example:Decide whether each is soluble

or not

NaNO3

Fe(C2H3O2)2

CaBr2

Ba(OH)2

Cu(OH)2

Soluble

Soluble

Soluble

Soluble

Not Soluble

Let’s Practice #2

Example:Write the

products for this reaction &

predict the precipitate

AgNO3 (aq) + NaCl (aq)

Remember to indicate compounds that dissolve with “aq” for “aqueous” and compounds that don’t dissolve with “s” for “solid”

Let’s Practice #2

AgCl (s) + NaNO3 (aq)

Example:Write the

products for this reaction

AgNO3 (aq) + NaCl (aq)

precipitate

Let’s Practice #3

Example:Write the

products for this reaction &

identify the precipitate

BaCl2 (aq) + K2CO3 (aq)

Remember to indicate compounds that dissolve with “aq” for “aqueous” and compounds that don’t dissolve with “s” for “solid”

BaCl2 (aq) + K2CO3 (aq)

Let’s Practice #3

KCl (aq) + BaCO3 (s)

Example:Write the

products for this reaction

precipitate

Net Ionic ReactionsShows the details of aqueous reactions that involve ions in aqueous solution

Molecular Equation: the typical equation you are use to writing keeping all molecules together

Complete Ionic Equation: shows all the particles in a solution as they really exist, as IONS or MOLECULES. Anything aqueous needs to be split apart into the cation and anionAnything solid stays intactCoefficients need to be multiplied by subscripts to determine the exact amount of each cation and anion.

Net Ionic Equation: the final equation showing the major players. All spectator ions have been removed.

Spectator ions: ions that do not participate in a reaction; they are identical on both sides of the equation & are crossed out!

NET IONIC REACTIONS for Precipitation Reactions

Molecular equation: KI(aq) + AgNO3(aq) AgI(s) + KNO3(aq)

Complete Ionic equation: K+1 + I-1 + Ag+1+ NO3

-1 AgI + K+1 + NO3-1

Net Ionic equation:

I-1 + Ag+1 AgI

Spectator ions: ions that do not participate in a reaction; they are identical on both sides of the equation & are crossed out!

NET IONIC REACTIONS for Precipitation Reactions

Molecular equation: 2 NaOH(aq) + CuCl2(aq) 2 NaCl(aq) + Cu(OH)2(s)

Complete Ionic equation:2 Na+1 + 2 OH-1 + Cu+2 + 2 Cl-1 2 Na+1 + 2 Cl-1 + Cu(OH)2

Net Ionic equation:

2 OH-1 + Cu+2 Cu(OH)2

Take Home Practice:Predict products and balance Iron (III) chloride reacts with sodium hydroxide

Molecular equation: 1 FeCl3(aq) + 3 NaOH(aq) 1 Fe(OH)3(s) +3 NaCl(aq)

Complete Ionic equation:3 Na+1 + 3 OH-1 + Fe+3 + 3 Cl-1 3 Na+1 + 3 Cl-1 + Fe(OH)3

Net Ionic equation:

3 OH-1 + Fe+3 Fe(OH)3

Section 6.5—Stoichiometry

How can we determine in a lab the concentration of electrolytes?

2 H2 + O2 2 H2O

2

No coefficient = 1

2For every 2 moles

of H2…

1 mole of O2 is need to react…

and 2 moles of H2O are produced

What do those coefficients really mean?

The coefficient of the balanced chemical equation tells how many moles of each substance is used in the reaction.

Mole RatioIs a conversion factor that relates 2 substances in moles; must use a balanced chemical equation to create it

2 H2 + O2 2 H2O

Examples of Mole Ratio’s

2mol H2 1 mol O2 2 mol H2O 1 mol O2 2 mol H2O 2 mol H2

What is stoichiometry?

Stoichiometry – Calculations using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation.

Example: What is the mole ratio of chlorine to sodium?

2 Na + Cl2 2 NaCl

2mol Na 1 mol Cl2 2 mol Na1 mol Cl2 2 mol NaCl 2 mol NaCl

Stoich (Mole-Mole) : 1 step problemusing the mole ratio

Example:If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are

needed?2 H2 + O2 2 H2O

Stoichiometry with Moles

4.2 mole H2

mole H2

mole O2 = ________ mole O2

2

12.1

From balanced equation: 2 mole H2 1 mole O2

Example:If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are

needed?2 H2 + O2 2 H2O

Stoich (Mole-Mole)

0.67 mole KNO3 Mole

KNO3

mole O2 = ________ mole O2

2

10.34

From balanced equation: 2 mole KNO3 1 mole O2

Example:If 0.67 moles of potassium nitrate

reacts, how many moles of oxygen are produced?

2KNO3 2KNO2 + O2

But we can’t measure moles in lab!

We can’t go to the lab and count or measure moles…so we need a way to work in measurable units, such as grams and liters!

Molecular mass gives the grams = 1 mole of a compound!

Stoich( Mole-Mass): 2 step problemuse mole ratio & then molar mass conversion factors

Example:How many grams of AgCl will be

precipitated if 0.45 mole AgNO3 is reacted as follows:2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2

From balanced equation: 2 mole AgNO3 2 mole AgCl

Stoichiometry with Moles & Mass

0.45 mole AgNO3

mole AgNO3

mole AgCl = ________ g AgCl

2

264

Molar Mass of AgCl:1 mole AgCl = 143.32 g

mole AgCl

g AgCl

1

143.32

Example:How many grams of AgCl will be

precipitated if 0.45 mole AgNO3 is reacted as follows:2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2

Stoich( Mass- Mol): 2 step problemuse molar mass & then mole ratio conversion factors

4.42 g H2

g H2

Mole H2 =1.46 mole NH3

2.02

1

From balanced equation: 3 mole H2 2 mole N2

Example:If 4.42 g of H2 reacts, how

many moles of NH3 are produced?

N2 + 3H2 2NH3

2 mole NH3

3 mole H2

Stoich( Mass-Mass): 3 step problemuse molar mass, then mole ratio & then molar mass conversion factors (Honors Only)

Example:How many grams Ba(OH)2 are

precipitated from 14.5 g of NaOH in the following reaction:

2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl

From balanced equation: 2 mole NaOH 1 mole Ba(OH)2

Stoichiometry with Mass (Honors)

14.5 g NaOH

g NaOH

mole NaOH

= ________ g Ba(OH)2

40.00

1

31.1

Molar Mass of NaOH:1 mole NaCl = 40.00 g

mole NaOH

mole Ba(OH)2

2

1

mole Ba(OH)2

g Ba(OH)2

1

171.35

Molar Mass of Ba(OH)2:1 mole Ba(OH)2 = 171.35 g

Example:How many grams Ba(OH)2 are

precipitated from 14.5 g of NaOH in the following reaction:

2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl

Stoichiometry with Mass (Honors)

Example:How many grams of HCl are needed to produce 65.0 g of magnesium chloride:__Mg + ____HCl ____MgCl2 + __H2

65 g MgCl21 mole MgCl2

g MgCl2

Balance the equation and fill in the missing information:

mole MgCl2

mole HCl

mole HCl

g HCl = g HCl

What about the stoichiometry of gases? Recall

Molar Volume of a Gas – at STP 1 mole of any gas = 22.4 liters

Stoichiometry with Gases 1 mol= 22.4 L @STP

Example:If you need react 1.5 g of zinc completely,

what volume of hydrogen gas will be produced at STP?

2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

From balanced equation: 1 mole Zn 1 mole H2

Stoichiometry with Gases

1.5 g Zn

g Zn

mole Zn

= ________ L H2

65.39

1

0.51

Molar volume of a gas:1 mole H2 = 22.4 L

mole Zn

mole H2

1

1

mole H2

L H2

1

22.4

Molar Mass of Zn:1 mole Zn = 65.39 g

Example:If you need react 1.5 g of zinc completely,

what volume of hydrogen gas will be produced at STP?

2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

From balanced equation: 4 mole H2O 5 mole O2

Stoichiometry with Gases

7.3 L O2

L O2

Mole O2

= _

22.4 L

1

0.26 mole H2O

Molar volume of a gas:1 mole O2 = 22.4 L

Mole O2

mole H2 O

5

4

Molar Mass of H:1 mole H2O = 18 g

Example: How many moles of water will be produced from the complete

combustion of 7.3 L of oxygen gas? Assume STP

C3H8 + 5O2 → 3CO2 + 4H2O

Keeping all these equalities straight!

TO GO BETWEEN USE THE EQUALITY

Grams & moles Molar mass (g)= 1 mole

Particles & Moles 1 mol = 6.02 x 1023 particles

Moles & liters of a gas at STP

1 mole = 22.4 L at STP

2 different chemicals in a reaction

Coefficient ratio(MOLE RATIO) from balanced equation

You Try!

Given the UNBALANCED EQUATION: __MgCO3 __MgO + __CO2, how many liters of CO2 gas are produced from the reaction of 15 grams of MgCO3? Assume STP!

Section 6.5b

Percent Yield

Percent YieldA “Yield” is a product

Actual Yield(A): the actual amount of product you produce in the lab

Theoretical Yield(T): the amount of product you should produce if nothing went wrong; use the balanced chemical equation to calculate this amount.

Percent Yield: ratio of actual yield to theoretical yield

Percent Yield

%yield = A x 100

TLets Practice in steps:

1a. If 4.20 moles H2 reacts completely with oxygen, how many grams of H2O are produced?

2 H2 + O2 2 H2O

This is a mol-mass problem.

Your answer is the theoretical yield of water?

If 4.20 moles H2 react completely with oxygen how many grams of H2O are produced?

2 H2 + O2 2 H2O

4.2 mol H2

2 moleH2O

2 mol H2 1 mole H2O

18.02 grams H2O

= ______g H2O

From balanced equation: 2 mole H2O 1 mole O2

Molar Mass of O2:1 mole H2O = 18.02g

This is the theoretical yield.

75.7 g

What is the percent yield if 60.0 grams of H2O are produced?

A= 60.0 g

T= 75.7 g

%yield = A x 100 T

%yield = 60.0x 100 75.7

79.3% yield

You have precipitated 8.50 g of Ba(OH)2. If you start with 4.57 grams of NaOH, what is the % yield. 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl

4.57 g NaOH 1 molNaOH

40.00g NaOH 2 mol NaOH

1 molBa(OH)2

= 9.79 g Ba(OH)2

From balanced equation: 2 mole NaOH 1 mole Ba(OH)2

Molar Mass of Ba(OH)2: 173.25 g

This is the theoretical yield.

Molar Mass of NaOH: 40.00 g

1mol Ba(OH)2

171.35g Ba(OH)2

If 9.78 grams are obtained in the experiment, what is the percent yield?

A= 8.50 g

T= 9.79 g

%yield = A x 100 T

%yield = 8.50 x 100 9.79

86.8% yield

Section 6.5c

Titrations

Titrations—Using Stoichiometry

Titration – A technique where the addition of a known volume of a known concentration solution to a known volume of unknown concentration solution to determine the concentration.

•Use a buret to titrate unknown concentration of solutions.

Titrations—Using StoichiometryThe titrant is the known concentration in the buret and the analyte is the unknown concentration in the flask.

Formula: nMaVa = nMbVb

na= number of H+ in the acidnb= number of OH- in the baseMa= molarity of acidMb= molarity of baseV= volume

End Point vs. Equivalence Point

Equivalence Point (or Stoichiometric Point) – When there are no reactants left over—they have all been reacted and the solution contains only products

-the point where the acid and the base are equal in equal moles moles acid = moles base

Importance of Indicators

Always select an indicator that has a pH value close to that of the pH of the equivalence point of the titration.

Indicators – Paper or liquid that change color based on pH level.

End Point: point at which the indicator in the solution changes colorIt signals the equivalence point and the stop of the titration

Titration Process

Titration Problem #1

How many liters of 0.10 M NaOH is needed to react with 0.125 L of 0.25 M HCl?

NaOH + HCl H2O + NaCl

Titration Problem #2

What is the molarity of a Ca(OH)2 solution if 30.0 ml of the solution is neutralized by 20.0 ml of a 0.50 M solution of HCl?

  Ca(OH) 2 + 2HCl 2H2O + CaCl2

Titration Problem #3

What volume of 2.0M solution of NH4OH is needed to neutralize 50.0 ml of a 0.50M solution of H2SO4?

2 NH4OH + H2SO4 2H2O + (NH4) 2SO4

Titration Curves

Shows the changes of pH during a titration

Identifies the pH of the equivalence point

Strong Base - Strong Acid

Weak Base - Strong Acid

Strong Base - Weak Acid Weak Base - Weak Acid

Titration curve for Titrating a strong acid with a strong base

pH is always = 7The titration curve graph shows the pH of the equivalence point. Take the vertical region and cut the length in half and then look to what pH value aligns to that point.

Titration curve for Titrating a strong base with an strong acid

pH is always = 7

Titration curve for Titrating a weak acid with an strong base pH is >7