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Reactions in Aqueous Solutions

A

solution

is a

homogenous

mixture of 2 or more substances.

Solvent:

the substance present in the

larger

amount.

Solute:

the substance present in the

smaller

amount.

Solution Soft drink (

) Milk (

) Air (

) Steel (

) Solvent H 2 O H 2 O Solute Sugar, CO 2 fat, protein, sugar N 2 Fe O 2 , Ar, CH 4 C aqueous solutions of KMnO 4

An

electrolyte

is a substance that, when dissolved in water, results in a solution that can conduct electricity.

nonelectrolyte

is a substance that, when dissolved, results in a solution that does not conduct electricity.

Strong

electrolyte

Weak

electrolyte

Non

electrolyte

Why do they conduct electricity in solution?

Ionic Compounds

dissociate

(break apart) to

form ions

when dissolved in water H 2 O NaCl (

) Na + (

) + Cl (

) Crystal Lattice

Ions

can pass along (

conduct

) free electrons (

electricity

)

Hydration

is the process in which an

ion

is surrounded by

water molecules

arranged in a specific manner.

www.youtube.com/watch?v=AN4KifV12DA Water is

polar

High e density near oxygen (-) d d + Low e density near hydrogen (+) -

Strong Electrolyte

– 100% dissociation

CaCl 2 H 2 O (

) Ca +2 (

) + 2Cl (

) (NH 4 ) 2 S (

H 2 O ) 2NH 4 + (

) + S -2 (

)

Weak Electrolyte

– do not

completely

dissociate

CH 3 COOH CH 3 COO (

) + H + (

) ~95% ~5% A

reversible

reaction. The reaction occurs in both directions.

Acetic acid is a

weak electrolyte

because its

ionization in water is incomplete

Nonelectrolytes

do not conduct electricity No cations (+) and anions (-) form in solution C 6 H 12 O 6 H 2 O (

) C 6 H 12 O 6 (

) (Molecules)

Precipitation Reactions

Precipitate – insoluble solid that separates from solution

Aqueous

: soluble

Precipitate

: insoluble solid PbI 2 Pb(NO 3 ) 2 (

) + 2NaI (

) PbI 2

Chemical equation

(

) + 2NaNO 3 (

) Pb 2+ + 2NO 3 + 2Na + + 2I PbI 2 (

) + 2Na + + 2NO 3 -

ionic equation

Pb 2+ + 2I PbI 2 (

)

net ionic equation

Na + and NO 3 are

spectator

ions 8

Precipitation of Lead Iodide

2NaI (

) + Pb(NO 3 ) 2 (

) Na + + 2I + Pb 2+ + 2NO 3 Pb 2+ + 2I PbI 2 (

) Na + Na + PbI 2 Ions precipitate due to a stronger

electrostatic

attraction than forces from H 2 O

Large charge

Small ion size

lead to greater attraction and make them less soluble.

Examples of Insoluble Compounds

CdS PbS Ni(OH) 2 Al(OH) 3 10

Solubility

is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.

Most are single charged Smaller charges form weaker ionic bonds and are easier to separate and NH 4 + ions 11

Example Soluble

(aq)

or Insoluble

(s)

Classify the following ionic compounds as soluble or insoluble using the Solubility chart (a) silver sulfate (Ag 2 SO 4 ) (a) Ag 2 SO 4

(s)

insoluble

(b) calcium carbonate (CaCO 3 ) (b) This is a carbonate and Ca is a Group 2A metal. Therefore, CaCO 3

(s)

insoluble

(aq)

soluble

(d) Barium Hydroxide (Ba(OH) 2 ) (d) Hydroxides are normally insoluble, but Barium is one of the exceptions along with all alkali metals making it

soluble (aq)

(e) Lead (II) Iodide (PbI 2 ) (e) Iodides are normally soluble, but it is

insoluble (s)

because bound to lead

Writing Net Ionic Equations

1. Write the

balanced

molecular equation with phase states.

Swap

the anions of reactants to get products. (re-criss-cross) Al 2 ( SO 4 ) 3(

) + 6 Na OH (

) → 2 Al(OH) 3(

) + 3 Na 2 SO 4 (

) *Do not carry the subscripts over (except for polyatomic ions) 2. Write the ionic equation showing the soluble product dissociated into ions, but the insoluble product left together.

2 Al +3 + 3 SO 4 -2 + 6 Na + + 6 OH → 2 Al(OH) 3(

) + 6 Na + + 3 SO 4 -2 3. Cancel the spectator ions on both sides of the ionic equation.

6 Na + + 3 SO 4 -2 4. Check that charges and number of atoms are balanced 2 Al +3 (

) Al +3 (

) + 6 OH (

) → 2 Al(OH) 3(

) + 3 OH (

) → Al(OH) 3(

)

Example Net Ionic Equation

Predict what happens when potassium phosphate (K 3 PO 4 ) solution is mixed with calcium nitrate [Ca(NO 3 ) 2 ] solution. Write the net ionic equation.

K + + NO 3 -1 Ca 2+ + Balance the reactants and products to get the chemical equation Use Solubility chart to determine (s) or (aq) for each

Example Solution

Step 2:

Separate the soluble (aq) compounds to their ions for the

ion

ic equation. Leave the insoluble (s) precipitate together *Remember:

Ions have charges

Show them

(K ≠ K + )

Step 3:

Cancel the spectator ions (K + and NO 3 -1 ) on each side of the equation, we obtain the net ionic equation:

Writing Net Ionic Equations Practice

Pb(NO 3 ) 2 and LiCl Al 2 (SO 4 ) 3 and NaBr Na 2 S and Ba(OH) 2 Li 2 CO 3 and Ca(NO 3 ) 2 Al 2 (SO 4 ) 3 and NH 4 OH H 3 PO Precipitation Reactions: Crash Course Chemistry #9 www.youtube.com/watch?v=IIu16dy3ThI 4 and Ba(OH) 2

Argyria: Silver ingestion

and NH 4 + ions and NH 4 + ions

Chemistry In Action:

An Undesirable Precipitation Reaction Ca 2+ (

) + 2HCO 3 -1 (

) CaCO 3 (

) + CO 2 (

) + H 2 O (

) Calcium Carbonate build up The fix? HCl reacts to form soluble CaCl 2 CaCO 3 (s) + 2HCl → CaCl 2 (aq) + H 2 O + CO 2 18

Properties of Acids

Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid.

Cause color changes in plant dyes.

React with certain metals to produce hydrogen gas.

2HCl (

) + Mg (

) MgCl 2 (

) + H 2 (

) React with carbonates and bicarbonates to produce carbon dioxide gas.

2HCl (

) + CaCO 3 (

) CaCl 2 (

) + CO 2 (

) + H 2 O (

) Aqueous acid solutions conduct electricity.

Properties of Bases

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Cause color changes in plant dyes.

Aqueous base solutions conduct electricity.

Examples: 20

Arrhenius

acid is a substance that produces H + (H 3 O + ) in water.

Arrhenius

base is a substance that produces OH in water.

Arrhenius Acid/Base definition is exclusive to aqueous solutions

Hydronium ion

, hydrated proton, H 3 O + Greater electron density H H O H Lower electron density

Brønsted acid

is a proton (H + ) donor A

Brønsted base

is a proton acceptor

“proton” - the H + ion consists of only a single proton Bronsted definition is not exclusive to aqueous solutions

base acid acid base A Brønsted acid must contain at least one

ionizable

Hydrogen!

Ionizable –

able to come off

Monoprotic

acids (1 H + ) HCl H + + Cl HNO 3 H + + NO 3 CH 3 COOH H + + CH 3 COO -

Diprotic

acids (2 H + ) H 2 SO 4 H + + HSO 4 HSO 4 H + + SO 4 2-

Triprotic

acids (3 H + ) H 3 PO 4 H 2 PO 4 HPO 4 2 H + + H 2 PO 4 H + H + + HPO 4 2 + PO 4 3 Strong electrolyte, strong acid Strong electrolyte, strong acid Weak electrolyte, weak acid Strong electrolyte, strong acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid 24

Strong Acid (HCl) Weak Acid (HF) Complete dissociation TedEd: The strengths and weaknesses of acids and base www.youtube.com/watch?v=DupXDD87oHc Partial dissociation

Generally follow the same solubility rules for other ionic compounds.

Completely soluble would indicate strong acid (unless organic acid) *

Most acids

tend to be weak with the short list of strong acids shown Organic acids possess the

Carboxylic acid

functional group and are

weak

H H C O C OH

Carboxylic Acid

Example Acid or Base

Classify each substance as a Brønsted acid or base in water: (a) HBr

Brønsted acid

(b)

Brønsted base

(c)

Acid & Base

(d) C 2 H 5 COOH

Propanoic Acid,

ionizable H + Organic acids will display the at the end (not grouped with other H’s) (e) Glucose (C 6 H 12 O 6 )

Neither

, Not an ionic compound nor is a H shown on the end

Neutralization Reaction

acid + base salt + water

*Know the products

HCl (

) + NaOH (

) NaCl (

) + H 2 O H + + Cl + Na H + + + + OH OH Na + + Cl + H 2 O H 2 O Net ionic equation If we start with equal molar amounts of acid and base we “neutralize” the solution and end up with water

Example Neutralization Reactions

Write molecular, ionic, and net ionic equations for each of the following acid-base reactions: (a) hydrobromic acid (

) + barium hydroxide (

) 2HBr(

) + Ba(OH) 2 (

) BaBr 2 (

) + 2H 2 O(

) (b) sulfuric acid (

) + potassium hydroxide (

) H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l)

Example Solutions

Solution

(a)Molecular equation: 2HBr (

) + Ba(OH) 2(

) → BaBr 2(

) + 2H 2 O (

) Ionic equation: 2H + (

) + 2Br − (

) + Ba 2+ (

) + 2OH − (

) → Ba 2+ (

) + 2Br − (

) + 2H 2 O (

) Net ionic equation: 2H + (

) + 2OH − (

) 2H 2 O (

) or H + (

) + OH − (

) → H 2 O (

) Both Ba 2+ and Br − are spectator ions.

Example Solutions

(b) Molecular equation: H 2 SO 4 (aq) + 2KOH(aq) K 2 SO 4 (aq) + 2H 2 O(l) Ionic equation: Net ionic equation: Note that because is a weak acid and does not ionize appreciably in water, the only spectator ion is K + .

Neutralization Reaction Producing a Gas

acid + Carbonate salt + water + CO 2 2HCl (

) + Na 2 CO 3 (

) 2NaCl (

) + H 2 CO 3 Carbonic acid degrades quickly 2HCl (

) + Na 2 CO 3 (

) 2NaCl (

) + H 2 O +CO 2 2H + + 2Cl + 2Na + + CO 3 2 2H + + CO 3 2 2Na + + 2Cl + H 2 O + CO 2 H 2 O + CO 2 CaCO 3 + 2HCl -> NaCl + CO 2 + H 2 O

Antacid

*Gastric juices include HCl

Acid-Base Properties of Water

Water is a weak electrolyte (1 H + in 10 7 H 2 O) H O H + H O H H 2 O (

) [ H O H H ] + H + (

) + OH (

) + H O -

Auto-ionization

Solution Is

neutral acidic basic [H + ] = [OH ] [H + ] > [OH ] [H + ] < [OH ] • Adding acid increases H + • Adding base increases OH 33

pH – A Measure of Acidity

pH describes the concentration of [H + ] on a

logarithmic scale.

pH = log [H + ] Remember: A log scale is

base 10

: pH 6 is 10x more acidic than pH 7 pH 5 is 100x more acidic than pH 7 pH 4 is 10 3 x more acidic than pH 7

Solution Is

neutral acidic basic [H + ] = [OH ] [H + ] > [OH ] [H + ] < [OH ] [H + ] = 1.0 x 10 -7 [H + ] > 1.0 x 10 -7 [H + ] < 1.0 x 10 -7 pH = 7 pH < 7 pH > 7

pH Meter H + is an electrolyte pH meters measure electrical conductance to determine concentration and then calculate pH

Acid-Base reactions are colorless and can be difficult to monitor.

pH Indicators

– substances that change colors at certain pH ranges and can be used for titrations The indicator changes color as pH changes

Phenolphthalein

Colorless in acid Pink in base

Chemistry In Action:

Antacids and the Stomach pH Balance NaHCO 3 (

) + HCl (

) NaCl (

) + H 2 O (

) + CO 2 (

) Mg(OH) 2 (

) + 2HCl (

) MgCl 2 (

) + 2H 2 O (

) Acid-Base Reactions in Solution: Crash Course Chemistry #8 www.youtube.com/watch?v=ANi709MYnWg

Reduction-Oxidation Reactions (Redox)

(electron transfer reactions) 2Mg + O 2 2MgO 2Mg 2Mg 2+ + 4e -

Oxidation

half-reaction (lose e ) O 2 + 4e 2O 2 2Mg + O 2 + 4e -

Reduction

half-reaction (gain e ) 2Mg 2+ + 2O 2 + 4e 39

Redox Reaction reactants

Its charge is “reduced” b/c it gained an e -

Neutral Zn is insoluble, but oxidized Zn +2 is soluble Neutral metals are insoluble, Ionized metals are soluble.

Write the Half-reactions of the given redox reaction.

0 +2 -2 Zn (

) + CuSO 4(

) +2 -2 0 ZnSO 4(

) + Cu (

) Zn (s) Zn 2+ + 2e Note: Lost e shown in products Zn is oxidized, lost electrons Zn is the

reducing agent

Cu 2+ + 2e Cu (s) Note: Gained e shown in reactants Cu 2+ is reduced, gained electrons Cu 2+ is the

oxidizing agent

Note: Sulfate (SO 4 -2 ) is neither oxidized or reduced so it is not shown It could be said that:“Zn

reduced

Cu” or “Cu

oxidized

Zn” Additional practice: 2Na (s) + Mg(NO 3 ) 2(aq) → Mg (s) + 2NaNO 3(aq) 4Fe + 3O 2 → 2Fe 2 O 3

Oxidation numbers (ON’s)

• Describes the charge of an atom, in a compound,

electrons were completely transferred between atoms (Both ionic and molecular compounds). • Used to predict how an atom will react with other atoms.

• For

ionic compounds

ON’s are equal to the ion charge

as determined by each atom’s group number • For

molecules

ON’s change

depending on what atoms they bond to and follow a set of rules.

Ex. Cl has an ON of -1 when bonded to a metal (NaCl), but not with non-metals. (Chlorate (ClO 3 ), Cl ON = +5)

*Note: An ON of +5 is not equal to a true ion of charge +5

Oxidation numbers (ON’s)

1. Free elements (uncombined state) and non-compounds have an oxidation number of Zero.

Na, Be, K, Pb, H 2 , O 2 , P 4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion.

Li + , Li = +1 ; Fe 3+ , Fe = +3 ; O 2 , O = -2 3. The oxidation number of oxygen is (In H 2 O 2 and O 2 2 it is –1 ).

usually -2

*you will not be given any of these exceptions on quizzes or tests.

4. Hydrogen: ON is

*Unless

bonded to metals in

binary

In these cases –1 compounds. (LiH, NaH, etc.).

*you will not be given any of these exceptions on quizzes or tests.

5. Alkali metals are +1 Alkaline Earth Metals are +2 Fluorine is always -1 . (Other Halogens can vary) 6. Sum of all ONs of each atoms in a compound/molecule (or ion) is equal to the net charge.

CaBr 2 +2 -1 Charge = 0 SO 4 -2 +6 -2 Charge = -2 ( +2  1) + ( -1  2) = 0 ( +6  1) + ( -2  4) = -2

Oxidation numbers of Transition metals

Alkali metals

always

have an oxidation number of +1 Alkaline Earth metals are always +2 • Transition metals (d group) can have several possible oxidation states that are stable Fe +3 Fe +2 can form FeCl 3 can form FeCl 2 Cu +2 Cu +1 can form CuCO 3 can form Cu 2 CO 3 • The oxidation state may change during a reaction.

K 2 Cr 2 O 7 is orange with Cr +6 Cr 2 (SO 4 ) 3 is blue with Cr +3

Redox Chemistry in Action: Breath Analyzer

+6 3CH 3 CH 2 OH + 2K 2 Cr 2 O 7 Ethanol + 8H 2 SO 4 Reduction of +3 dichromate to Cr +3 3CH 3 COOH + 2Cr 2 (SO 4 ) 3 + 2K 2 SO 4 + 11H 2 O Acetic acid 47

Assign oxidation numbers to

all the elements

in the following compounds and ion: +1 -2 (a) Li 2 O Oxygen is always -2 and Li is +1 in a compound (+1)•2 + (-2)•1 = 0 +1 +5 -2 (b) HNO 3 +6 -2 (c) Oxygen is always -2 and H is +1 in a compound.

(+1)•1 + (N) •1 + (-2)•3 = 0 ; N must = +5 Oxygen is always -2 and the sum must equal -2 (Cr)•2 + (-2)•7 = -2 ; Cr must = +6 +1 +6 -2 (d) HSO 4 -1 Oxygen is always -2 and H is +1 in a compound.

(+1)•1 + (S) •1 + (-2)•4 = -1 ; S must = +6 +2 +5 -2 (e) Ca 3 (PO 4 ) 2 Oxygen is always -2 and Ca is +2 in a compound.

(+2)•3 + (P) •2 + (-2)•8 = 0 ; P must = +5

The Oxidation Numbers of Elements in their Compounds Red # signifies most common oxidation state in compound 49

Types of Oxidation-Reduction Reactions

Combination Reaction A + B C 0 0 2Al + 3Br 2 +3 -1 2AlBr 3 Decomposition Reaction C A + B +1 +5 -2 2KClO 3 +1 -1 0 2KCl + 3O 2 50

Types of Oxidation-Reduction Reactions

Combustion Reaction A + O 2 B 0 0 S + O 2 +4 -2 SO 2 0 0 2Mg + O 2 +2 2MgO -2 Derailed sulfur train video 51

Types of Oxidation-Reduction Reactions

Displacement Reaction A + BC AC + B 0 +1 +2 Sr + 2H 2 O Sr(OH) 2 0 + H 2 +4 TiCl 4 0 0 +2 + 2Mg Ti + 2MgCl 2 Hydrogen Displacement Metal Displacement 0 Cl 2 -1 -1 0 + 2KBr 2KCl + Br 2 Halogen Displacement How do we know what can displace what?

Activity Series

The

Activity Series

for Metals

Shows the tendency for an atom to give away its electrons (reduce) another atom (which ones is on top reduces the other).

Al has greater reduction potential Hydrogen Displacement Reaction M + BC MC + H 2 M is metal BC is acid or H 2 O Ca + 2H 2 O Ca(OH) 2 + H 2 Ag + H 2 O AgOH + H 2 Metal Displacement Reaction Al + FeCl 3 Fe + AlCl 3 AlCl 3 FeCl 3 + Fe + Al 53

The Activity Series for Halogens F

> Cl

> Br

> I

2 I 2 can reduce F 2 , but F 2 can’t reduce I 2 Halogen Displacement Reaction 0 Cl 2 -1 -1 0 + 2KBr 2KCl + Br 2 I 2 + 2KBr 2KI + Br 2 54

Types of Oxidation-Reduction Reactions Disproportionation Reaction The same element is simultaneously oxidized and reduced.

reduced Example: 0 Cl 2 + 2OH +1 ClO -1 + Cl + H 2 O oxidized Elements most likely to disproportionate: each have at least 3 oxidation states

Redox Reactions Problems

Classify the following redox reactions and indicate changes in the oxidation numbers of the elements: (a) (b) (c) (d) (e) CH 4 (g) + O 2 (g) → CO 2 (g) + H 2 O(l) Redox Reactions: Crash Course Chemistry #10 www.youtube.com/watch?v=lQ6FBA1HM3s

Redox

Solutions

(a) This is a

decomposition

reaction because one reactant is converted to two different products. The oxidation number of N changes from +1 to 0, while that of O changes from −2 to 0.

(b) This is a

combination

reaction (two reactants form a single product). The oxidation number of Li changes from 0 to +1 while that of N changes from 0 to −3.

metal displacement

reaction. The Ni metal replaces (reduces) the Pb 2+ ion. The oxidation number of Ni increases from 0 to +2 while that of Pb decreases from +2 to 0.

(d) The oxidation number of N is +4 in NO 2 and it is +3 in HNO 2 +5 in HNO 3 . Because the oxidation number of the

same

element both increases and decreases, this is a

disproportionation

and reaction.

(e)

Combination

and

combustion

Summary of Reactions Types

Precipitation: Two aqueous solutions mixing to forming a solid product.

Na 2 CO 3 (aq) + Ca(NO 3 ) 2 (aq) → 2NaNO 3(aq) + CaCO 3 (s) Acid-Base Reactions: Transfer of H Neutralization: Reaction of Acid & + and/or OH Base ions in H 2 O solution.

to form a salt and water 2 H NO 3 + Ba( OH ) 2 → Ba(NO 3 ) 2 + 2H 2 O (l) Acid + Metal : Oxidize to dissolve metal 2 2 H H Cl + + (aq) + Mg Mg (s) → MgCl 2 → Mg +2 (aq) and produce Hydrogen gas + + H H 2 2(g) (net ionic equation) Redox: Reaction involving the

transfer of electrons

between atoms Combination: Combustion : 0 0 4Fe + 3 O 2 +3 → 2Fe 2 -2 O 3 Decomposition: CaCO 0 Displacement: Na (s) 3 +2 -2 +4 -2 → CaO + CO 2 +1 -1 + AgCl (aq) +1 -1 → NaCl (aq) 0 + Ag (s)

Collision Theory

of Chemical Reactions

• The rate of a reaction increases as the number of molecular collisions increases.

• Ways to increase collisions: – Increase dynamic interactions (

Liquid

> Gas > Solid) – Increase Concentrations (closer proximity) – Increase Temperatures (faster movements) • Any molecule in motion possesses Kinetic energy, the faster it moves, the greater KE • High energy collisions are needed to break original bonds (E a :Activation energy) Ted-Ed: How to speed up chemical reactions (and get a date) https://www.youtube.com/watch?v=OttRV5ykP7A

Disposing of excess WWII Sodium metal in 1947

2Na + 2H 2 O → 2NaOH + H 2(g) 60

Solution Stoichiometry

The

concentration

of a solution is the amount of solute present in a given quantity of solvent or solution.

molarity

= moles of solute liters of solution 61

Measures of Concentration

The

concentration

of a solution is the amount of solute present in a given quantity of solvent or solution.

Molarity (M)

moles of solute

= liters of solution

Includes solute volume Because density (volume) can change with temperature it is helpful to express solvent by mass when sample undergoes temperature changes

Molality (m)

moles of solute

= mass of solvent (kg)

Excludes solute mass No volumetric measurements needed; all mass

(grams)

= moles of solute liters of solution (mL) moles = Liters x Molarity

= moles of solute Molarity 63

Problem

Calculate the Molarity and Molality of a H 2 SO 4 solution containing 24.4 g of sulfuric acid in 198 g of water at 50 °C.

= moles of solute liters of solution H 2 O at 70 °C, 0.97 g/mL

= 0.249 moles 0.204 Liters of solution = 1.22 M H 2 SO 4

Molarity/Molality Problem

The definition of molality (

) is The mass of water is 198 g, or 0.198 kg. Therefore, Molarity will equal Molality at 4 °C, but Molarity will decrease as temperature goes up and density goes down

Preparing a Solution of Known Concentration from solids Volumetric Flask Mix till dissolved Bring to desired volume

Molarity Problem

What is the Molar concentration of the Sodium ion [Na + ] when 23.4 g of NaCl and 34.1 g of Na 2 O are dissolved in 0.60 L H 2 O?

23.4 g NaCl 1 mol 58.5 g NaCl = 0.40 mol NaCl = 0.40 mol Na + 34.1 g Na 2 O 1 mol 62.8 g Na 2 O = 0.54 mol Na 2 O = 1.08 mol Na + (doubles from subscript) 0.40 mol Na + + 1.08 mol Na + = 1.48 mol Na +

= 1.48 mol 0.60 L = 2.5 M Na + 67

Molarity Problem

How many grams of potassium dichromate (K 2 Cr 2 O 7 ) are required to prepare a 250-mL solution whose concentration is 2.16

mol

mol

 L L

Dilution

is the procedure for preparing a less concentrated solution from a more concentrated solution (

stock

Dilution Add Solvent Moles of solute before dilution (1)

M 1

V 1

=

Moles of solute after dilution (2)

2 V 2 Number of moles

does not

change 69

Dilution Practice

Describe

how you would prepare 500 mL of a 1.75

H

SO

solution, starting with an 8.61

stock solution of H

SO

.

Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same.

M

V

= M

V

Dilution Solution

Solution

We prepare for the calculation by tabulating our data:

1 = 8.61

1 = ?

M M

2 = 1.75

= 500 mL Thus, we must

dilute 102 mL of the 8.61 M H 2 SO 4 water to give a final volume of 500 mL solution with

Dilution Problem

Describe

how you would prepare 300 mL of a 0.4

H

PO

solution, starting with an 1.5

stock solution of H

PO

.

M

V

= M

V

Bell Ringer

You have 250 mL of a 3.0

Ba(OH)

solution. What is the concentration if we

add

150 mL of water to the solution?

M 1 = 3.0 M V 1 = 250 mL M 2 V 2 = ?

= 250 mL + 150 mL = 400 mL Crash Course: Water and Solutions for Dirty Laundry www.youtube.com/watch?v=AN4KifV12DA

Bell Ringer

1) How many

grams

of solid NaNO 3 125 mL of a 0.85 M NaNO 3 are needed to produce solution? 2) What is the Molar concentration of the Sodium ion [Na + ] when 2.8 g of Na 3 PO 4 85 mL?

and 4.5 g of Na 2 CO 3 are dissolved in 3) How would you prepare 250 mL of a 0.65 M H 2 SO 4 solution from a stock of 6.5 M H 2 SO 4 solution?

4) You have 50 mL of 6.0 M NaF and 450 mL water are

added

, what is the new Molarity?

Titrations

In a

titration,

a solution of accurately

known

concentration is added gradually to another solution of

unknown

concentration until the chemical reaction between the two solutions is

complete

Standard solution

– solution with

known concentration

to be precisely added for comparison

Equivalence point

– the point at which the

reaction is complete

example) 1 mol H 2 SO 4  2 mol NaOH (

obtained from balanced equation

) 75

Titrations

Indicator

– substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 76

Titrations can be used in the analysis of: Acid-base reactions

H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4

Redox reactions

5Fe 2+ + MnO 4 + 8H Mn 2+ + + 5Fe 3+ + 4H 2 O *Can involve color changes without indicator 77

Titration Steps

1) Write Balanced equation for stoichiometry (mole-to-mole ratio) 2) Determine moles of Standard solution used.

(mol = M x L) 3) With stoichiometry, convert moles standard (train tracks) to moles unknown 4) Determine unknown Molarity using given volume (L) (M = mol/L) Alternative Equation:

M

s 

V

= M

u 

V

u Coefficient

Coefficient

Titration Problem #1

It takes 32 mL of 2.0M HCl standard to neutralize a 500. mL solution of Ba(OH) 2 . What is the concentration of Ba(OH) 2 ? 1) 1 Ba(OH) 2 + 2 HCl → BaCl 2 + 2H 2 O (Reacts 1:2) 2) mol HCl = 2.0M x 0.032 L = 0.064 mol HCl 3) 0.064 mol HCl 1 mol Ba(OH) 2 = 0.032 mol Ba(OH) 2 2 mol HCl 4) M Ba(OH) 2 = 0.032 moles 0.500 L Alternative: M H  V H = M OH  V OH = 0.064 M Ba(OH) 2.0

 32 = M OH  500 2 Coefficient # Coefficient # 2 1 M OH = 0.064

Titration Problem #2

How many milliliters (mL) of a 0.610

M

NaOH solution are needed to neutralize 20.0 mL of a 0.245

M

H

SO

solution?

2 NaOH + H 2 SO 4 2 H 2 O + Na 2 SO 4 1) For every 2 moles base added, it neutralizes 1 mole acid

Titration #2 Solution

2) Next we calculate the number of moles of H 2 SO 4 20.0 mL solution: Moles = M x L in a 0.245 M x 0.0200 L = 0.00490 mol H 2 SO 4 3) From the Balanced Equation: 1 mol H 2 SO 4  2 mol NaOH. 4.9 x 10 -3 mol H 2 SO 4 2 mol NaOH 1 mol H 2 SO 4 = 9.80 × 10 -3 mol NaOH 4) L = 9.80 x 10 -3 mol 0.61 M NaOH = 0.0161 L or 16.1 mL NaOH

Titration: Finding the Molar Mass of an Unknown

Lauric Acid

is a short-chain fatty acid that is solid at room temperature and

monoprotic

. We dissolve 0.022 grams into 500. mL of water and then titrate it with 0.010 M NaOH. If it takes 11.0 mL of NaOH to neutralize the fatty acid, what is the Molar Mass of Lauric Acid?

•

Molar mass

has the units

grams/mole

. We weighed out the mass of the solid acid in grams. Titration can tell us how many moles of acid are present in the same sample.

0.022 grams Moles H + = ? Molar mass • Because it is

monoprotic

, it will

react 1:1

with NaOH.

Molar Mass Titration Solution

1) Given information states it reacts 1:1 2) (0.010 M NaOH) x (0.0110 L) = 1.1 x 10 -4 mol NaOH 3) 1.1 x 10 -4 mol NaOH 1 mol Lauric acid 1 mol NaOH = 1.1 x 10 -4 mol Lauric Acid 4) Molar Mass = 0.022 grams 1.1 x 10 -4 moles C 11 H 23 COOH = 200 g/mol Large component of coconut oil ~ 3-6% of milk

Example: Redox Titration

A 16.42-mL volume of 0.1327

KMnO 4 solution is needed to oxidize 25.00 mL of a FeSO 4 solution in an acidic medium. What is the concentration of the FeSO 4 solution in molarity? The net ionic equation is want to calculate need to find given

Redox Titration Solution

Solution

The number of moles of KMnO 4 (in 16.42 mL) = M x L (0.1327 M KMnO 4 ) x (0.01642 L) = 2.179 x 10 -3 mol From the net ionic equation we see that 5 mol Fe 2+  1 mol MnO 4 -

= moles of solute liters of solution = 1.090 x 10 -2 mol = 0.436 M FeSO 4 0.025 L

Chemistry in Action: Metals from the Sea

Many metals are found in the earth’s crust, but it is

cheaper

to “mine” from the sea CaCO 3 (

) CaO (

) + CO 2 Precipitation (

) CaO (

) + H 2 O (

) Ca 2+ Slightly soluble (

) + 2OH Mg 2+ (

) + 2OH (

) Mg(OH) 2 Precipitation (

aq (s

) ) Mg(OH) 2 (

) + 2HCl (

) MgCl 2 (

) + 2H 2 O (

) 1.3 g of Magnesium/ Kg seawater Electrolysis of MgCl 2 ( redox ) Mg 2+ + 2e Mg MgCl 2 2Cl (

) Cl 2 + 2e Mg (

) + Cl 2 (

) 86



Gravimetric Analysis

Analytical technique based on the

measurement of mass

•

Precipitation

: the

analyte

is precipitated out of solution by adding another reagent to make it insoluble. Then filtered and weighed.

•

Volatilization

: the

analyte

is converted to a gas and removed. The loss of mass from the starting material indicates the mass of gas.

Gravimetric Analysis 1. Dissolve unknown substance in water (if not already dissolved) 2. React unknown with

precipitating reagent

to form a solid precipitate

Reagent: chemical added to another substance to bring about a change

3. Filter, dry, and weigh precipitate.

4. Use chemical formula and mass of precipitate to determine amount of unknown

analyte

(chemical of interest in experiment) 88

Gravimetric Analysis Problem #1

A sample of an unknown soluble compound contains Ba +2 and is dissolved in water and treated with excess sodium phosphate. If 0.411 g of Barium phosphate precipitates out of solution, what mass of Barium in found in the unknown compound?

We need to find 1 st find the Mass % of the

analyte

compound using their respective molar masses in the precipitated 3 x 137.3 g Ba +2 601.9 g Ba 3 (PO 4 ) 2 x 100% = 68.4% Ba in Ba 3 (PO 4 ) 2 0.684 x 0.411 g = 0.253 g Ba +2 *It is

not

mandatory to convert to percentage form. The

mass fraction

can be used directly. 411.9 g Ba +2 601.9 g Ba 3 (PO 4 ) 2 x 0.411g Ba 3 (PO 4 ) 2 = 0.253 g Ba +2

Gravimetric Analysis Problem #2

We have 250 mL of a Copper (Cu +1 ) aqueous solution. We add excess sodium carbonate (Na 2 CO 3 ) to precipitate out 3.8 g of Cu 2 CO 3 . What is the [Cu +1 ] Molarity of the solution?

Since we need to find moles of Cu +1 , it will be quicker to

use train-tracks

instead of % composition (either would work). 3.8 g Cu 2 CO 3 1 mol Cu 2 CO 3 2 mol Cu +1 187.0 g Cu 2 CO 3 1 mol Cu 2 CO 3 = 0.041 mole Cu +1

= 0.041 mol Cu + = 0.16 M Cu +1 0.250 L

Gravimetric Analysis Problem #3

A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO 3 . If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound?

35.45 g Cl 143.4 g AgCl x 1.0882 g AgCl = 0.269 grams Cl

H

O reactions Exam Review Problems

1)Write the net ionic equation for: (NH 4 ) 2 S and MgSO 4 2)Write the

labeled

half reactions: 2N 2 O → 2N 2 + O 2 3)How many grams NH 4 NO 3 needed for 250 mL of 3.5M?

4) Describe preparation of 300 mL 0.5M HCl (5M stock) 5)120 mL of H 3 PO 4 is titrated with 35 mL 2.4 M standard KOH, what is concentration of H 3 PO 4 ? 6) 53.87 g mixture contains a quantity of lead (Pb). When dissolved and treated with excess NaCl a precipitate of PbCl 2 is formed. What is the %Pb in the original unknown if 45 grams PbCl 2 is collected?