Chapter 4. Aqueous reactions and Solution Chemistry.

Chemical reactions in water are important, as the chemistry of life occurs in water, and chemical reactions in the environment occur mainly in water.

δ+

δ+δ -

partial charges on awater molecule

4.1. General Properties of Aqueous Solutions.

Electrolytic Properties:

NaCl dissolved in water produces a solution that conducts electricity, whereas table sugar (sucrose, C12H22O11) does not. This can be demonstrated with an electric light-bulb that lights up only with ionic solutions.

Demonstration of electrical conductivity of an electrolyte:

bulb glowsin presenceof electrolytebut not ofpure water

Circuit passes through electrodesin solution in beaker

ElectrolyteSolutionIn beaker

How circuit on previous slide works:

Distribution of charge on the water molecule:

The water molecule is polar, which in the case of water means that the hydrogens have partial positive charges, while the oxygen has a partial negative charge:

δ+

δ+

δ-

The partial positive charges (δ+) are attracted to negativecharges, while the partial negative charges(δ-) are attracted topositive charges

H

H

O

The structure of water around anions and cations: Each anion is surrounded by partial positive

charges from waters, and each cation is surrounded by partial negative charges from waters:

δ+

δ-

δ-

δ-

δ+ δ+

δ+δ+

δ-

δ-

δ-

δ+ δ+

δ+

δ+anion

cation

Ionic CompoundsWhen ionic compounds dissolve in water, they tend to dissociate completely:

--

-

--+

+

+

+

-

-

+

++ H2O

NaCl (s) → Na+ (aq) + Cl- (aq) solid cation anion

Na+ cation

Cl- anion

watermolecules

Molecular Compounds

Most molecular compounds do not dissociate when they dissolve in water:

+ H2O

Methanol(molecular compound)

methanol dissolved in water

Methanol molecule not ionized

water

Molecular Compounds that ionize in water:

Some molecular compounds dissociate (ionize) in water (mainly acids)

HCl (aq) → H+ (aq) +Cl- (aq)

Weak acids, such as acetic acid, dissociate only partially (double arrow shows equilibrium does not lie completely to the right):

HC2H3O2 (aq) H+ (aq) + C2H3O2- (aq)

acetic acid acetate ion

Strong acids, such as hydrochloric acid,dissociate completely (single arrow shows equilibrium lies well to the right):

Strong and Weak electrolytes:

Strong electrolytes are completely ionized in water. Weak electrolyes only partially ionize in water. A prime example of a weak electrolyte is acetic acid. The fact that it only ionizes to a small extent is indicated by a double arrow:

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

A single arrow is used to indicate complete ionization of strong electrolytes, e.g. NaOH:

NaOH (aq) → Na+ (aq) + OH- (aq)

4.2 Precipitation reactions:A precipitate is an insoluble solid formed by a reaction in solution.

Bottom ofTest-tube

yellow precipitateof silver iodide (AgI)

mix clear solutions of AgNO3 andNaI to producea precipitate of AgI(s)

Soluble Ionic Compounds:

Compds containing: Exceptions

NO3- none

C2H3O2- none

Cl- Ag+, Hg22+, Pb2+

Br- “

I- “

SO42- Sr2+, Ba2+, Hg2

2+, Pb2+

(note: you will be given the information on this sheet in exams/tests)

Insoluble Ionic Compounds:

Compds containing exceptions

S2- alkali metal cations, NH4+,

Ca2+, Sr2+, Ba2+

OH- “

CO32- alkali metal cations, NH4

+

PO43- “

(note: you will be given the information on this sheet in exams/tests)

To use the Tables on the two preceding slides, consider the following examples:

Which of the following compounds are insoluble in water?

a) BaSO4 b) K2SO4 c) PbCl2 d) (NH4)2CO3

a) If we look at the tables, we see that all sulfates are soluble except Ba2+ (plus others) insoluble

b) Tables show all sulfates are soluble, and alkali metal ions are not exceptions, so is soluble

c) All chlorides are soluble, except Pb2+ (plus others)

so is insoluble

d) All carbonates are insoluble, except NH4+ (ammonium

ion) plus others, so is soluble

Ionic EquationsMolecular equation: (do not show as ions)

Pb(NO3)2 + 2 KI → PbI2(s) + 2 KNO3

Complete Ionic equation (show all ions):

Pb2+ + 2 NO3- + 2 K+ + 2 I- →

PbI2(s) + 2 K+ + 2 NO3-

Net ionic equation (omit ions that do not change during the reaction, NO3

- and K+):

Pb2+(aq) + 2 I-(aq) → PbI2(s)

Ca(NO3)

2 (aq) + NaC

2H

3O

2 (aq) →

?

Spectator ions: ALL !

If all salts are soluble, no precipitation reaction will take place

Note:

Ca(NO3)

2 (aq) + Na2CO

3 (aq) →

CaCO3(s) + 2 NaNO3(aq)

Write out the complete ionic equation and then remove spectator ions (ions that do not change):

Ca2+ + 2 NO3- + 2 Na+ + CO3

2- → CaCO3(s) + 2 Na+ + 2 NO3

-

Ca2+ (aq) + CO3

2- (aq) → CaCO3 (s)

Spectator ions: 2 Na+, 2 NO3

-

Net ionic equation:

Example: What is the net ionic reaction that corresponds to the molecular reaction below?

4.3 Acid-Base Reactions.Acids are substances that ionize in aqueous solutions to form protons, i.e. H+(aq) ions.HCl, HNO3 are monoprotic acids. i.e. they ionize to give only one proton per acid molecule:

HNO3(aq) H+(aq) + NO3- (aq)

H2SO4 is a diprotic acid, and gives two protons.

H2SO4(aq) 2 H+(aq) + SO42-(aq)

(Note that for H2SO4 only the first ionization is complete.)

Bases:Bases are substances that accept protons. A base is a proton acceptor. Common bases are OH- compounds of group 1 cations (NaOH, KOH) or heavier Group 2 (Ba(OH)2). These produce the hydroxide (OH-(aq)) ion in solution.

NaOH (aq) Na+ (aq) + OH- (aq)

bases will react with a proton (this is the net ionic reaction below):

H+ (aq) + OH- (aq) H2O (l)acid base water

Ammonia – a weak baseMany bases do not contain OH-, such as NH3 (ammonia).

NH3(aq) + H+(aq) NH4+(aq) + H2O

or NH3(aq) + H2O NH4+(aq) + OH-(aq)

ammonia ammonium

The latter equilibrium lies well to the left, so

NH3 is a weak base. A solution of NH3 in water contains only a tiny fraction as NH4

+ ions (< 1%).

bigger arrow shows equilibriumlies to the left

The difference between ammonia and ammonium:

To remember the difference between ammonia and ammonium, remember that ammonium is a cation like sodium or potassium.

NH

H

H NH

H

H

H

+ H+

+

ammonia ammonium

NH3 (neutral NH4+ (a cation)

molecule)

Strong and Weak Acids and Bases.

Acids and bases that are STRONG ELECTROLYTES. i.e. completely ionize in aqueous solution, are strong acids and bases. Those that are WEAK ELECTROLYTES i.e. only partly ionize, are weak acids and bases. For example, HF is a weak acid, since it is only partly ionized. However, it is very reactive since the F-(aq) ion will attack glass!

Common Strong Acids and Bases (Table 4.2)

Acids:

HCl, HBr, HI, HClO3, HClO4, HNO3, H2SO4.

Bases:

Group 1. LiOH, NaOH, KOH, RbOH, CsOH

Heavy group 2: Ca(OH)2, Sr(OH)2, Ba(OH)2.

Weak acids and bases:

If not listed as strong, it is probably weak. Some examples:

Weak acids:

HF, H3PO4, H2SO3, HC2H3O2,

(hydrofluoric, phosphoric, sulfurous, acetic acids)

Weak bases:NH3 (ammonia)

Neutralization reactions: (p 134)

Acids = sour taste (e.g. vinegar), Bases = bitter (e.g. soap).

HCl(aq) + NaOH(aq) → NaCl (aq) + H2Oacid base salt water

Net ionic reaction:

H+ (aq) + OH- (aq) H2O acid base water

The Na+ and Cl- arespectator ions

Balancing acid-base reactions

Note: to balance these equations, just make the OH- and H+ balance out

2 HCl(aq) + Ca(OH)2(aq) → CaCl2 (aq) + 2 H2O

3 HCl(aq) + Al(OH)3(aq) → AlCl3 (aq) +

3 H2O

Two protons needed for two hydroxides

4.5. Concentration of Solutions

Molarity. This is used to express the concentrations of solutions:

Molarity = moles solute_______

Volume of solution in liters

Units = mol/liter (moles per liter)

A 1.00 molar (written 1.00 M) solution has one mol of substance dissolved in every

liter of solution:

1 liter flask 1 mole 1.00 M

(empty) CuSO4 (159.5 g) CuSO4

Dissolve the1 mole of CuSO4 in the flask

fill theflask tothe 1 litermark withwaterline on

neck offlask shows1 liter mark

stopper

Examples:What strength solution do we obtain if we dissolve:

a) 0.25 mol CuSO4 in a 250 ml flask:

Conv.: 1 liter = 1000 ml

250 ml = 250 ml x 1liter = 0.25 liter

1000 ml

Molarity = ____ moles of solute_____

volume of solution in liters

= _______0.25 moles_______

0.25 liters

= 1.00 M (or moles/liter)

Examples:

What molarity solution do we obtain if we dissolve:

b) 4.82 g CuSO4 in a 200 ml flask:Note: 200 ml = 0.200 liter.

Molecular mass CuSO4 = 159.5 g/mol

moles CuSO4 = 4.82 g x 1mole = 0.0302 moles 159.5 g

molarity = 0.0320 moles = 0.151 M 0.200 liter

Expressing the concentration of an Electrolyte:

A 1.00 M solution of NaOH is also 1.00 M in Na+ ions and OH- ions.

NaOH (aq) Na+ (aq) + OH- (aq)

1 molar 1 molar 1 molar

A 1.00 M solution of H2SO4 is 1.00 M in SO42-, but 2.00

M in H+.

H2SO4 (aq) 2 H+ (aq) + SO42- (aq)

1 molar 2 molar 1 molar

Just pay attention to the coefficents in the balanced equation

A 1.00 M solution of Na3PO4 is 1.00 M in PO43-,

but 3.00 M in Na+. What about a 0.1 M solution?

Na3PO4 (aq) 3 Na+ (aq) + PO43- (aq)

1 3 11 M 3 M 1 M

or 0.1 M 0.3 M 0.1 Mcoefficients in balanced equation

You notice again the power of the balanced equation. You can multiply all the coefficients by 1.0 M or 0.1 M.

Interconverting Molarity, Moles, and Volume:

If we know any two of the above, we can calculate the third quantity.

e.g. calculate no. of moles HNO3 in 2.0 L of 0.200 M HNO3. Molarity = moles/liters

0.200 M = x moles/2.0 L

x moles = 0.200 mol x 2.0 liter 1 liter

= 0.4 moles

Sample exercise:

How many grams of Na2SO4 are required to make 0.35 L of 0.500 M Na2SO4?Moles = 0.35 L x 0.500 moles/liter = 0.175 moles.

O.175 moles Na2SO4 (MM = 142.0 g/mol)Grams = 0.175 moles x 142.0 g/mol

= 24.9 g

Dilution (p. 148).

When we dilute something, we take a volume of solution and place it in a larger container and make it more dilute by adding water:

1 liter flask 2 liter flask 2 liters of solution

1 M CuSO4

solution

transfersolutionto larger flask

make upto mark with water

0.5 MCuSO4

solution

Dilution (p. 148).

Molarity = moles liter

So moles = molarity x litersIn diluting a solution, the number of

moles of dissolved substance stays the same. Moles before dilution = moles after dilution, so:

before dilution after dilution

M(conc) x V(conc) = M(dil) x V(dil)

Exercise 4.14:How many milliliters of 3.0 M H2SO4 are needed to make 450 ml of 0.10 M H2SO4?

3.0 M x V(conc) = 0.10 M x 450 ml

V(conc) = 0.10 M x 450 ml 3.0 M

= 15.0 ml

Note: make sure units of volume are the same on both sides: liters and liters, or ml and ml.

M(conc) X V(conc) = M(dil) x V(dil)

Problem:

What volume of 2.50 M Pb(NO3)2 contains 0.0500 mol of Pb?

M = moles/volume

So volume = moles/M

= 0.0500 mol/2.50 M

= 0.0500 mol x 1 liter

2.50 mol

= 0.02 L or 20 ml

4.6. Solution stoichiometry.

Here we need to be able to convert from grams to moles, and then use M = moles/volume to convert from moles to molarity. Thus for example:

79.8 g of CuSO4 1 liter of 0.5 M CuSO4

0.5 mol

0.5 mol x 1 liter 1 liter

79.8 g x 1 mole 159.5 g

Problem:

How many grams of Ca(OH)2 are needed to neutralize 25.0 ml of 0.100 M HCl?

Balanced equation:

2HCl + Ca(OH)2 → CaCl2 + 2 H2O

2 moles 1 mole 2 moles 2 moles

so will need 1 mole calcium hydroxide to neutralize 2 moles HCl.

Moles HCl = 0.100 M x 0.025 L = 0.0025 moles

2HCl + Ca(OH)2 → CaCl2 + 2 H2O2 moles 1 mole 2 moles 2 moles

0.0025 moles

Factor = moles we have/moles in equation= 0.0025/2 = 0.00125

So moles of Ca(OH)2 = 0.00125 x 1 = 0.00125 mol

Convert moles to grams:

Mol. Mass Ca(OH)2 = 74.1 g/mol

Therefore grams = 74.1 g x 0.00125 mol

1 mole

= 0.0926 g

Solution Stoichiometry and Chemical Analysis

What volume of a 0.30 M HCl solution is needed to completely react 3.5 g of Ca(OH)2 ?

2 HCl (aq) + Ca(OH)2 (aq) → 2 H2O (aq) + CaCl2 (aq) 2 1

The question is how many moles of Ca(OH)2 do we have, and how many moles of HCl will we need to react with it?

No. of moles Ca(OH)2 = 3.5 g x 1 mole = 0.035 mol 100.0 g

From the balanced equation we see that we need 2 mols of HCl for each mol Ca(OH)2:

2 HCl (aq) + Ca(OH)2 (aq) → 2 H2O (aq) + CaCl2 (aq)

2 1

0.070 mol 0.035 mol

Problem then is what volume of 0.30 M HCl will provide 0.070 mol?

Molarity = moles/volume

0.30 M = 0.070

volume

Volume = 0.070 mol x 1 liter = 0.23 L

0.30 mol or 230 mL

What is the concentration of a solution that is made by adding 0.3L of water to 15mL of a 0.65M solution?

M conc × V conc = M dil × V dil

Vconc = 15mL = 0.015LMconc= 0.65M

Vdil = 0.3L + 15mL

= 0.3L + 0.015L = 0.315L

dil

concconcdil V

VMM

LLM

M dil 315.0015.065.0 M031.0

rearrange