9 Aqueous Solutions

Water as a Polar Solvent

The water molecule is polar. That is it has a slightly positive end and a slightly negative end.

HH

-

+O

+

+

Oxygen attracts electrons and becomes slightly negatively charged the hydrogen atoms become slightly positively charged

HH

-

++O

The shape of the molecule is important to its polarity. It must have a positive and negative end.

When a polar substance (e.g. an ionic compound) is dissolved in water the ions are attracted to the water molecules

HH

-

++O

Na+

Cl-

Na+

Na+

Cl-

Cl-

Cl-

Water molecules separate, surround and disperse the ions into the liquid.

Some ionic compounds are only slightly (sparingly) soluble in water. e.g. silver chloride. This is because the electrostatic attractions within the compound are greater than the attraction between the ions and the water molecules

All ionic compounds will dissolve to a certain extent even

though we call them insolubleSolubility of NaCl in water at 20oC = 365 g/L Solubility of AgCl in water at 20oC = 0.009 g/L

Ethanol – a polar molecule

C CH

H

H

H

H

H

O+

Polar substances will dissolve in ethanol which has a polar and a non-polar part. Water and ethanol are therefore miscible. Note the ethanol and water do not dissociate into ions!

HH

-

O-

+

+

-

AcidsAcids are an important group of covalent

compounds that dissolve in water.

e.g. HCl H2SO4 HNO3

These all interact so strongly with water that the molecules dissociate into ions

HCl H+ + Cl-

In fact we do not get a lone H+. The H+. Is attracted to the water molecule to give H3O+

So we should really write the equation as follows

HCl + H2O H3O+ + Cl-

HH

-

++O

H+

..

Writing Ionic equationsA molecular equation

AgNO3(aq) + NaCl (aq) AgCl (s) + NaNO3(aq)

Total Ionic equation

Ag+(aq) + NO3

-(aq + Na+

(aq) + Cl-(aq) AgCl (s)+ NaNO3(aq)

Sodium nitrate is soluble in water so no reaction occurs between sodium and the nitrate ions. The ions remain dissociated in solution as solvated ions. These ions are called spectator ions.

Silver chloride is insoluble so reaction occurs between silver ions and chloride ions

the net ionic equation shows the actual chemical change taking place

Ag+(aq) + Cl-(aq)

AgCl (s)

Solubility

See page 139 for solubility of compounds in water

In order to predict whether a precipitate occurs we need to know the solubility of that compound in water

Write ionic equations including state symbols for the following reactions

Barium nitrate with sodium carbonate

Sodium chloride with lead (II) nitrate

Aluminium nitrate with sodium phosphate

Precipitation Reactions

In precipitation reactions two soluble compounds react to form an insoluble product, a precipitate. Precipitates form because the electrostatic attraction between the ions outweighs the tendency for the ions to become solvated (surrounded by the solvent molecules) and move randomly through the solution.

You can predict whether a precipitate occurs by considering the solubility of the products.

In the reaction between barium nitrate and sodium carbonate an insoluble precipitate of barium carbonate forms.

Acid-Base Reactions

An acid is a substance that produces H+ ions when dissolved in water

H2O

HX H+(aq) + X-

(aq)

A base is a substance that produces OH- ions when dissolved in water

H2O

MOH M+(aq) + OH-

(aq)

Strength of an acid or base

Strong acids or bases dissociate completely into ions when they dissolve in water

Weak acids or bases dissociate so little that most of their molecules remain intact

Strong acids and bases are therefore electrolytes (conduct a current)

Weak acids and bases are very weak electrolytes

For a strong acid or base dissociation into ions is virtually 100% so the conc.

If we dissolve 1mol of solid sodium hydroxide in water we will get 1mol of OH- ions

NaOH Na+ + OH-

1mol 1mol 1mol

This is not true of a weak acid or base

As noted earlier an acid produces H+ ions

when dissolved in water

So for example HCl is a covalent gas and does not behave as an acid. Only when it is dissolved in water does it dissociate into ions.

HCl(g) + H2O(l) H3O+(aq) + Cl-(aq)

Writing the net ionic equation for an acid base reaction

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

NaCl will be dissociated but water will be undissociated

The net ionic equation is

H+ (aq) + OH-(aq) H2O(l)

Strong acids Most inorganic acids e.g.

sulphuric, nitric, hydrochloric

Strong bases sodium hydroxide, calcium hydroxide

Weak acids ALL organic acids e.g. acetic (ethanoic acid)

Weak bases ammonia

Ammonia

Not all bases actually contain OH-.Ammonia NH3 is a covalent gas that reacts with water to produce a basic solution. It is a weak base i.e. the following reaction does not go to completion but is in equilibrium. Both forward and backward reactions are occuring at the same rate.

NH3 + H2O NH4+ + OH-

Ammonia removes a proton from water to leave OH- ion

Because it is a weak base and doesn’t dissociate fully in water the no of moles of OH- produced will be much less than the actual concentration of the ammonia solution.

This is also true of weak acids.

The pH scale

pH is the negative log of H+ concentration

pH = - log [H+]

for a 0.1molar solution of HCl (a strong acid)

pH = -log 0.1 = 1

What is the pH of a 0.01 mol solution of HCl?

-log 0.01 = 2

What is the pH of a 0.1 molar solution of sulphuric acid?

H2SO4 2H+ + SO42-

[H+] = 2 x 0.1 = 0.2 mol/l

And pH = -log 0.2 = 0.7

What is the pH of the following?• 0.005 mol/l HNO3

• 0.05 mol/l H3PO4

• [H+] = 0.005 mol/L pH = -log 0.005pH = 2.3

b) [H+] = 3 x 0.05 mol/L pH = -log 0.15pH = 0.82

Finding pH from [H+]What is the [H+] of a solution of HCl with a pH of 3?

Antilog -3 = 0.001[H] = 0.001

What is the [HCl]?

From the equation HCl H+ + Cl- we can see that the ratio of [H+] to [HCl] = 1:1

[HCl] = 0.001 mol/L

What is the [H+] of a solution of H2SO4 with a pH of 2.4?

Antilog -2.4 = 0.004 mol/L

[H+] = 0.004 mol/L

H2SO4 2H+ + SO42-

[H2SO4] = 0.004/2 = 0.002 mol/L

What is the [H2SO4]?

Finding pH from [OH-]

pOH = -log[OH-]

and

pH +pOH = 14

What is the pH of a 0.1 mol/L solution of sodium hydroxide?

pOH = -log 0.1 = 1 pH = 14 – 1 = 13

Find pH of the following solutions

a) 0.001 mol/l potassium hydroxide

b) 0.05 mol/L calcium hydroxide

a) pOH = -log 0.001 = 3 pH = 14 -3 = 11

b) [OH-] = 2 x 0.05 pOH = -log 0.1 = 1pH = 14 -1 = 13

Calculate the pH of of a solution made by dissolving 1.00g of calcium hydroxide in 500ml of water.

RMM Ca(OH)2 = 40 + (2 x 16) + (2 x 1) = 74

1.00g = 1.00 = 0.0135mol 74

Moles [OH-] = 2 x 0.0135 = 0.027 Moles per litre = 2 x 0.027 = 0.054

pOH = -log 0.054 = 1.27 pH = 14 – 1.27 = 12.3