9 Aqueous Solutions
Transcript of 9 Aqueous Solutions
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