Lecture 4: Aqueous solution chemistry
Lecture 4 Topics
1. Solutes & solvents
Brown, chapter 4
4.1
• Electrolytes & non-electrolytes
• Dissociation
2. Solution concentration & stoichiometry
• Molarity & interconversion
• Dilution
Types of aqueous chemial reactions
3. Precipitation reactions
• Complete ionic equations
4. Neutralization reactions
• Acids & bases
• Neutralization reactions
• Non-hydroxide bases produce gases
• Titration
4.5 – 4.6
4.2
4.3
4.6
• Summary of complete ionic equations
5. Reduction & oxidation reactions
• Oxidation numbers
• Oxidation of metals by acids & salts
• Activity series
4.4
Neutralizations create salt & water
by reacting acids & bases
Acids donate protons.
Bases accept protons.
Sulfur & carbonate bases produce gases.
Titrations determine concentrations.
Acids & bases
Acids Compounds that ionize in solution to form a H+ ion; proton donors
So acids increase the [H+] and lower pH.
Acids taste? Sour
Bases Compounds that accept a H+ ion; proton acceptors
So acids decrease the [H+] and increase pH.
Bases taste? Bitter
pH measures the concentration
of free H+ on a log scale.
Inverse: high [H+] = low pH
Coca-cola - pH 1-2
Bleach & ammonia - pH 13-14
Both acids & bases exist in forms that can donate or accept multiple H+:
Monoprotic acid: HCl, HBr, HI, HF
These acids can donate as many protons as
Diprotic acid: H2SO4
Triprotic acid: H3PO4
they have: 1 for mono; 2 for di; 3 for tri.
Usually only the first H+ is released easily, so
the acids ‘strength’ is not additive.
Similarly bases can be mono-, di-, or tri- basic: NaOH, Ca(OH)2, Al(OH)3
So monobasics accept 1 H+, dibasics accept 2 H+ & tribasics accept 3 H+.
Unusual base? (non-OH) NH3 + H+ --> NH4+ Note that NH4+1 can act as an acid.
Ammonia & ammonium are weak bases & acids.
Strong acids & bases?
There are 7 strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4
Strong bases? If the metal is from column 1A or Ca, Sr, Ba
Remember that all strong acids & bases are strong electrolytes
p.128-31
Acid-base conjugate pairs
Are pairs of molecules that differ only by a single hydrogen.
The shift from acid to base is
due to transfer (aka donation)
of a H+1 ion (or proton) from
the acid to the base.
• The acid (H2O) becomes its
conjugate base (OH-1).
• The base (NH3) becomes its
conjugate acid (NH4+1).
Two conjugate pairs are shown here:
Top:
OH-1 is the conjugate base of water.
Bottom: NH4+1 is the conjugate acid of ammonia.
p.129
Neutralizations: acid-base reactions
Reactions between acids & base:
Are exchange reactions.
Produce water and a ‘salt’.
Produce a neutral pH (~7).
HCl + NaOH  H2O + NaCl
1-2
acid
12-13
7
adding base
NA
 pH values
neutralized
p.132-4
1.
2.
3.
Neutralization details
HCl + NaOH  H2O + NaCl
1-2 12-13
7
NA  pH values
Neutralization rxn’s can also be analyzed as complete & net ionic equations:
HCl + NaOH  H2O + NaCl
H+ + Cl- + Na+ + OH-  H2O (pure liquid) + Na+ + Cl--H+
+ Cl- + Na+ + OH-  H2O (pure liquid) + Na+ + Cl-
Water does not dissociate; it’s not an ionic, but a molecular compound.
Since all neutralization rxns produce water the net ionic equation of all
neutralizations is H+ + OH-  H2O.
More neutralization examples: products, balance, complete & net
Mg(OH)2(ppt) + 2 HCl(aq)  Mg+2 + 2Cl-1 + 2H2O
H2SO4(aq) + 2 KOH(aq) 
2K+1 + SO4-2 + 2H2O(l)
4H3PO4(aq) + 6 Ca(OH)2(aq)  2Ca3(PO4)2 (ppt) + 12H O(l)
2
p.132-4
Non-hydroxide bases form gases
When an acid reacts with a non-hydroxide base a gas is formed
rather than water  gas + salt
• CO3-2 ion
• S-2 ion
ions that form non-hydroxide bases
2 HCl(aq) + Na2S(aq) 
H2S(g) + 2NaCl(aq)
2H+1(aq) + 2Cl-1(aq) + 2Na+1(aq) + S-2(aq)  H2S(g) + 2Na+1(aq) + Cl-1(aq)
HCl(aq) + NaHCO3(aq) 
H2CO3(aq) + NaCl(aq)
H2O(l) + CO2(g)
H+1(aq) + Cl-1(aq) + Na+1(aq) + HCO3-1(aq)  Na+1(aq) + Cl-1(aq) + H2O(l) + CO2(g)
So here the net reaction is the production of water and CO2 gas.
Note: neither gases nor water (a pure liquid) dissociate.
The gas actually just bubbles out of solution.
Why does H2CO3 dissociate?
It’s a weak but unstable acid & spontaneously decomposes to water & CO2
p.134-5
Recap: complete ionic equations
When writing complete ionic equations for precipitation or neutralization
reactions there are three types of products (or chemical species) that
NEVER DISSOCIATE:
1.
Solids/precipitates - so strongly attracted that water can’t break them up
2. Water (or other pure liquids) - are molecular (or non-electrolytes) and so dilute,
but do not dissociate
3. Gases - are not soluble in water (or aqueous solutions) and bubble away
- Gases are also molecular.
Titration
An experimental method for determining the concentration of a chemical.
Requirements:
1. A chemical that will react with the unknown & whose concentration is known
2. A balanced chemical equation
3. M & volume of the standardized solution
4. An indicator that tells you when the reaction is complete; pH-sensitive dye @ pH 7.0
You can titrate 10.0 mL of unknown HCl solution with 43.2 mL of a 0.15 M
NaOH solution. What’s the concentration of the HCl?
Steps:
HCl + NaOH --> H2O + NaCl equivalence point ~pH 7.0
10.0 mL 43.2 mL
?? M
0.15 M
0.0432 L 0.15 mol NaOH 1 mol HCl
= 0.648 M HCl
1L
1 mol NaOH 0.0100 L HCl
1. Bal equation
2. Assoc. data
3. Standard?
4. Moles standard
5. A -> B moles
6. Divide by L B
How many grams of Cl-1 in a water sample if 20.2 mL of 0.100 M Ag+1 is
needed to titrate?
Ag+1 + Cl-1 --> AgCl(s)
20.2 mL
0.100 M
0.0202 L 0.100 mol Ag+1 1 mol Cl-1 35.45 g Cl-1 = 0.0717 g Cl-1
1L
1 mol Ag+1 1 mol
p.150-3
Putting it all together!
A 70.5 mg sample of potassium phosphate is added to 15.0 mL of
0.050 M silver nitrate, resulting in precipitation.
1. Write the balanced chemical equation
2. What is the limiting reactant?
3. Calculate theoretical yield (g) of precipitate.
K3(PO4) + 3Ag(NO3)  Ag3(PO4)ppt + 3K(NO3)
0.0705 g 15.0 mL
0.050 M
0.0705 g 1 mole K3(PO4) 1 mole Ag3(PO4) 418.59 g Ag3(PO4) = 0.139 g
212.24 g
1 mole K3(PO4)
1 mole Ag3(PO4)
0.0150 L 0.050 mole Ag(NO3) 1 mole Ag3(PO4)
1L
3 mole Ag(NO3)
418.59 g Ag3(PO4) = 0.105 g
1 mole Ag3(PO4)
So Ag(NO3) is limiting, and the theoretical yield of ppt is 0.105 g
p.153