1. Lecture 4: Aqueous solution chemistry
Lecture 4 Topics Brown, chapter 4
1. Solutes & solvents
Electrolytes & non-electrolytes
Dissociation
2. Solution concentration & stoichiometry – 4.6
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
Summary of complete ionic equations
5. Reduction & oxidation reactions
Oxidation numbers
Oxidation of metals by acids & salts
Activity series

2. Neutralizations create salt & water by reacting acids & bases
Acids donate protons.
Bases accept protons.
Sulfur & carbonate bases produce gases.
Titrations determine concentrations.
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Acids & bases Acids Bases
Compounds that ionize in solution to form a H+ ion; proton donors So acids increase the [H+] and lower pH.
Acids taste? Sour
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
Bases
Compounds that accept a H+ ion; proton acceptors
So acids decrease the [H+] and increase pH.
Bases taste? Bitter
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 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.
Diprotic acid: H2SO4
Triprotic acid: H3PO4
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+ --> NH 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

4. 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

5. 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 NA  pH values
p.132-4
acid
adding base
neutralized

6. Neutralization details
HCl + NaOH  H2O + NaCl 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) + HCl(aq) 2
 Mg+2 + 2Cl-1 + 2H2O
H2SO4(aq) + KOH(aq) 2
 2K+1 + SO H2O(l) 4
H3PO4(aq) + Ca(OH)2(aq) 6
 2Ca3(PO4)2 (ppt) + 12H2O(l)
p.132-4

7. 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

8. 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. 2. 3.
Solids/precipitates - so strongly attracted that water can’t break them up
Water (or other pure liquids) - are molecular (or non-electrolytes) and so dilute, but do not dissociate
Gases - are not soluble in water (or aqueous solutions) and bubble away - Gases are also molecular.

9. Titration
An experimental method for determining the concentration of a chemical. Requirements: 3. 4.
A chemical that will react with the unknown & whose concentration is known A balanced chemical equation M & volume of the standardized solution An indicator that tells you when the reaction is complete; pH-sensitive 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:
1. Bal equation 2. Assoc. data 3. Standard? 4. Moles standard 5. A -> B moles 6. Divide by L B
HCl + NaOH --> H2O + NaCl equivalence point ~pH 7.0
10.0 mL mL ?? M M
L mol NaOH 1 mol HCl = M HCl L mol NaOH L HCl
How many grams of Cl-1 in a water sample if 20.2 mL of M Ag+1 is needed to titrate? Ag+1 + Cl-1 --> AgCl(s)
20.2 mL M
L mol Ag mol Cl g Cl-1 = g Cl L mol Ag mol
p.150-3

10. Putting it all together!
A 70.5 mg sample of potassium phosphate is added to 15.0 mL of M silver nitrate, resulting in precipitation.
Write the balanced chemical equation
What is the limiting reactant?
Calculate theoretical yield (g) of precipitate.
K3(PO4) + 3Ag(NO3)  Ag3(PO4)ppt + 3K(NO3)
g mL
0.050 M
g 1 mole K3(PO4) 1 mole Ag3(PO4) g Ag3(PO4) = g
g mole K3(PO4) mole Ag3(PO4)
L mole Ag(NO3) 1 mole Ag3(PO4) g Ag3(PO4) = g
1 L mole Ag(NO3) mole Ag3(PO4)
So Ag(NO3) is limiting, and the theoretical yield of ppt is g
p.153