calculating pH of a weak acid

Strong vs weak acid....... dilute vs concentrated acid

Acids and conjugate base pairs

Acids and their conjugate bases and bases and their conjugate acid differ by one H

examples:

acids and conjugate bases.

HCl / Cl-

CH3COOH / CH3COO-

H2SO4 / HSO4-

bases and conjugate acids

NH3 / NH4+

CH3NH2 / CH3NH3+

Acids and bases

Strong acids
Completely dissociate in water.
Are mineral acids (any acid that formula starts with an H)
HNO3
HCl
H2SO4
H3PO4

Weak acids
Partially dissociate in water
Are organic acids that have a COOH group
CH3COOH (ethanoic acid)

Strong bases
Any metal hydroxide
NaOH

Weak bases
NH3
CH3NH2

Aqueous Chemistry - Solubility Summary Hints

Calculate the solubility of.....

Here you are being asked to calculate the [ion] concentration in solution in molL-1 you do this by solving for [x] using Ks.
For a 1:2 ratio salt, the concentration of 2mol ion will be 2x.


The concentration of Cl ions in solution is 0.1 molL-1, calculate Ks for a solution of MgCl2.......

Here you are being asked to calculate Ks or IP if the solution is not a saturated solution. To do this you use the ion concentration you have been given.

Calculate Known
Mg2+ + 2Cl-
0.1/2 mol/L 0.1 mol/L

If you are given the conc of the 2 mol [ion] in solution, then the 1 mol [ion] in solution will be 1/2 times that value.

Then plug these values into the expression for Ks.

Ks = [mg2+] [Cl-]2


When two solutions are mixed that have a Common ion ...........

These questions require you to plug calculated [ion] into the IP expression and then compare to Ks to determine if a precipitate is formed.


What is the maximum [ion] need to get a precipitate.......

These questions are asking you to use the Ks expression and Ks value to solve for [x]

Here if the 2mol [ion] is known, you /2 to get the 1mol [ion]. But is the 1 mol [ion] is known then you multiply by 2 to get the 2mol [ion].

Calculating Ks.........

Calculating the ion concentration on solution using Ks....

Equilibrium

Equilibrium

When a salt dissolves an equilibrium is set up.

example: PbI2(s) --> Pb2+(aq) + 2I-(aq)

The equilibrium constant Kc is a ration of [products]/[reactants] at equilibrium

When you write a Kc expression, you never include solids or water when it is present as a solvent.

Qs is the reaction quotent for the dissolving process when a solution is formed that is NOT a saturated solution.

Ks is the solubility product and is for the dissolving process when a solution is formed that IS saturated.

Ks(BaSO4) = [Ba2+] [SO42-]

Titration Hints

When you are doing a titration you must always do the following:

A) Recording results:

1. Always record the first titration as rough and never use this volume in the average titre calcualtion.

2. Record the titre volumes to 0.05mL

3. You need at least 3 titres within 0.2mL of eachother.

4. Always record your results in a table form as follows:


rough 1 2 3 4

Final volume (mL)
initial volume (mL)
Volume used (mL)

B) Experimental technique

1. Rinse the conical flasks with distilled water.

2. Rinse the burette with the test solution you are going to use in it.

3. Rinse the pipette with the solution you are going to measure in it.

Disproportionation

This occurs when a reactant is both reduced and oxidised in the same reaction.

eg: H2O2 --> H2O + O2

here H2O2 is reduced to water and oxidised to O2

H2O2 + 2H+ + 2e- --> 2H2O reduction

H2O2 --> O2 + 2H+ + 2e- Oxidation

Another example:

Cl2 --> HCl + HOCl

Here the oxidation number of Cl goes from 0 to -1 in HCl, indicating a gain of electrons = reduction
And from 0 to +1 in HOCl, indicating a loss of electrons = oxidation