|Lecture Notes: 8 December||
Weak acid dissociations involve equilibria. The equilibrium constants for weak acids have a specific symbol = Ka = Keq for the dissociation of an acid.
Example. What is the pH of a 0.10 M solution of acetic acid. Ka = 1.8 x 10-5
One of the most frequent calls for calculating acid equilibria is calculations involving buffers. What is a buffer?
With this in mind let's do some examples.
Example. Calculate the pH of a "buffer" (a solution which resists changes in pH) made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. Ka = 1.8 x 10-5
Let's try the example again using the Henderson Hasselbalch equation, pH = pKa + log[A-] / [HA]
Henderson-Hasselbalch equation: pH = pKa + log([A-] / [HA])
Example: Calculate the pH of a "buffer" made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. pKa = 4.74. Notice that we are already past the "reaction" stage, so just set up for "@ Equilibrium." let's do the problem.
Notice that only the ratio matters, the actual concentrations of the acid and salt are not that important for determining pH!
Example. What ratio of acetate ion to acetic acid would you need to make up an acetate buffer with a pH of 5.25.? pKa = 4.74. Easiest way to approach this is to use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc-
Example: How many moles of sodium acetate must be added to 0.120 moles of acetic and sufficient water to make a liter if we want a buffer with a pH of 4.50? pKa = 4.74.
Example: What is the pH of a solution made up by mixing 250.0mL of 0.10M sodium hydroxide with 750.0mL of 0.050M acetic acid? pKa = 4.74
First need to do CHEMISTRY, that is need to react the acid with the base to find out how much acid and salt will remain afterwards. Sodium hydroxide is a strong base, so write as hydroxide ion, while we write out the formula for acitic acid since irt is weak and largely in the unionized form:
CH3COOH + OH1- CH3COO1- + H2O Before reaction (0.7500L)(0.050M)/1.000L = 0.0375M (0.2500L)(0.1M)/1.000L = 0.025M 0 - After reaction 0.0125M 0 0.025M -
Now we can enter the proper values for HA and A- into the Henderson-Hasselbalch equation:
pH = pKa + log[A-] / [HA] = 4.74 + log(0.0250)/(0.0125) = 4.74 +M log 2 = 4.74 + 0.301
pH = 5.04
For simple elemental ions it is easy to determine the charge on an atom, but in many other circumstances this is not the case. In order to name compounds we frequently need this information which is obtained from oxidation numbers.
Oxidation numbers are in essence an electronic accounting method in which electrons are assigned to a particular atom in a bond or interaction. As such they give an approximate picture of where electrons actually reside in compounds. We will find this information very useful later when we look at particular types of chemical reactions. Oxidation numbers are essential for nomenclature.
Oxidation numbers are most readily assigned using a simple set of rules:
Finally, note that in writing formulae, the element with the more positive oxidation number comes first. There are, of course, a few exceptions, the most well known being ammonia: NH3 (by the rules it should be H3N).
© R A Paselk
Last modified 9 December 2009