Acids and bases
H2O(l) H+(aq) + OH-(aq)
Kw=[H+][OH-] The value of Kw is 1 x 10-14 at 25ºC but varies with temperature.
pH = -log[H+] (pH is the negative log of the concentration of H+ ions),
pOH = -log[OH-],
pKw = -log([H+][OH-]) ( is equal to 14 at 25ºC)
Use the above equations to calculate other values.
Note that for strong acids and bases [H+] or [OH-] are directly related to the concentration of the acid/base....therefore doubling the concentration of the acid will double [H+] and halve [OH-] (and the reverse is true for bases).
HA(aq) H+(aq) + A-(aq)
B(aq) + H2O(l) BH+ + OH-(aq)
Ka = ( [H+][A-] / [HA] )
Kb = ( [BH+][OH-] / [B] )
Ka is a constant which describes the ionisation of an acid (ie how strong it is) and Kb does the same for bases. pKa is the log form of Ka, defined as pKa = -log(Ka) and pKb = -log(Kb). Like previously with the pH scale, a 1 fold change in pKa will signify a ten fold change in Ka and the same for Kb.
Ka x Kb = Kw (ie they equal 1 x 10-14 at 25ºC)
pKa + pKb = pKw (ie 14 at 25ºC)
Strong acids have weak conjugate bases. Strong bases have weak conjugate acids. A strong acid has a large Ka value (or a small pKa value). A strong base, likewise, has a large Kb value and a small pKb value.
All the above equations need to be applied as appropriate given the required input data.
18.2 - Buffer solutions
A buffer solution is composed of a weak acid/base and it's conjugate base/acid. (assuming weak acid for what follows, reverse for base). A solution of weak acid is made, and this forms a equilibrium with the water as follows
HA + H2O A- + H3O+
To this solution, some of the acid's conjugate added (A-) in the form of the salt of the acid - usually the sodium or potassium salt, resulting in an increase in the concentration of A-, some of which reacts with the H3O+. The result of this is, when equilibrium is reestablished, there is a considerable amount of both HA and A- present in the solution, in an dynamic equilibrium. If some other acid is added, this will react with the A-, but this causes the equilibrium to shift to the right, almost completely counteracting any pH change. The addition of a base, which reacts with the HA, cause the equilibrium to be shifted to the left, again resulting in very little pH change. This continues until one of the two components, either HA or A- are completely used up, at which time the pH then changes normally.
Given the concentration of both the Acid and its conjugate base, and the Ka value of the acid, the [H+] can be calculated and this can then be converted into a value for pH.
18.3 - Salt Hydrolysis
A salt formed by reaction between a strong acid and a weak base will be acidic by hydrolysis - think of it like the strong acid's properties shining through. The correct explanation can be found here.
Similarly a salt formed originally by the reaction between a strong base and a weak acid will be basic by hydrolysis.
Salts formed from stong acid - strong base pairs will be neutral by hydrolysis
18.4 - Acid-base titrations
Strong acid, strong base...The curve starting off very low is initially very flat until at equivalence it is almost vertical then very flat again (starting and finishing very low and high respectively one the graph due to low and high pH of strong acid and base respectively.
Weak acid, strong base...The curve begins comparatively high on the graph, and rises sharply initially. after a period it reaches a region where the solution acts as a buffer, still rising continually, but not as steep. the curve then then turns up sharply at equivalence and then tapers off to the strong base's pH value.
Strong acid, weak base...identical to the strong acid strong base curve only the eventual point is lower since the weak base will have a lower pH.
Weak acid, weak base... the graph starts sharply up, but then tapers off, reaching only a somewhat steep section in center, before flattening off to the weak base pH. There is no steep section and so it is not possible to find a suitable indicator.
(One other thing...at the point halfway to equivalence pH=pKa or pOH=pKb...and pH+pOH = 14...so you can find pKa or pKb from the curve...it can be derived, but it's easier to remember it)
18.5 - Indicators
Indicators work by setting up a weak acid/base equilibrium where the acid and its conjugate base have different colors...
HIn(aq) H+(aq) + In-(aq)
Where HIn is one color and In is the other. This equilibrium can be adjusted by the concentration of H+ being through the addition of acids or bases.
The pH range of the indicator falls around it's pKa value, and so to be useful, the pKa must fall within the inflection of the titration curve.
The value of pKa for the indicator must fall around the equivalence point of the titration to work effectively.
Notes and definitions
In the case of an ionic substance when it dissolves the ions completely separate from the lattice and become solvated by the water molecules (ie attached to, and surrounded by the water molecules) return