Ka can be calculated from equilibrium amounts.
Approximations: Equation above can be simplified to equation below by assuming two things…
Dissociation of Water: There is a small concentration of H+ from dissociation of water which is
neglected. Therefore when HA dissociates, H+ and A- ions formed in equal quantities.
Exception- Approximation above does not work for very weak weak acids or dilute weak acids. If
pH more than 6, then [H+(aq)] from dissociation of water more significant compared to
dissociation of weak acid. More dilute solution means more water.
Decrease in Concentration: The equilibrium concentration of HA is smaller than undissociated
concentration of HA as [HA (aq)]eqm = [HA (aq)]start – [H+ (aq)]eqm. Can neglect any decrease
in concentration of HA from dissociation as very small in weak acids.
Exception- Approximation above does not work for stronger weak acids and for dilute solutions.
[HA (aq)]start – [H+ (aq)]eqm difference is more significant as [H+ (aq)] more significant.
Therefore can simplify to…
Ka = [ H +(aq)]
2
[HA (aq)]
Calculate pH using Ka:
[ H +(aq )]=√(Ka x [ HA (aq )]) Make sure to include brackets around the expression.
All aqueous weak acids contains hydroxide ions as well since water dissociates. H2O ⇌ H+
+ OH-
.
Therefore if know [H+], can find concentration of hydroxide ions in an acid.
[OH-] = Kw
H+¿ ¿