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+¿ ¿