| = | Ka | |||||
| 10 |
This is because, as shown at the right, the system is no longer at equilibrium if the dilution factor is applied to all concentrations.
The direction and extent of the change in [H3O+] depends on which member of the conjugate pair is the major species.
Important distinction: [A ] is the equilibrium concentration of A. This is the concentration after the system reaches equilibrium.
c(A) is the concentration of A just after mixing before any reaction occurs. This is how much A is available for reaction.
| Ka = | [H3O+]2 | |
| c(acid) |
c(acid) in any solution and [H3O+] are related as shown at the right because:
To reach equilibrium the reaction below occurs to a small extent.
acid(aq) + H2O
conjugate base(aq) + H3O+Thus: [conjugate base] = [H3O+] and [acid] ≈ c(acid)
If c(acid) decreases, so does [H3O+], but [H3O+] changes less because it is squared.
| Ka = | c( base)[H3O+]2 | |
| Kw |
c( base) and [H3O+] are related as shown at the right because:
To reach equilibrium, the reaction below occurs to a small extent:
conjugate base(aq) + H2O
acid(aq) + OH–[conjugate base] ≈ c( base) and [acid] = [OH–] = Kw/[H3O+]
If c(base) increases, [H3O+] decreases, but [H3O+] changes less because it is squared.
| Ka = | c( base) | ×[H3O+] | |
| c( acid) |
Consider a solution prepared by dissolving both members of the conjugate pair.
As shown at the right, [H3O+] depends on the ratio of the initial concentrations. Because reaction of each member of the conjugate pair produces the second member and the extent of both reactions is very small, there is little change in the ratio of their concentrations (and in [H3O+]) when equilibrium is re-established.
As shown at the right, [H3O+] depends on the ratio of the initial concentrations. Because reaction of each member of the conjugate pair produces the second member and the extent of both reactions is very small, there is little change in the ratio of their concentrations (and in [H3O+]) when equilibrium is re-established.