1. If an acid has a Ka value of 1.5x10 -5 , what will be the value of the equili

Question

1. If an acid has a Ka value of 1.5x10-5, what will be the value of the equilibrium constant for the following reaction?

HA(aq) + OH-(aq) <-------> A-(aq) + H2O(aq)

HINTS: Reactions to use:

HA(aq) + H2O(l) <----> A-(aq) + H3O+(aq)
H2O(l) + H20(l) <----> H3O+(aq) + OH-(aq)

HINTS: Equations to use,
Ka=[A-][H3O+]/[HA]
Kw=[H3O+][OH-]

2. Will this reaction be reactant or product favored? What does this tell you about the "completeness" of the titration reactions used in this laboratory?

Explanation / Answer

1. For the given reaction K = Ka/ Kw = (1.5*10^-5 ) /10^-14

                  K = 1.5*109

2. this reaction will be product favoured .

we can say that reaction completes rapidly , so it will be difficult to obtain the exaact equivalence point.

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