Write the dissociation reaction and the corresponding K a equilibrium expression

Question

Write the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (For the dissociation reaction, include states-of-matter under the given conditions in your answer. Use the lowest possible whole number coefficients. Concentration equilibrium expressions take the general form: Kc = [HCl]2 / [H2] . [Cl2].)

(a)    HC5H9O2(aq)
dissociation reaction:

equilibrium expression:

(b)    Cr(H2O)63+
dissociation reaction:

equilibrium expression:

(c)    C3H7NH3+
dissociation reaction:

equilibrium expression:

Explanation / Answer

Answer – We are given the acid and need to dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. We know acid donate the H+ and for mthe conjugate base.

a)HC5H9O2(aq)

Dissociation reaction:

HC5H9O2(aq) + H2O -----> H3O+ + C5H9O2-(aq)

equilibrium expression:

Ka = [H3O+] [C5H9O2-(aq)] / [ HC5H9O2(aq)]

b) Cr(H2O)63+ (aq)

Dissociation reaction:

[Cr(H2O)6]3+(aq)+ H2O -----> H3O+ + [Cr(H2O)5 OH]3+(aq)

equilibrium expression:

Ka = [H3O+] [Cr(H2O)5 OH]3+/ [Cr(H2O)6]3+

c) C3H7NH3+

Dissociation reaction:

C3H7NH3+(aq) + H2O -----> H3O+ + C3H7NH2 (aq)

equilibrium expression:

Ka = [H3O+] [C3H7NH2] / [C3H7NH3+]

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