Write the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer. Omit water in your answer.)

Question

Write the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer. Omit water in your answer.) 
(a) HC_2H_3O_2 
(b) Co(H2O)_6 3+ 
(c) CH_3NH3 +1

please help, so confused. It first asks for the reaction and then the equilibrium for each...?

DrBob222 · Answer

No reason to be confused.
K = right side/left side
coefficients become exponents.
all are concns in moles/L.
For example,
CH3NH3^+ + HOH ==> CH3NH2 + H3O^+
Ka for CH3NH3 = Kw/Kb for CH3NH2
Ka = (H3O^+)(CH3NH2)/(CH3NH3^+)