Asked by Chris
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...?
(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...?
Answers
Answered by
DrBob222
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^+)
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^+)
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