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...?

Answer 1

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^+)

Answer 2

[AI-generated answer — Step-by-Step Bot]

Yes, you are correct. The question asks for the dissociation reaction and the corresponding Ka equilibrium expression for each acid.

(a) HC₂H₃O₂ (acetic acid):
Dissociation reaction: HC₂H₃O₂ ⇌ H⁺ + C₂H₃O₂⁻
Ka equilibrium expression: Ka = [H⁺][C₂H₃O₂⁻]/[HC₂H₃O₂]

(b) Co(H₂O)₆³⁺ (cobalt(III) hexaaquacomplex ion):
Dissociation reaction: Co(H₂O)₆³⁺ ⇌ Co³⁺ + 6H₂O
Ka equilibrium expression: Ka = [Co³⁺][H₂O]⁶/[Co(H₂O)₆³⁺]

(c) CH₃NH₃⁺ (methylammonium ion):
Dissociation reaction: CH₃NH₃⁺ ⇌ CH₃NH₂ + H⁺
Ka equilibrium expression: Ka = [CH₃NH₂][H⁺]/[CH₃NH₃⁺]

Note: SATP conditions typically refer to standard ambient temperature and pressure conditions, where the temperature is 25°C and the pressure is 1 atm.

Answer 3

[AI-generated answer — Explain Bot]

Yes, you are right! The question asks you to provide the dissociation reaction and the corresponding equilibrium expression (Ka) for each acid listed.

Let's go through each acid one by one:

(a) HC2H3O2 (acetic acid):
When acetic acid (HC2H3O2) dissolves in water, it dissociates into its ions:
HC2H3O2 (aq) ⇌ H+ (aq) + C2H3O2- (aq)

Now, let's write the Ka equilibrium expression for this dissociation reaction:
Ka = [H+][C2H3O2-] / [HC2H3O2]

(b) Co(H2O)6 3+ (hexaaquacobalt (III) ion):
The hexaaquacobalt (III) ion, Co(H2O)6 3+, does not dissociate in water because it is already a fully hydrated complex ion. Hence, there is no dissociation reaction for this compound, and its Ka equilibrium expression is not applicable.

(c) CH3NH3+1 (methylammonium ion):
When the methylammonium ion (CH3NH3+1) dissolves in water, it can undergo a partial dissociation reaction:
CH3NH3+1 (aq) ⇌ H+ (aq) + CH3NH2 (aq)

Now, let's write the Ka equilibrium expression for this partial dissociation reaction:
Ka = [H+][CH3NH2] / [CH3NH3+1]

Remember, for all the expressions above, the state-of-matter (aq) indicates that the species is dissolved in water.

I hope this explanation helps to clarify the process for writing the dissociation reaction and equilibrium expression (Ka) for each acid. Let me know if you have any further questions!