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...?

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

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

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!