TBR Chem Chapter 4 # 46

[LuminousTruth]

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If a weak acid is titrated with enough strong base so that [A-] > [HA], then for the resulting compound:
A) [H3O+] < Ka
B) pH<pKa
C) [H3O+] > [A-]
D) [HA] < [H3O]

After reading the explanation, I think I understood why A is correct. I initially chose D and I would like to know why it is clearly wrong (I don't quite get the TBR explanation for D.)

My reasoning: Since there is more conjugate base than the acid, that means the reaction looks something like this: HA -> [A-] + [H3O+]. I would eliminate C since [H3O+] should be EQUAL to [A-], even if it is a weak acid since any dissociation would form both parts. (Did I make a false assumption here?)

I thought since there was more [A-], there had to be an equal amount of H3O+ present due to the A-, therefore, the H3O+ would be in greater quantity than HA.

Was there holes in my reasoning?

 

[Saint Fu]

Remember that this is in the context of a titration with a strong base. Addition of the base causes acid dissociation, but the H+ don't just become free floating. They associate with OH- from the base to form water. Since it's a weak acid, the amount of free H+ (and corresponding H3O+) is always going to be less than the amount of HA.

Edit: Maybe I should have said 'excess' H+ instead of 'free', but either way I hope it makes sense.

 

[MangoPlant]

Actually you can do this question analytically!

Your equation is right. HA --> A- + H3O+ From this let us write the equilibrium expression.

Ka = [H3O+][A-]/[HA]

Now the question tells us that [A-] > [HA] which (dividing both sides by [HA] implies that [A-]/[HA] > 1.
Notice how [A-]/[HA] is in our Ka expression. Thus our Ka expression can be rewritten as:

Ka = [H3O+]*some number greater than one

Clearly, if Ka is [H3O+] times a number greater than one, than Ka > H3O+

Let me know if you, or anyone else needs any clarification!

 

[LoLCareerGoals]

Actually you can do this question analytically!

Your equation is right. HA --> A- + H3O+ From this let us write the equilibrium expression.

Ka = [H3O+][A-]/[HA]

Now the question tells us that [A-] > [HA] which (dividing both sides by [HA] implies that [A-]/[HA] > 1.
Notice how [A-]/[HA] is in our Ka expression. Thus our Ka expression can be rewritten as:

Ka = [H3O+]*some number greater than one

Clearly, if Ka is [H3O+] times a number greater than one, than Ka > H3O+

Let me know if you, or anyone else needs any clarification!

I like your explanation, but I am still put off by the fact that there is titration happening here and we didn't really model it into the reaction other than using the final inequality.
Also words "for resulting compound" are a bit confusing.
I expected having to actually write out some sort of a:
NaOH + HA -> NaA + H20 and then using NaA as a "resulting compound".


But just so I am not confused by this, could you confirm the following reasoning is correct:
First NaOH + HA -> [Na+] + [A-] + H2O(l) partial titration happens (fewer moles of NaOH were added than moles of there were moles of HA so there is some HA left) such that inequality above is now true.
Regular dissolution equation for the acid is: HA + H2O(l) -> [A-] + [H3O+] and then we follow your arithmetic. Looks correct?

 

[MangoPlant]

I like your explanation, but I am still put off by the fact that there is titration happening here and we didn't really model it into the reaction other than using the final inequality.
Also words "for resulting compound" are a bit confusing.
I expected having to actually write out some sort of a:
NaOH + HA -> NaA + H20 and then using NaA as a "resulting compound".


But just so I am not confused by this, could you confirm the following reasoning is correct:
First NaOH + HA -> [Na+] + [A-] + H2O(l) partial titration happens (fewer moles of NaOH were added than moles of there were moles of HA so there is some HA left) such that inequality above is now true.
Regular dissolution equation for the acid is: HA + H2O(l) -> [A-] + [H3O+] and then we follow your arithmetic. Looks correct?

I agree that "for resulting compound" is confusing. I think it would make more sense if it said "for resulting mixture" or "for resulting solution."

The inequality by itself actually incorporates the titration into our equation. Remember, the inequality we have is a RESULT of the titration that we performed.

Your reasoning seems correct. Also remember that pure liquids/solids are not incorporated into the Ka expression.