Equilibrium Constant

[thais]

Hello All,
I am a bit confused about this question on Equilibrium constant.

The question asks for the equilibrium constant of the neutralization reaction between oxalic acid (Ka=5.9 x 10^-2) and sodium hydroxide?
I get it that Keq = Ka x Kb.
However, I assumed that Kb of NaOH is 1 since it is completely dissociated in H2O, but this is not the right reasoning .
Why is NaOH Kb = 1 x 10^14?

Thank you very much!

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[MT Headed]

Kb is just a K value like all the others. It's products divided by reactants.

By suggesting that NaOH has a Kb of 1, you are implying that in water it splits up only half way, so the concentration of product (OH) = the concentration of reactant remaining (NaOH). But you know that NaOH is a strong base and splits up completely, so it will have a lot more products than reactants and the Kb must be much larger than 1.

 

[thais]

So is it safe to assume that a strong base Kb always = 10^14, and for a strong acid Ka always = 10^14?
Thank you!

 

[MedPR]

So is it safe to assume that a strong base Kb always = 10^14, and for a strong acid Ka always = 10^14?
Thank you!

No.

Are there more details to your original question?

 

[ljc]

This isn't my strong suit, but I'd guess you're not supposed to actually be doing any calculations on this problem. What are the answer choices?

 

[thais]

True... The answer choices are
A) 5.9x10^-2
B) 5.9x10^12
C) 5.9x10^14
D) infinity

On the explanation the reaction for Na is shown as:

H3O+ + NaOH -> 2H2O + Na+ (1/Kw = 1*10^14)

I get it that Kw would equal 10^-14 hence 1/Kw=10^14 however, I'm a bit confused when I should take a strong acid and base as this and when I shouldn't (I hope my question is making sense - thanks for all the help).

 

[MedPR]

True... The answer choices are
A) 5.9x10^-2
B) 5.9x10^12
C) 5.9x10^14
D) infinity

On the explanation the reaction for Na is shown as:

H3O+ + NaOH -> 2H2O + Na+ (1/Kw = 1*10^14)

I get it that Kw would equal 10^-14 hence 1/Kw=10^14 however, I'm a bit confused when I should take a strong acid and base as this and when I shouldn't (I hope my question is making sense - thanks for all the help).

I don't understand your question. They tell you that you are using a strong acid (oxalic acid) and a strong base (NaOH).

I'm guessing that the answer is 1*10^14 because you are neutralizing a strong acid with a strong base. Both will completely dissociate in solution, so your net ionic equation will be H+ + OH- --> H2O. Kw = 1*10^-14 for the dissociation of water, so your Keq here is 1/Kw.

Regardless, you should be able to immediately rule out answers A and D.

 

[MT Headed]

I don't understand your question. They tell you that you are using a strong acid (oxalic acid) and a strong base (NaOH).

Oxalic acid, with a pKa1 of 1.25, is not a strong acid. It's weaker than hydronium (-1.74).

 

[MT Headed]

Okay, I'm not 100% on this, but I think I got it:

The net reaction is:
HOxalate + OH- ---> Oxalate- + H2O (Keq1=???)

This reaction is the sum of the following two reactions:
HOxalate + H2O ---> Oxalate- + H3O+ (Keq2=5.9x10^-2)
H3O+ + OH- ---> H2O (Keq3=1x10^14)

If you add the two reactions, you multiply the two K values, and arrive at:
Keq3 = Keq1 x Keq2 = 5.9x10^12

The whole sodium hydroxide thing is kind of a red herring... NaOH is a strong base and dissociates completely into spectator Na+ ions and reactant OH- ions.

 

[MedPR]

Oxalic acid, with a pKa1 of 1.25, is not a strong acid. It's weaker than hydronium (-1.74).

Well damn.

Disregard my previous post then.

 

[thais]

Thanks guys!

 

[lkthlttr]

Okay, I'm not 100% on this, but I think I got it:

The net reaction is:
HOxalate + OH- ---> Oxalate- + H2O (Keq1=???)

This reaction is the sum of the following two reactions:
HOxalate + H2O ---> Oxalate- + H3O+ (Keq2=5.9x10^-2)
H3O+ + OH- ---> H2O (Keq3=1x10^14)

If you add the two reactions, you multiply the two K values, and arrive at:
Keq3 = Keq1 x Keq2 = 5.9x10^12

The whole sodium hydroxide thing is kind of a red herring... NaOH is a strong base and dissociates completely into spectator Na+ ions and reactant OH- ions.

Strong work

 

[MedPR]

I thought the answer was 5.9*10^14?