TPR Equilibrium question

[663697]

Ten moles of a compound were added to make a liter of solution, and the resulting pH was 5.7. What is the relationship between the Ka for the protonated form of the compound and Kb for the deprotonated form?

A. Ka < Kb
B. Ka = Kb
C. Ka > Kb
D. Cannot be determined

Can somebody please explain why the answer is A? I thought that if pH < 7, it is more acidic than basic and thus Ka > Kb.

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[desperadoflakes]

You are correct in thinking that since our resulting pH is below 7, then the solution we added must be an acid. We can refer to the acid as HA. You are right, if the Kb of HA was greater than the Ka of HA, than HA would have to be a base. However, the question is not asking you to compare the Ka of HA with the Kb of HA. HA is an acid, and the Ka of HA is greater than the Kb of Ha.
No, you are comparing the Ka of HA with the Kb of A-. The question is asking you to compare the strength of the acid HA with the strength of its conjugate base A-. How do we do that? Well, we know that weak acids tend to have stronger conjugate bases.

So the question that we have to answer is "Is HA a weak acid or a strong acid? Is A- a weak base or a strong base?"

This question can be solved with math, or with intuition that comes from practicing these problems. If you added 10 mol of HA to 1 L, than [HA] = 10M. Increasing the concentration to 10M only brought the pH down by 1.3 units (from pH = 7). That's not very much at all. This is a weak acid, and it has a stronger conjugate base.

What does it mean to have a stronger conjugate base? It means that if we initially added 10 mol of A- (instead of HA) to water, than the pH would go up by greater than 1.3 units.

We can also solve this with math.

For a weak acid, we can use the equation:

[H+] = sqrt(Ka * [HA])
TBR also has a really nice shortcut equation for weak acids:
pH = 1/2pKa - 1/2log[HA]

We know [HA] = 10mol/L = 10M
1/2 * log (10) = 1/2 * 1 = 1/2. And pH = 5.7

5.7 = 1/2pKa - 0.5
pKa = 12.4
pKa of HA + pKb of A- = 14
pKb = 1.6
Ka = 1 * 10^-12.4
Kb = 1 * 10^-1.6

Kb > Ka.

So we are using a weak acid. What do we know about weak acids? Their conjugate bases are stronger.

What we added to the solution was 10 mol of HA. The resulting pH has to be below 7. If it was above 7, that would mean that adding HA would take H+ out of the solution.

 

[663697]

You are correct in thinking that since our resulting pH is below 7, then the solution we added must be an acid. We can refer to the acid as HA. You are right, if the Kb of HA was greater than the Ka of HA, than HA would have to be a base. However, the question is not asking you to compare the Ka of HA with the Kb of HA. HA is an acid, and the Ka of HA is greater than the Kb of Ha.
No, you are comparing the Ka of HA with the Kb of A-. The question is asking you to compare the strength of the acid HA with the strength of its conjugate base A-. How do we do that? Well, we know that weak acids tend to have stronger conjugate bases.

So the question that we have to answer is "Is HA a weak acid or a strong acid? Is A- a weak base or a strong base?"

This question can be solved with math, or with intuition that comes from practicing these problems. If you added 10 mol of HA to 1 L, than [HA] = 10M. Increasing the concentration to 10M only brought the pH down by 1.3 units (from pH = 7). That's not very much at all. This is a weak acid, and it has a stronger conjugate base.

What does it mean to have a stronger conjugate base? It means that if we initially added 10 mol of A- (instead of HA) to water, than the pH would go up by greater than 1.3 units.

We can also solve this with math.

For a weak acid, we can use the equation:

[H+] = sqrt(Ka * [HA])
TBR also has a really nice shortcut equation for weak acids:
pH = 1/2pKa - 1/2log[HA]

We know [HA] = 10mol/L = 10M
1/2 * log (10) = 1/2 * 1 = 1/2. And pH = 5.7

5.7 = 1/2pKa - 0.5
pKa = 12.4
pKa of HA + pKb of A- = 14
pKb = 1.6
Ka = 1 * 10^-12.4
Kb = 1 * 10^-1.6

Kb > Ka.

So we are using a weak acid. What do we know about weak acids? Their conjugate bases are stronger.

What we added to the solution was 10 mol of HA. The resulting pH has to be below 7. If it was above 7, that would mean that adding HA would take H+ out of the solution.

I guess when I was doing this question under timed conditions I threw out a lot of my knowledge for some reason.. since we add 10 mol to a 1 L solution, just to be sure could we check the theoretical pH if this was a strong acid and compare it to the given pH, thus confirming that it is a weak acid? From there, we could look at the Ka and Kb expressions: Ka = [H3O+][A-]/[HA] , Kb = [HA][OH-]/[A-]; and notice that since this is a weak acid HA >> A- therefore Ka < Kb?

I was sort of confused by the TBR equation you used as I'm using TPR materials and haven't come across that haha

 

[desperadoflakes]

I guess when I was doing this question under timed conditions I threw out a lot of my knowledge for some reason.. since we add 10 mol to a 1 L solution, just to be sure could we check the theoretical pH if this was a strong acid and compare it to the given pH, thus confirming that it is a weak acid? From there, we could look at the Ka and Kb expressions: Ka = [H3O+][A-]/[HA] , Kb = [HA][OH-]/[A-]; and notice that since this is a weak acid HA >> A- therefore Ka < Kb?

I was sort of confused by the TBR equation you used as I'm using TPR materials and haven't come across that haha

That TBR equation is super super helpful for quickly figuring out either the pH or the pKa of a weak acid, I would recommend learning it.

Really, here's what it is. You know that:
pKa (HA) + pKb (A-) = 14.

If pKa of HA is greater than 7, that means pKb of A- will be less than 7.
This is the same thing as saying that a weak acid or base will have a stronger conjugate. Meaning that an acid with a low Ka will have a conjugate base with a high Kb.