TBR CBT 1 Henderson-Hassebalch question

[Tokspor]

What is the pH of a solution of 25 mL 0.10 M H3CCO2H (pKa = 4.74) after 10 mL of 0.10 M KOH(aq) has been added to it?

A. pH = or < 3.74
B. 4.74 > pH > 3.74
C. 5.74 > pH > 4.74
D. pH = or > 5.74

The correct answer is B, and I understand how to estimate this answer. However, the solution to this question provides a way to solve this exactly. You would use the Henderson-Hasselbach equation, but I am not sure how to get the numbers for it.

In this situation, pH = pKa + log [H3CCO2-]/[H3CCO2H].

When you add 10 mL of KOH like the question stem states, you have titrated 10 mL of the acid H3CCO3H. So you have 10 mL of H3CCO2-. However, the text says you will have 15 mL of H3CCO3H. This means the [H3CCO2-]/[H3CCO2H] ratio is 10x/15x, and you can use the equation in this way:

pH = pKa + log(10/15)

However, shouldn't some H3CCO3H dissolve spontaneously in solution since the pKa is 4.74? So you would have less than 15x of the H3CCO3H.

Is it that the concentration of H3CCO3H dissolving spontaneously is so insignificant that 15x is about the correct number?

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[loveoforganic]

The acid is always dissolved. Assuming you mean dissociate, yes. And the conjugate base is floating around spontaneously picking up protons as well. The HH equations accounts for that with the pKa value, I'm pretty sure.

 

[boaz]

The acid is always dissolved. Assuming you mean dissociate, yes. And the conjugate base is floating around spontaneously picking up protons as well. The HH equations accounts for that with the pKa value, I'm pretty sure.

The HH equation is just rearranged form of the equilibrium expression, with logs. When we use the HH equation we make the assumption that the acid form doesn't dissociate and the base doesn't get protonated. In our case, we assume that it's just log(10/15) even though it's not completely accurate. It works for our purposes because the insignificant difference is such that the log of that number is basically the same regardless. Log(10.01/14.99) is about equal to log(10/15).
The fact that it's not completely accurate can be shown with the case of complete titration of acid until there is "no more" acid left. If we plug "0" into the HH equation it breaks down because any number divided by zero is not a number.
This is totally beyond the MCAT, but the takehome message is that when using the HH equation we assume there's no base-protonation/acid-deprotonation.

 

[loveoforganic]

TY for clearing it up.

 

[Postictal Raiden]

just wondering, if a similar question appeared on the mcat, how would I be able to solve it without a calculator?

 

[boaz]

just wondering, if a similar question appeared on the mcat, how would I be able to solve it without a calculator?

pH=pKa + log(10/15)

=4.74 + log(1/3)

Another way of saying log(1/3) is:

"What power do we need to take 10 to 1/3?"

Well... we know that to get from 9 to 1/3 we do this: 9^-.5 = 1/3

So we can assume that to get from 10 to 1/3 we'll need "a little bit less than" 0.5. So pH is about .5 units away from the pKa. It's lower than the pKa because the log is negative.

 

[csiplon]

just wondering, if a similar question appeared on the mcat, how would I be able to solve it without a calculator?

Looking at the HH equation pH=pKa when your Log term = 0. How do you get your log term to = 0? your acid and it's conjugate base concentration have to equal each other. How could I go about that by adding a strong base of equal molarity? Adding half of the amount of strong base to the weak acid that the solution began with. But the question states that we added LESS than half (10ml) of strong base to the amount of weak acid (25ml). So your pH should be a little less than 4.74

 

[boaz]

Looking at the HH equation pH=pKa when your Log term = 0. How do you get your log term to = 0? your acid and it's conjugate base concentration have to equal each other. How could I go about that by adding a strong base of equal molarity? Adding half of the amount of strong base to the weak acid that the solution began with. But the question states that we added LESS than half (10ml) of strong base to the amount of weak acid (25ml). So your pH should be a little less than 4.74

👍 Yea, given the choices, this is a shorter and faster way of getting to the answer than the way I showed above. Since we haven't yet reached the pH=pKa point we know that the pH will definitely be less than the pKa. B is the only such choice.

 

[equityrange]

However, the text says you will have 15 mL of H3CCO3H.

How come its 15mL? was it given to you?

 

[loveoforganic]

Reaction of a weak acid + strong base.

 

[equityrange]

i understand that you would need more volume of a weak acid but is there a way to calculate to 15mL?

 

[csiplon]

i understand that you would need more volume of a weak acid but is there a way to calculate to 15mL?

I believe 15mL is what's left in the protonated form after you titrated 10mL to it (started w/25mL).