Help with confusing biochem question

[rsweeney]

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I am practicing some biochem and I am stuck:

You start with a 1M solution of acid HA (Ka = 10^-4M). What would be the pH of the solution after you have added 0.5 equivalents of NaOH?

Thanks

 

[shawred]

I am not that great in biochem, but I would guess that the pH=pKa in this instance. A half equivalent of base would react with half the acid leaving an equal amount of acid and conjugate base. Thus, according to Henderson Hasselbalch, the ph would be equal to the pKa. In this case, the answer would be 4. (Carboxylic Acid is a weak acid with a pKa of 4).

I am practicing some biochem and I am stuck:

You start with a 1M solution of acid HA (Ka = 10^-4M). What would be the pH of the solution after you have added 0.5 equivalents of NaOH?

Thanks

 

[rsweeney]

I am not that great in biochem, but I would guess that the pH=pKa in this instance. A half equivalent of base would react with half the acid leaving an equal amount of acid and conjugate base. Thus, according to Henderson Hasselbalch, the ph would be equal to the pKa. In this case, the answer would be 4. (Carboxylic Acid is a weak acid with a pKa of 4).

ok I see. Since NaOH will fully dissociate, 0.5 equivalents of NaOH will react with half (or 0.5M) of the 1M HA leaving you with 0.5M HA unreacted. Hence, you have 0.5M HA and 0.5M A- left. Thus 0.5/0.5 is 1 and the log of 1 is zero. Thus pH = pKa.

The problem I was having was that I was expecting an equation to solve the problem for me. At first I thought volume was missing until now. I see that this question is more conceptual. Thank you!

 

[rsweeney]

I see how the first question makes sinse, but what about when you are not given the volume of base added in this question? You really can't conceptualize this one.
Suppose you were to start with 1.0 l of a 1
M solution of a weak acid such as acetic
acid (pKa = 4.8). What is the pH of your
solution after you have added 0.09 moles (0.09 equivalents)
amounts of sodium hydroxide (strong base)

I figure that all of the 0.09 moles will react with the acid leaving 0.91M of the acid unreacted and all of the 0.09moles will have formed the conjugate base at the end of the titration. Thus, pH=pKa + log(0.09/0.91) = 3.8

New pH = 3.8

Does this look correct?

If this is right then I have it!

 

[romed81]

This is what I think:
First of, you don't need to volume, since you have molarity.
HA dissociates to give you H3O and A
At the beginning, [HA] is 1, [H3O] is 0, and [A] is 0
Assume the change is x
At eq you have Ka= (x*x)/(1-x). Assume x is much smaller than 1, so you have x^2=10^-4, so x=10^-2
This is the concentration of your [H], which will react in one-to-one relationship with your base, which means that at eq (since it is a strong base, and strong acid) you will have 1/2 of the concentration of the [H] left, which is 5*10^-3. Take neg log of that, and you have your pH.
Hope that's right.

 

[spiridon]

Also, sometimes you have a triprotic acid H3A with three pKa's. In that case, if you add 0.5 equivalents of NaOH, the pH will be pKa1, if 1.5 equivalents - pKa2, if 2.5 equivalents - pKa3. It actually helps if you draw the titration curve to visualise this.