[Lucas07]

So I had this lab where we prepared a buffer solution using an unknown monoprotic weak acid and 0.1M NaOH, and the objective of the lab is to calculate the Ka of the solution.

The data I got from the experiment was that I used 17.85mL of NaOH to titrate 15mL of unknown weak acid (or HA). I then added another 15mL of the same unknown weak acid to the already titrated solution.

I then measured the pH of this final solution using an electronic pH meter and found it to be 4.72.

My questions are:

What is the initial concentration of unknown weak acid (or HA)?

What are the final concentrations of HA and A-?

What is the final concentration of H+ in the solution?

And what is the Ka of the solution?

My calculations give me:

Initial[HA]=0.054M

Final[H+]=1.9x10^-5M

Final[HA]=0.054M - 1.9x10^-5M

Final[A-]=0.054M + 1.9x10^-5M

Ka=1.9x10^-5



Sorry if this is a really long question, but the lab is due tomorrow and I really can't figure this out by myself.

[Astrokel]

Could you show your working because i find your answers incorrect.

[Lucas07]

Sure.


For the initial concentration of HA I know how many moles I needed to neutralize it by doing 0.1M NaOH x 0.0175L NaOH = 0.00175L which I then divided by the final volume which is 0.03285L

For HA and A- I assumed that the initial concentrations for them were the same as the initial for HA, and I knew what the final concentration in the solution of H+ was 1.9x10^-5M from the equation pH=-log([H+]). So I just subtracted it from HA and added it to A- (if you set up an ICE box you would get the final concentration of H+ as being the variable since you have 0 for the initial concentration).

And for Ka I just did Ka=[A-][H+]/[HA]

Sorry if this is confusing.

[Astrokel]

For the initial concentration of HA I know how many moles I needed to neutralize it by doing 0.1M NaOH x 0.0175L NaOH = 0.00175L which I then divided by the final volume which is 0.03285L

You don't have to divide by final volume, the question is asking initially even before neutralization.

For HA and A- I assumed that the initial concentrations for them were the same as the initial for HA,

It is actually the 15ml you have added in the end so you are right it will be the same as [HA]_initial. And as for [A^-]_initial, remember you are doing a buffer solution, where does source of A^- comes from?

And for Ka I just did Ka=[A-][H+]/[HA]

"So I had this lab where we prepared a buffer solution"

[Borek]

And for Ka I just did Ka=[A-][H+]/[HA]

Funny thing about this question is that you don't need to calculate anything.

You started with some unknown amount of acid. Your neutralized it. Then you added exactly the same amount of acid. Thus in your solution there are identical amounts of both acid HA and its conjugate base A^-. That means pKa = pH

Note, that what you did - adding and subtracting H⁺ concentrations - has not changed anything, as concentration of H⁺ is sevral orders of magnitude smaller than concentrations of both HA and A^-. Try to calculate ratio [A^-]/[HA] and ([A^-]+H⁺])/([HA]-H⁺]) and compare these numbers.