[lazipeter]

Q1. For a 0.1M HCOOH, I wanna know whether 0.1 means the concentration of HCOOH initially(unionized) or the concentration of HCOOH at equilibrium??

and what's the problem with this calculation?

Q: CALCULATE THE VOLUME OF 0.1M NaOH that must be added to 100cm^3 of 0.1M HCOOH to obtain a solution of pH 4.0 at 298 K (Given Ka=1.78*10^-4)
First for the equilibrium 1,

Assume the volume of NaOH add is y dm^3,

the equilibrium 2 is:


Thank you very much!!

[rabolisk]

For any acid solution, the concentration refers to the initial, unionized concentration. For example, if you had 1M HCl, you really mostly have H₃O⁺ and Cl^-, and only trace amount of HCl. Weak acids barely dissociate but regardless, the concentration given refers to the formal (initial) concentration, rather than equilibrium concentration.

You don't need to do equilibrium 1. Yes, that would be the equilibrium state of the solution before NaOH is added. But once the strong base is added, reaction 1 is driven to the right as acid-base neutralization occurs. Acids and bases react practically to completion because  H₃O⁺ + OH^-    H₂O lies so far to the right. Essentially, what is happening is that OH^- reacts with the trace amounts of H₃O⁺ that initially exist, but as the H₃O⁺ is used up, equilibrium 1 shifts to the right, until all the OH^- has been used. But remember that reaction 1 itself produces the conjugate base of formic acid, formate. Instead of thinking of these as one reaction driving the other reaction, just combine them to give HCOOH + OH^-  HCOO^- + H₂O. The reaction practically goes to completion.

The answer can be calculated using the Hendersen-Hasselbach equation. I'm not exactly sure what you did for equilibrium 2. If you wanted to solve it using equilibrium, you would have some concentration of HCOOH remaining, as well as some concentration of HCOO^- due to the neutralization. Then you would solve for H₃O⁺ concentration using Ka.

[lazipeter]

Thanks for answering.
I see your point and I understand it now.
but I have a further question.

Why, for a 0.1 M HCOOH,

the real value of Ka is Ka=x^2/(0.1)       (0.1 is the concentration of HCOOH)
rather than Ka=x^2/(0.1-x) ??
(because the answer I found for this question is Ka=x^2/(0.1) but not Ka=x^2/(0.1-x) )

THANKS!

[AWK]

This is a sufficient approximation though with Ka=x^2/(0.1-x) you will get a slightly better result (this is also approximation, the unabridged equation is cubic)