Deriving Formula for Concentration of Hydronium at Equivalence Point?

[eventob]

Homework Statement

This is a part of the lab report for a practical I did last week. I need to derive the following formula for the concentration of hydronium at the equivilance point of a strong base weak acid titration (acetic acid with sodium hydroxide).

Homework Equations

M=V/c

Formula for acid constant.

The Attempt at a Solution

My attempt at a solution is attached (the last portion of the attached image, after the text). Not sure how to move on from this, to get to the expression with the radical sign and Kw (product of the acid and base constants).

[PLAIN]https://files.itslearning.com/data/his/46492/1.png

Thanks in advance.

 

[Borek]

No idea what you are doing here. At the equivalence point you just have a solution of a salt of a weak acid - that's equivalent to the solution of a conjugate base...

 

[eventob]

I am asked to derive the formula at the top of the picture, for the concentration of hydronium. Thought that the equivalence point was the point where number of moles of CH3COOH was equal to the number of moles og NaOH, so I tried to derive the formula from that equiality. :)

Could I write:

expression for Ka*expression for Kb=Kw, and derive the formula from this?

Thanks.

 

[Borek]

Start with an approximated formula for the OH^- concentration in the solution of a weak base.

pH a weak acid or a weak base.

 

[eventob]

Thank you.


Tried to use the approximated formula now, and here is what I got. Think I got it right so far, but how should I proceed to solve it in terms of [H3O+]? Tried to multiply both sides with [H3O+]/[OH-], but I don't think that's right.

[PLAIN]https://files.itslearning.com/data/his/46492/2.jpg

 

[Borek]

You can assume [HA] is so low C_b has not changed. Scroll down to the bottom of the page I linked to.

You can use LaTeX to post formulas:

[noparse]K_b = \frac {[OH]^2} {C_b-[HA]}[/noparse]

gives

K_b = \frac {[OH]^2} {C_b-[HA]}

 

[eventob]

Thank you very much. Solved it now. :)

Didn't know that this forum supported TEX/LaTEX, much more convenient than my copy/paste method i guess.