[reyaz01]

Hey guys, I am really hitting my head to figure out this question, can anyone help me?

Calculate the pH change if to 100cm3 of a buffer solution made of 0.5 moles of ethanoic acid and 0.5 moles of Ch₃CooNa is added 10cm3 of 0.1 M NaOH.

They gave us moles, not concentration, how do you calculate the concentrations of the ethanoic acid and the ethanoate to substitute into the Henderson Hasselbach? Do I just divide each by 0.1 L ? Or what? Thanks!!

And one more thing, is the concentration of the ethanoate (CH₃COO^-) ion the sum of the ethanoate ions produced by both the ethanoic acid as well as the sodium ethanoate?

[Borek]

They gave us moles, not concentration, how do you calculate the concentrations of the ethanoic acid and the ethanoate to substitute into the Henderson Hasselbach? Do I just divide each by 0.1 L ? Or what?

They haven't told you what concentration is? Not even a definition? And they have not taught you how to calculate concentration given number of moles and volume? Sue them. Sue the school system, sue the state, sue the bus driver, and sue the girl's parents.

[reyaz01]

That doesn't exactly help me. 

[Borek]

What is concentration definition?

[reyaz01]

I know where your going with this, but the solution my lecturer gave me had the following statement, which confuses me:
Since the ethanoate and the ethanoic acid are in the same volumes, their molar quantities can be taken to represent their respective concentrations.

He then used the 0.5 moles as if they were concentration and substituted them into the Henderson Hasselbach.

Also another important question, what is the concentration of teh ethanoate ion and why? (is it just the concentration of the sodium ethanoate or the sum of the ethanoate from sodium ethanoate and the ethanoate ion formed from the dissociation of the acetic acid?

[Borek]

I know where your going with this, but the solution my lecturer gave me had the following statement, which confuses me:
Since the ethanoate and the ethanoic acid are in the same volumes, their molar quantities can be taken to represent their respective concentrations.

He then used the 0.5 moles as if they were concentration and substituted them into the Henderson Hasselbach.

Ca = na / V
Cb = nb / V

Ca/Cb = na / V : nb / V = na/nb

Volume cancels out.

Also another important question, what is the concentration of teh ethanoate ion and why? (is it just the concentration of the sodium ethanoate or the sum of the ethanoate from sodium ethanoate and the ethanoate ion formed from the dissociation of the acetic acid?

It should be sum of acetate from both sources, but in the presence of ethanoate dissociation can be safely neglected.

[reyaz01]

THANK YOU THANK YOU THANK YOU! ORRR i get it now... because there is the ethanoate there is a large amount of CH₃COO^- ions and according to lechatelier's principle, the equilibrium will lie far to the right and there fore there will be large amount of undisociated CH₃COOH but just enough H⁺ to make the system acidic, hence acidic buffer system!! OMG i love you!