[ves]

I found this a very interesting solution!!! My friend and I are doing a prac that we devised ourselves... (besides doing several other things, we did this... but got into a tangle writing about it)

I have added 10mL of 5M acetic acid (therefore 0.05 moles) and 0.05 moles of sodium thiocyanate to 490mL of water (giving a 500mL solution).

Now what exactly happens to this solution and how can I find the pH???

important theory:
pKa of acetic acid = 4.76
pKa of hydrogen thiocyanic acid = 0.85

I have a few ideas: (and some involve quadratics when solving for 'x')

 the acetic acid slightly dissociates giving a certain amount of H+ , whilst the salt hydrolyzes in water to give
SCN- + H20 <--> HSCN + OH-
We would require a pKb for this so since pKa = 0.85, pKb = 10^(-14) / 0.85 right?
And since that is small, the amount of HSCN and OH- is very small...

but then do u have to consider HSCN dissociating (since it is strong) to give more H+ blah blah blah all that sorta thing

or does the salt simply dissociate leaving SCN- ions floating around??? I have no idea...

After THIS, then I started adding 0.5M HCl (2ml at a time and then calculating the pH)

oh i know this sounds daunting and requires too much brain but i KNOW u guys have it in you! C'mon you geniuses! You'll get banned for mental laziness and laxity!!! hehehhe

peace out moles!
ves

[Donaldson Tan]

	HSCN		CH3COOH		H+		SCN-		CH3COO-
Initial (mol)	0		0.05		0		0.05		0
Change (mol)	+y		-x		+x-y		-y		+x
Equilibrium (mol)	y		0.05-x		x-y		0.05-y		x

I have set up the ICE table for your question above.

Use it to solve for the pH of of your solution.

[ves]

wow u the man!!! butttt um not quite sure what u've done there (at all!! lol) or what an ICE is

But were my conjectures correct? does the acetic acid dissociate on its own whilst the salt hydrolyzes?

oh and if so, what happened to the OH-?

[AWK]

We would require a pKb for this so since pKa = 0.85, pKb = 10^(-14) / 0.85 right?

pKb=14-pKa=13.15

But were my conjectures correct? does the acetic acid dissociate on its own whilst the salt hydrolyzes?
oh and if so, what happened to the OH-?

We have three equilibrium in the solution: Ka, Kb and Kw

[ves]

oops!
I meant Kb = 10^(-14)/10^(-0.85)

We have three equilibrium in the solution: Ka, Kb and Kw

eh? lol 

im still a bit lost ever since geodome bamboozled me lol

[AWK]

At the fisrt crude approximation you can calculate just pH of  acetic acid because acetic acid is sufficiently weak and do not replace HSCN in its salt. When you use additional HCl - treat your solution as buffer (when HCl is a limiting reagent) but use a quadratic equation instead of Henderson-Hasselbach equation, or as a mixture of strong acids (sum of HCl + HSCN)

[ves]

ok just for demonstration purposes:

before adding 2mL 0.5M HCl:

u just consider the pH of acetic acid because.. there wont be much H+ or OH- from SCN-? (please confirm your reasoning)

so Ka expression: 10^(-4.76) = (x^2)/(0.1-x)            (0.1 is the molarity of acetic acid in the solution whilst x is the change etc)
we can ignore the -x so its ~0.0013 = [H]
therefore pH = -log 0.0013 ~ 2.89
are u sure thats right....?

now wen i add 2mL of 0.5M HCl then what happens there numerically and chemically?

cheers.. if it makes it easier : add me on mr_roast_chicken@hotmail.com (don't ask about the name i made it in year 5 lol keeping the legend going )

[AWK]

2 ml of 0.5 M HCl contains 0.001 mole of HCl. It replaces HSCN in salt hence you have 0.049 moles of NaSCN and 0.001 moles of HSCN in 502 ml.
K_HSCN=[H+][SCN-]/[HSCN]=[H+][0.049/0.502+[H+]]/[0.001/0.502-[H+]]
Note - I am taking into account a dissociation of HSCN that leads to quadratic equation. Be extremely careful with discriminant of equation - take as much significant digits as possible (because of possible numerical instability of equation solution).
Compare this solution to the pH of acetic acid itself

[ves]

so once i find [H+] from the HSCN etc i add the number of moles of H+ to the number of moles of H+ given from acetic acid and then find its conc then pH?

(btw in your calcs, ud have to divide thru by 0.5L for conc ... right?)

[AWK]

so once i find [H+] from the HSCN etc i add the number of moles of H+ to the number of moles of H+ given from acetic acid and then find its conc then pH?

(btw in your calcs, ud have to divide thru by 0.5L for conc ... right?)

Yes, in equilibrium equations you should put  concentrations or activities, but the stioichiometries we calculate in moles. There is another possibility - for  volume of 1 liter only, concentrations and moles can be used interchangeably.

[ves]

hey AWK or ne other fellow geniuses, if possible and for saving time - kindly add me on mr_roast_chicken@hotmail.com

greatly appreciated

[Borek]

Heh, while I was looking for something else I have located this thread. I was on my bike trip at the time, that's why I have missed it then...

Did I told you I am working on the next program?