FOr you chemistry nerds out there

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Posted by stk on Apr-15-2004 05:40:

FOr you chemistry nerds out there

Lets say you have a pH of a 0.10 M HCN solution , which is 5.2

What is the [H+]?
What is the [CN-]?

How can you find the concentration of H+ when you dont have a ph for that? I know that pH+pOH = 14, but what the heck are u suppose to do when pH+pCN = ???

Posted by TAnt on Apr-15-2004 05:51:

When you put together pH + pCN you get 1...I dont know!?

Posted by montie on Apr-15-2004 05:52:

first you need the Ka value for HCN, it hsould be in your book
and with the dissaciation equation for HCN

HCN-->H+ + CN-
in a 1L soln. you have 0.10M of HCN

which dissasciates

into x amount of H+ and x amount of CN-

so using the equilibrium equation and the Ka value you will have

[H+][CN-]/[HCN]=Ka which is (x)(x)/(0.1-x)=Ka

now since the amount of HCN that dissaciates is negligable to the change in its concentration you can throw out the -x in (0.1-x)

which gives you x^2/0.1=Ka

plug and chug and you get your value for [H+] and [CN-]

Posted by montie on Apr-15-2004 05:53:

oh wait you have pH

haha forget what i just said

pH = -log(H+)

solve that and then you have [H+] and [CN-]

Posted by stk on Apr-15-2004 06:01:

it says that the ph of .10 M HCN, not the [H+], is 5.2

Does that still mean the pH of the .10 M HCN is still equal to -log([H+])?

Posted by Dmatrox on Apr-15-2004 06:09:

quote:
Originally posted by stk
it says that the ph of .10 M HCN, not the [H+], is 5.2

Does that still mean the pH of the .10 M HCN is still equal to -log([H+])?

no i dont think the pH of HCN is equal to -log([H+])

i thought you have to use the henderson hasslebach equation for weak acids, but may be im wrong, can remember, it was a year ago

pH = pka + log([CN]/[H+])

Posted by montie on Apr-15-2004 06:11:

quote:
Originally posted by Dmatrox
no i dont think the pH of HCN is equal to -log([H+])

i thought you have to use the henderson hasslebach equation for weak acids, but may be im wrong, can remember, it was a year ago

pH = pka + log([CN]/[H+])

yeah yo ucan use that too

pH is just equal to -log(H+) of the soln.

Posted by Dmatrox on Apr-15-2004 06:12:

bah, dont ppl use google anymore?

http://employees.oneonta.edu/kotzjc...Ch17_Weak_3.pdf

http://www.google.ca/search?q=weak+...F-8&hl=en&meta=

Posted by stk on Apr-15-2004 06:25:

thx ..

but i have another problem

"formic acid, Hfor, has a Ka value of 1.8e-4. You need to prepare a buffer having a pH of 3.4 from 0.10 Hfor and 0.1 M NaFor Solution. How many mL of the NaFor solution should you add to 20 mL of the 0.10 M HFor to make your buffer?"

So from this question..i have no idea what the equation is?

Is is something like Hfor + Na+ --> H+ + NaFor
<--

If it's correct, then [H+][NaFor] / [Na+][Hfor] = 1.8e-4

K now i am lost, where do i go from here?

And what is a buffer? I couldnt find it anywhere in my chem textbook..