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 = ???

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

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-]

oh wait you have pH

haha forget what i just said


pH = -log(H+)

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

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+])?

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+])

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.

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=

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..