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FOr you chemistry nerds out there
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| stk |
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 = ??? |
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| A.T. |
| When you put together pH + pCN you get 1...I dont know!? :crazy: |
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| montie |
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-] |
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| montie |
oh wait you have pH
haha forget what i just said
pH = -log(H+)
solve that and then you have [H+] and [CN-] |
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| 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+])? |
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| Dmatrox |
| 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 :D
pH = pka + log([CN]/[H+]) |
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| montie |
| 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 :D
pH = pka + log([CN]/[H+]) |
yeah yo ucan use that too
pH is just equal to -log(H+) of the soln. |
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| stk |
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.. |
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