Posted by **Chemistry** on Tuesday, April 7, 2009 at 9:04pm.

so i got a test and like usual my teacher didn't teach me half of the stuff that's going to be on it so he gave us a pretest of what is going to be on it and an answer key and i got several questions

13.) what is the pH of a formic acid if the concentration is .00055 M?

ok so here's what my teacher wrote can you please help me understand?

[H^+]^2/.00055 = 1.78E-4

ok so what formula is this??? aparently concentration goes on the bottom and I have no idea were he got 1.78E-4 from...

after that he went and found what

[H^+]^2 that was which I completley understand if you know 1.78E-4 which I still have no idea were he got from or how to get it or what formula it is...

then he found the [H^+] which I get how to do once you find [H^+]^2

then he found the Ph which I now how to do once you know the [H^+] just need help feeling in the holes

and then based off a problem after it that is similar just asking for a different variable I can tell from my teacher's work that the formula is

([H^+]^2)/(HA) = something not sure what this something is but aparently it's given some were sense my teacher showed no work on how to get this answer or what it equalls anywere even though it's not given... were does the solution come from? Once I know were that number comes from I can do the rest of the math

15) What is the pOH of .05M solution of pryidine

so we know the pOH and then my teacher finds the OH which I understand how to do and gets 8.66E can't read what comes after the E but I understand how to find it

then if I have the right order my teahcer did (OH raised to something I can't read) squared and got 7.5E-11 which I have no idea why he needed this

then he put this formula

Kb = ([A^+][OH^-])/(AOH)

ok so aparently we were given AOH in the begining of .05M and we solved form OH already and got 8.66E something

1.5E-9=(OH^-)2/.05

so apparently it's the same formula from before

I don't see why he went thorugh all of this when he could of just done

POH=-log[OH^-]

to get the pOH after he solved for the [OH^-]???????////

so if you are lost I am to please help me understand what's going on

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