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
I understand what is going on. You are lost, and do not understand Ka or equilibrium.
Start here:
http://library.thinkquest.org/C006669/data/Chem/equilibrium/ka.html
I understand you're struggling to understand the steps your teacher took to solve the problem. Let's break down each question and the steps involved.
13.) What is the pH of a formic acid if the concentration is 0.00055 M?
To find the pH of a solution, we need to first find the concentration of the hydrogen ion ([H+]). Your teacher used the formula [H+]^2/.00055 = 1.78E-4. This equation is derived from the equilibrium constant expression for the ionization of a weak acid. The equation assumes that [H+] comes from the dissociation of a weak acid, in this case, formic acid (HCOOH).
To find [H+]^2, you need to solve the equation [H+]^2/.00055 = 1.78E-4 for [H+]^2. The value 1.78E-4 is given in the answer key (probably obtained from a reference source). Now you can solve for [H+] by taking the square root of 1.78E-4.
Once you have [H+], finding the pH is straightforward. The pH is calculated as the negative logarithm (base 10) of [H+]: pH = -log[H+].
Regarding your second question about the formula ([H+]^2)/(HA), it seems like you may be referring to the Henderson-Hasselbalch equation, used to calculate pH or pOH for a weak acid or base solution. However, it doesn't appear directly related to the first problem you mentioned.
15.) What is the pOH of a 0.05 M solution of pyridine?
To find the pOH, you can use the equation pOH = -log[OH-]. Your teacher went through additional steps by finding the concentration of hydroxide ion ([OH-]), squaring it, and using the equation Kb = ([A+][OH-])/(AOH), which seems unrelated to finding the pOH.
Using the formula pOH = -log[OH-] directly would be simpler and yield the desired result. You can calculate the pOH by plugging in the concentration of OH- (8.66E-something) into the equation.
In summary, it seems there may have been a mix-up or unnecessary steps in your teacher's approach. It's important to clarify with your teacher or consult additional resources to ensure accuracy and understanding.