I keep getting wrong answers for this question.
Determine the pH of an HF solution of each of the following concentrations.
a) .280 M
b) 5.3*10^-2 M
c) 2.50*10^-2 M
No Ka has been given. I haven't attempted b or c yet because I want to at least solve a first. So far I got a answer of 3.75 and 1.88 and both are incorrect. I think I know how to solve for this. Using Ka and approximations. I just don't understand where I am going wrong. Thanks in advance!
4 answers
These can't be solved with a Ka value. Why don't you post your work for a and let me look at it. I'm sure I'll see immediately what is going on.
Oh okay I thought you used Ka to solve these. That's what I'm doing wrong. I tried looking up the Ka for HF and multiplying it to .280 then take the -log of that number. The reason why I had to different numbers is because I found two different Ka's. If I can't use Ka do I just take the -log of .280?
First I made a typo with my response. You certainly DO need Ka and that's the only way you can do it. I just didn't type in with instead of without.
My text hqas 7.2E-4 for HF.
...........HF ==> H^+ + F^-
initial.0.280.....0.....0
change....-x......x.....x
equil....0.280-x...x....x
Ka = 7.2E-4 = (x)(x)/(0.280-x)
Solve for x and convert to pH.
Probably you can get by without a quadratic equation for the 0.280but it may take one for the others, especially the very weak one.
Using this value I get 1.85 for the pH of the 0.280 M soln. Your text may give a different value for Ka.
My text hqas 7.2E-4 for HF.
...........HF ==> H^+ + F^-
initial.0.280.....0.....0
change....-x......x.....x
equil....0.280-x...x....x
Ka = 7.2E-4 = (x)(x)/(0.280-x)
Solve for x and convert to pH.
Probably you can get by without a quadratic equation for the 0.280but it may take one for the others, especially the very weak one.
Using this value I get 1.85 for the pH of the 0.280 M soln. Your text may give a different value for Ka.
the pH is NOT 1.85
2 answers