Determine the pH of an HF solution with a 4.70x10^-2M and 2.20x10^-2M.
I tried this by looking up the Ka and putting it over molarity-ka squared.
Am I doing something wrong? Can someone please show me the proper way to solve this problem?
.........HF => H^+ + F^-
I.......4.7E-2..0....0
C.......-x......x....x
E.....4.7E-2....x....x
Ka = (x^2)/(4.7E-2 - x)
Solve for x = (H^+) and convert to pH.
If you are having trouble obtaining the correct answer you may need to solve the quadratic; i.e., don't neglect the x in 4.7E-2 -x.