Asked by KC
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?
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?
Answers
Answered by
DrBob222
.........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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.