This is a common ion problem. The ionization of HF, a weak acid, is reduced (suppressed somewhat) by the addition of the F^- from the NaF. The F^- is the common ion.
......HF --> H^+ + F^-
I...0.15.....0.....0
C.....-x.....x.....x
E...0.15-x...x.....x
The NaF is 100% ionized; therefore, the F^- from that is 0.26M
Ka = (H^+)(F^-)/(HF)
Plug in the numbers and solve.
(HF) = 0.15-x
(H^+) = x
(F^-) = 0.26+x. The 0.26 is from the NaF and x from the HF.
Solve for x = (H^+) and convert to pH.
(F^-) =
Calculate the ph of a solution containing 0.15M HF (ka= 7.1x10^-4) and 0.26M NaF.
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