Determine the equilibrium [F-] of the following solution with initial concentrat

Question

Determine the equilibrium [F-] of the following solution with initial concentrations of [HF] = 1.296M, [NaF] = 1.045M. (Ka for HF is 6.6x10^-4). PLEASE PROVIDE DETAILED SOLUTION FOR GET 5/5 RATING...Thanks!

Explanation / Answer

Ka = [F-][H-] / [HF] HF -> F- + H+ Initial 1.296 -> 1.045 + 0 Change => -x --> + x + x equib 1.296-x 1.045+x, x 6.6x10-4 = (1.045+x)x / 1.296-x since change in HF will be small [HF] is approximately 1.296 and F- will remain approximately the same at 1.045M 6.6x10-4 = 1.045x / 1.296 6.6x10-4 *1.296 / 1.045 = x = 0.000819 = [H+] approx. ph = -log(0.000819) = 3.09

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