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Calculate the pH, pOH, [H + ], and [OH - ] for a 0.0040 M solution of HNO 3

Calculate the pH, pOH, [H + ], and [OH - ] for a 0.0040 M solution of HNO 3. [H + ] = 0.0040 M. pH = - log [0.004] = 2.40. pOH = 14 - 2.40 = 11.60. [OH - ] = inverse log (-11.60) = 2.51. X 10 -12 M.

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Calculate the pH, pOH, [H + ], and [OH - ] for a 0.0040 M solution of HNO 3

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  1. Calculate the pH, pOH, [H +], and [OH -] for a 0.0040 M solution of HNO3 [H +] = 0.0040 M pH = - log [0.004] = 2.40 pOH = 14 - 2.40 = 11.60 [OH - ] = inverse log (-11.60) = 2.51. X 10 -12 M

  2. Calculate the pH, pOH, [H +], and [OH -] for a 1.0 M solution of hydrofluoric acid if the Ka for hydrofluoric acid is 7.2 x 10 -4 HF  [ H +] + [F -] Ka = [ H +] [ F -] [HF] HF  [ H +] + [F -] Initial (M) 1.0 0 0 Change (M) - x x x Equil (M) 1.0 - x x x 7.2 x 10-4= [ x] [x] x = 2.7 x 10 -2 M = [H +] [ 1.0] pH = - log 2.7 x 10 -2 = 1.57 pOH = 14 – 1.57 = 12.43 [OH - ] = inverse log (-12.43) = 3.7 x 10 -13 M

  3. Calculate the Ka of formic acid (HCHO2) if a 0.10 M solution has a pH of 2.38 at 25 C. HCHO2 [ H +] + [CHO2-] Ka = [ H +] [[CHO2-] [HCHO2 ] [H +] = inverse log (-2.38) = 0.0041687 M HCHO2 [ H +] + [CHO2-] Initial (M) 0.10 0 0 Change (M) Equil (M) 0.10 – 0.004168 0. 004168 0.004168 0.09583 Ka = (0.004168) (0.004168) = 1.8 x 10 -4 0.09583

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