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# 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 -

## Presentation on theme: "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 -"— Presentation transcript:

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

Calculate the pH, pOH, [H +], and [OH - ] for a 1.0 M solution of hydrofluoric acid if the K a for hydrofluoric acid is 7.2 x 10 -4 HF  [ H + ] + [F - ] K a = [ 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

Calculate the K a of formic acid (HCHO 2 ) if a 0.10 M solution has a pH of 2.38 at 25  C. HCHO 2  [ H + ] + [CHO 2 - ] K a = [ H + ] [[CHO 2 - ] [HCHO 2 ] [H + ] = inverse log (-2.38) = 0.0041687 M HCHO 2  [ H + ] + [CHO 2 - ] Initial (M) 0.10 0 0 Change (M) Equil (M) 0.10 – 0.004168 0. 004168 0.004168 0.09583 K a = (0.004168) (0.004168) = 1.8 x 10 -4 0.09583

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