University College - Chemistry Chem 112A Lecture Notes - Lecture 9: Buffer Solution, Ph
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13 Dec 2016
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[ ]0(cid:4667: buffers, conjugate acid-base equilibria, hno2(aq) + h2o(l) h3o+ (aq) + no2 (aq), ka = 4. 6 x 10 -4. 0. 25 x: ka =[ h3o+][ no2, using successive approximation, x = 4. 6 x 10-4, -log[x] = 3. 34, here ph = pka. H2o(l) (aq) (aq) + hno2(aq), kb = 2. 2 x 10-11. 0. 25 + x (aq) + h2o(l) oh- +x x: kb = x(0. 25+x)/(0. 25-x), using successive approximations x = 2. 2 x 10-11, -log[x] By adding the 0. 01 m strong acid, it gets consumer by 0. 01 m of a-, adding. Ka = x(0. 24+x)/(0. 26-x), using successive approximations x = 5. 0 x 10-4, gives us a ph of 3. 30. Notice ph hasn"t changed much, the solution is very slightly more acidic: add 0. 01 m strong base (assume no volume change) and we make the initial assumption that added base is consumed by reaction with ha.
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