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Published byCecily Jones Modified over 9 years ago
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1 Example Calculate the volume of 14.8 M NH 3 and the weight of NH 4 Cl (FW = 53.5) you would have to take to prepare 100 mL of a buffer at pH 10.00 if the final salt concentration is to be 0.200 M. k b = 1.75x10 -5 Solution The key to solving any problem is the equilibrium of substances in solution.
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2 Here, we have ammonia and ammonium which are combined in the equation: NH 3 + H 2 O NH 4 + + OH - K b = [NH 4 + ][OH - ]/[NH 3 ] We are aware of the pH which means that we can find [OH - ] and we are given the concentration of NH 4 + as 0.200 M. Therefore: pOH = 14 – 10 = 4 [OH - ] = 10 -4 M Now we can solve the equilibrium relation to find [NH 3 ] 1.75x10 -5 = (0.200x 10 -4 )/[NH 3 ] [NH 3 ] = 1.14 M
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3 We need 100 mL of 1.14 M to be prepared from 14.8 M so we have mmol ammonia needed = 1.14x100 = 114 mmol mL ammonia = mmol/molarity = 114/14.8 = 7.7 mL Or simply, M i V i = M f V f 14.8* V mL = 1.14 * 100 V mL = 7.7 mL The weight of NH 4 Cl can also be found as the volume and molarity are given in the problem Mmol NH 4 Cl = 0.200 x 100 = 20.0 mmol Mg NH 4 Cl = 20.0 x 53.5 = 1070 mg
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4 Example How many g ammonium chloride (FW = 53.5) and how many mL of 3.0 M NaOH should be added to 200 mL water and diluted to 500 mL to prepare a buffer at pH 9.5 and a salt concentration of 0.10 M. Solution Once again, the key to solving this problem is the equilibrium of substances in solution.
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5 Here, we have ammonia and ammonium which are combined in the equation NH 3 + H 2 O NH 4 + + OH - K b = [NH 4 + ][OH - ]/[NH 3 ] We are aware of the pH which means that we can find [OH - ] and we are given the concentration of NH 4 + as 0.10 M. Therefore: pOH = 14 – 9.5 = 4.5 [OH - ] = 10 -4.5 = 3.2x10 -5 M
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6 Now we can solve the equilibrium relation to find [NH 3 ] 1.75x10 -5 = 0.10 x 3.2x10 -5 /[NH 3 ] [NH 3 ] = 0.18 M mmol NH 3 = 0.18 x 500 = 90 mmol mmol NaOH = mmol ammonia V mL NaOH = mmol/molarity = 90/3.0 = 30 mL mmol NH 4 + = 0.10 x 500 = 50 mmol Total mmol salt = mmol ammonia + mmol ammonium = 90 + 50 = 140 mmol mg NH 4 Cl = 140 x 53.5 = 7.49x10 3 mg = 7.49 g
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7 Solutions of Polyprotic Acids Polyprotic acids are weak acids, except sulfuric acid where the dissociation of the first proton is complete, which partially dissociate in water in a multi step equilibria where hydrogen ions are produced in each step. Examples include carbonic, oxalic maleic, phosphoric, etc. A general simplification in the calculation of pH of such acids is to compare k a1 and k a2 where,usually k a1 /k a2 is a large value (>10 4 ) and thus equilibria other than the first dissociation step can be ignored.
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8 Let us look at the following example: H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 [H + ] = [H + ] H3PO4 + [H + ] H2PO4 - + [H + ] HPO4 - Looking at the values of the acid dissociation constants for the three steps, it is obvious that the first step occurs about 10 6 times greater than the second and thus the amount of protons in the second step is negligible ( [H + ] H2PO4 - ) compared with the first.
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9 In addition, the third step comes from the second step and since the second step contributes a negligible amount of H + we can also neglect the third step ([H + ] HPO4 - ) or any other consecutive steps. Therefore, only the first equilibrium contributes to the H + concentration.
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10 Example Find the pH of a 0.10 M H 2 SO 4 (k a2 = 1.7*10 -2 ) Solution The first dissociation is 100% complete, therefore, we have: [H + ] = [HSO 4 - ] = 0.10 M from first dissociation. The second dissociation is as follows: HSO 4 - H + + SO 4 2-
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11 1.7*10 -2 = x(0.10 + x)/(0.10 – x) assume that 0.10 >> x x = 1.7*10 -2 Relative Error = (1.7x10 -2 /0.10) * 100% = 17% Therefore, the assumption is invalid and the equation must be solved by the quadratic equation.
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12 Example Find the pH of a 0.10 M H 3 PO 4 solution. Solution H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 Since k a1 >> k a2 (k a1 /k a2 > 10 4 ) the amount of H + from the second and consecutive equilibria is negligible if compared to that coming from the first equilibrium.
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13 Therefore, we can say that we only have: H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2
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14 K a1 = x * x/(0.10 – x) Assume 0.10>>x since k a1 is small (!!!) 1.1*10 -2 = x 2 /0.10 x = 0.033 Relative error = (0.033/0.10) x 100 = 33% The assumption is invalid and thus we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.028 Therefore, [H + ] = 0.028 M pH = 1.55
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15 Example Find the pH of a 0.10 M H 2 CO 3 solution. K a1 = 4.3x10 -7, k a2 = 4.8x10 -11 Solution We have the following equilibria H 2 CO 3 H + + HCO 3 - k a1 = 4.3 x 10 -7 HCO 3 - H + + CO 3 2- k a2 = 4.8 x 10 -11 Since k a1 is much greater than k a2, we can neglect the H + from the second step
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16 H 2 CO 3 H + + HCO 3 - k a1 = 4.3 x 10 -7 K a1 = x * x/(0.10 – x) Assume 0.10>>x since k a1 is small 4.3*10 -7 = x 2 /0.10 x = 2.1x10 -4 Relative error = (2.1x10 -4 /0.10) x 100 = 0.21% The assumption is valid and [H + ] = 2.1x10 -4 M pH = 3.68
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17 If we would like to calculate the amount of H + coming from the second equilibrium ([H + ] second step = [CO 3 2- ]) we substitute 2.1x10 -4 for [H + ] as follows: K a2 = [H + ][CO 3 2- ]/[HCO 3 - ] But from the first step we have [H + ] = [HCO 3 - ] k a2 = [CO 3 2- ] = 4.8x10 -11 = [H + ] second step
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18 Therefore we are justified to omit second dissociation where the hydrogen ion concentration obtained from the first step (2.1x10 -4 M) is much greater than the [H + ] obtained from the second dissociation (4.8x10 -11 M).
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