A-B Titrn_Ans_06

# A-B Titrn_Ans_06 - Revised Spring'06 Answers to Acid Base...

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Revised Spring '06 Answers to Acid Base Titration Supplementary Sheet 1. Calculate the pH of a solution prepared by mixing 200 mL of 0.10 M NH 3 (aq) with 300 mL of 0.10 M NH 4 Cl. (K b for NH 3 = 1.8x10 -5 ) This problem involves the formation of a buffer by mixing a weak base (NH 3 ) and the salt of its conjugate acid (NH 4 Cl). Since there is dilution of both solutions taking place, this must be accounted for before the pH of the solution can be determined by the Henderson-Hasselbalck equation. The dilution is accounted for using the equation M conc V conc = M dil V dil . NH 4 Cl NH 4 1+ + Cl 1- NH 3 + H 2 O NH 4 1+ + OH 1- [NH 4 1+ ] dil = (300 mL)(0.10 M)/(500 mL) = 0.060 M {500 mL = total volume = 300 mL + 200 mL} [NH 3 ]dil = (200 mL)(0.10 M)/(500 mL) = 0.040 M One can use a number of routes to get to the answer. The calculation can simply be done using dependent variables (x’s). [NH 3 ] [NH 4 1+ ] [OH 1- ] initial 0.040 0.060 0 change -x +x +x let x = [OH 1- ] equil 0.040 - x 0.060 + x +x [NH 4 1+ ][OH 1- ] (0.060 + x)(x) K b = ----------------------- = --------------------------------- = 1.8x10 -5 [NH 3 ] (0.040 - x) ignoring the added and subtracted x’s yields x = [OH 1- ] = 1.2x10 -5 pOH = 4.92 pH = 9.08 Alternatively, one can use either version of the Henderson-Hasselbalck Equation: [NH 4 1+ ] (0.060) pOH = pK b + log ------------- = - log(1.8x10 -5 ) + log ------------ = 4.74 + 0.18 = 4.92 pH = 9.08 (NH 3 ) [NH 3 ] (0.040) or pK a = 14 - pK b = 14 - 4.74 - 9.26 (NH 4 1+ ) (NH 3 ) [NH 3 ] (0.040) pH = pK a + log ------------------- = 9.26 + log --------------- = 9.26 + (-0.18) = 9.08 (NH 4 1+ ) [NH 4 1+ ] (0.060)

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Acid Base Titration Supplementary Sheet Answers Page 2 2. Calculate the pH of a solution formed by mixing equal amounts of 0.10 M NaOH and 0.10 M HCN. (Ka for HCN = 4.0x10 -10 ) Since equal volumes of equal concentrations of strong base and weak acid are mixed, this represents the endpoint of a titration and all of the HCN has been converted to CN 1- . The concentration of the CN 1- has to be corrected for dilution.
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