# MCDB 108A Quiz3_Answer_Key - MCDB 108A Duane W. Sears...

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MCDB 108A Duane W. Sears FALL 2010 Online Quiz #3- ANSWER KEY 9/9/2010 Online Quiz3 Answer Key.doc page 1 of 9 pages For Questions 1-2 and 4-11 you are asked to calculate accurate numerical answers to within a specified decimal range as indicated for each question. Follow the examples in parentheses. Problem (1) Calculate the pK dn of a 0.01 M weak acid solution in pure water if pH = 4.20 after equilibrium has been achieved. NOTE : Your numerical answer should be entered to 2 decimal points and credit will be given for any answer within +/-0.05 of the best answer. ! pK dn = ______. (For example: 0.01, 2.34, 56.78, etc.) Answer for Problem #1 : pK dn = 6.40 Equilibrium reaction: [HA] ! [H + ] + [A - ] K dn = [H + ] * [A - ] / [HA] Conditions and givens: pH = 4.20 [H + ] = 10 -4.20 [HA] + [A - ] = 0.01 M Approximations and logic: ** [H + ] = 10 -4.20 ** [H + ] = [A - ] because one A - ion is generated for every proton dissociated from the acid. ** [HA] = 0.01 = 10 -2 (approximately) because, by definition, weak acids dissociate only a small fraction of their protons in the absence of base. Suggested solution: Solve for Kdn . K dn = 10 -4.20 * 10 -4.20 / 10 -2 = 10 (-4.20 - 4.20 + 2) = 10 -6.40 pK dn = -log(K d ) = 6.40. Problem (2) Calculate the decimal fraction - i.e., Yd - of 0.1 M formic acid that is dissociated in pure water at equilibrium if pK dn = 3.75 . NOTE : Your numerical answer should be entered to 3 decimal points and credit will be given for any answer within +/-0.0015 of the best answer. ! Y d = ______. (For example, 0.001, 2.305, 56.785, etc.) Answer for Problem #2 : Yd = 0.041 +/- 0.0015 [HA] ! [H + ] + [A - ] K dn = [H + ] * [A - ] / [HA] pK dn = 3.75 K dn = 10 -3.75 [HA] + [A - ] = 0.1 ** [H + ] = [A - ] because one A - ion is generated for every proton dissociated from the acid.

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MCDB 108A Duane W. Sears FALL 2010 Online Quiz #3- ANSWER KEY 9/9/2010 Online Quiz3 Answer Key.doc page 2 of 9 pages ** [HA] " 0. 1 M . By definition , weak acids dissociate only a small fraction of their protons in the absence of strong base. In other words, [HA] >> [A - ] " 0. 1 M. Suggested solution: Solve for [A - ] or [H + ], and then solve for Y d . K dn = [H + ] * [H + ] / 0.1 = [H + ] 2 / 0.1 Solving for[H + ], [H + ] = SQRT{0.1 * 10 -3.75 } = SQRT{10 -1 * 10 -3.75 } = SQRT{10 (-1-3.75) } [H + ] = SQRT{10 -4.75 } = 10 -2.375 . Thus, pH = 2.375 Solving for Y d : Yd = [A - ] / ([A - ] + [HA]) = 10 -2.375 / 10 -1 = 10 -2.375 + 1 = 10 -1.375 Yd = 0.0422 2 nd use Yd = 1 / (1 + 10 pK-pH ) see Problem 3 Yd = 1 /(1 + 10 3.75 –2.375 ) = 1 / (1 + 10 1.375 ) = 0.0405 Yd = 0.0405 Problem (3) From the following possibilities, select the equation that correctly defines the fraction dissociated , or the dissociation fraction ( Yd ), for a weak acid with dissociation constant , K dn , at the equilibrium H ion concentration , [H + ]. (a)
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## This note was uploaded on 10/29/2010 for the course MCDB MCDB 108A taught by Professor Sears during the Fall '09 term at UCSB.

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MCDB 108A Quiz3_Answer_Key - MCDB 108A Duane W. Sears...

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