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PS9_Key

PS9_Key - Chemistry 2080 Spring 2009 Problem Set#9 due date...

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Chemistry 2080 Spring 2009 Problem Set #9 - due date: Friday, April 17th at 2 pm Name:________________ Lab TA Name:__________________ Lab Day/Time:______________________ Practice problems (not graded). Chapter 18 : 2, 4, 7, 9, 20, 49 1) What is the minimum concentration of ammonium chloride (NH 4 Cl) that would just prevent a precipitate forming in a solution that is 0.015 M in MgCl 2 and 0.20 M in ammonia (NH 3 )? [State what value you are using for the K sp of Mg(OH) 2 ]. (This is very like Example 18-9, page 760 of Petrucci et al.) The maximum value of the ion product (IP) before precipitation occurs is 1.8x10 -11 (i.e. K sp for Mg(OH) 2 ). Since [Mg 2+ ] is 0.015 M, the maximum value possible for the [OH - ] (and not quite get a ppt) can be calculated from the expression: IP = [Mg 2+ ][OH - ] 2 = [0.015][OH - ] 2 = 1.8x10 -11 thus: [OH - ] 2 = 1.2x10 -9 and so: [OH - ] = 3.46x10 -5 So now we need to calculate what concentration of NH 4 + would give this concentration of OH - in the presence of 0.20 M ammonia. We have the equation: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) K b = 1.8x10 -5 So that K b = [NH 4 + ][OH - ] / [NH 3 ] = [NH 4 + ](3.46x10 -5 ) / 0.20 = 1.8x10 -5 thus: [NH 4 + ] = (1.8x10 -5 )(0.20) / (3.46x10 -5 ) = 0.10 M

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PS9_Key - Chemistry 2080 Spring 2009 Problem Set#9 due date...

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