4 + SO 4 NH 4 NH3 + H + SO H 2 O HSO OH - H 2 O H + + OH - NaNO 3 Na + -

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1 Chemistry Spring 014 Exam : Chapters 6-7 Name 80 Points Complete problem 1 and four of problems -6. CLEARLY mark the problem you do not want graded. You must show your work to receive credit for problems requiring math. Report your answers with the appropriate number of significant figures. Bonus (4 points): One day last week, Dr. Lamp told those in class a ph at 8:30 AM as he started class and said that the ph would be an answer to one of the exam questions. What was that ph? 3.68 You MUST complete problem 1. (16 pts.) 1. When ammonium sulfate dissolves, both the anion and the cation can participate in acidbase equilibria. Considering all the equilibria active in solution, write enough equations so that you could solve for the concentration of each species in a solution that is saturated with ammonium sulfate that also contains 0.10 M sodium nitrate. You must write the charge balance expression and at least one mass balance. Identify all unknowns and write enough explicit, independent equations so that only algebra remains to solve for the unknowns. A numerical answer is not necessary. Do not consider activities. (NH 4 ) SO 4 (s) K sp = 76 + NH 4 K a = 5.7 x SO 4 - K b = 9.8 x Reactions (Unknowns are in Bold): (NH 4 ) SO 4 (s) NH SO 4 + NH 4 NH3 + H + SO H O HSO OH - H O H + + OH - NaNO 3 Na NO 3 So, we need a total of eight equations for our eight unknowns. Charge Balance: [NH 4 + ] + [H + ] + [Na + ] = [SO 4 - ] + [HSO 4 - ] + [OH - ] + [NO 3 - ] Mass Balance: total concentration of ammonia = (total concentration of sulfur (sulfate)) [NH 3 ] + [NH 4 + ]= [SO 4 - ] + [HSO 4 - ] Concentrations of Na + and NO 3 are constant [Na + ] = 0.10 M [NO 3 - ] = 0.10 M Equilibrium constant expressions: K a = [NH 3 ][H + ] K sp = [NH + 4 ] [SO - 4 ] K w = [H + ] [OH - ] [NH + 4 ] K b = [HSO 4 - ][OH - ] [SO 4 - ] We now have enough equations to solve for all of the unknowns. 1

2 Do four of problems -6. Clearly mark the problem you do not want graded. (16 pts. ea.). A saturated solution of BaSO 4 (K sp = 1.1x10-10 ) that originally held a volume of 1.00 L is allowed to evaporate until the solution volume is L. How does the new concentration of Ba + compare to the concentration in the original solution? Clearly justify your response, using calculations were appropriate. Do not consider activities. Since the solution was initially saturated, a decrease in solution volume will force an increase in the concentrations of Ba + and SO 4 - in solution, leading to supersaturation. This increase will cause the K sp for barium sulfate to be exceeded, resulting in the precipitation of BaSO 4, and a decrease in the concentration of Ba + and SO 4 - back to their original values. Therefore, the [Ba + ] after evaporation will be the same as its initial concentration.

3 3. Is it possible to perform a % complete separation of barium and silver by precipitation with sulfate if both Ba + and Ag + are present initially at 0.00 M? Justify your decision. Do not consider activities. BaSO 4 K sp = 1.1 x Ag SO 4 K sp = 1.4 x 10-5 Two equilibria to consider: BaSO 4 = Ca SO 4 Ag SO 4 = Ag SO 4 K sp = 1.1 x K sp = 1.4 x What [SO 4 - ] is needed to lower each ion's concentration to 0.01% of its initial value? Target [Ba + ] = (0.00M) =.0 x 10-6 M, Target [Ag + ] =.0 x 10-6 M Ba + : [SO - 4 ] = K sp = 1.1 x = 5.5 x M SO 4 [Ba + ].0 x 10-6 M Ag + : [SO - 4 ] = K sp = 1.4 x 10-5 = 3.5x 10 6 M SO - 4 (wow!) [Ag + ] (.0 x 10-6 M) So, Ba + will precipitate first.. Will Ag + precipitate if [SO 4 - ] = 5.5 x 10-5 M? Q = [Ag + ] [SO 4 - ] = (0.00 M) (5.5 x 10-5 M) =. x 10-8 Since Q<K sp for Ag SO 4, Ag + will not precipitate before [Ba + ] =.0 x 10-6 M. Separation is feasible. 3

4 4. Consider the table of activity coefficients on the last page of this exam. As you move from left to right across any row on the table, the values for activity coefficient decrease. As you move down in a given column, the activity coefficient also decreases. Clearly describe the phenomena that cause these trends. Do not simply point out the trends; you must explain why the trends exist. No calculations are necessary. The cause for both trends is rooted in electrostatic interactions between the ion of interest and other materials in solution. Moving left to right across a row in the table, the ionic strength of the solution increases. As increases, there is a greater probability that the ion of interest will undergo an interaction with a third-party ion in solution, thus decreasing its activity in the reaction of interest. Moving down in the table, the charge on the ion is increasing. This increase in charge provides for stronger electrostatic interactions with all other ions in solution, also reducing activity. 4

5 5. Using activities, find the silver concentration of a solution of M Mg(NO 3 ) saturated with Ag CrO 4 (K sp for Ag CrO 4 is 1. x ). Compare this result to that obtained if you were to ignore activities. Let's set up the system first: Ag CrO 4 = Ag CrO 4 i c - +x +x e - x x Ksp = (A Ag+ ) A CrO4- = ( Ag+ [Ag + ]) CrO4 [CrO 4 - ] = ( Ag+ [x]) CrO4 [x] = 4( Ag+ ) CrO4 (x) 3 x = [CrO 4 - ] = K sp 1/3 4( Ag+ ) CrO4 [Ag + ] = x To find activity coefficients, we need to calculate ionic strength: [Mg + ] = M [NO 3 - ] = (0.0167M) = M = 1/([Mg + ] (+) + [NO 3 - ](-1) ) = 1/(0.0167(4) (1)) = M For = M, Ag+ = 0.80, CrO4- = [Ag + ] = x = K sp 1/3 4( Ag+ ) CrO4 Substituting into the equation above gives [Ag + ]=.03 x 10-6 M Ignoring activities, [Ag + ] = x = K sp 4 1/3 Which gives [Ag + ]= 1.34 x 10-6 M 5

6 6. Ethylamine (CH 3 CH NH ) is a monobasic weak base with a pk b of Calculate the ph of a solution prepared by mixing 0.0 ml of M HCl with 50.0 ml of M ethylamine (CH 3 CH NH ) and diluting the resulting solution to ml. Do not consider activities. What s left after HCl and ethylamine react? HCl + B BH + + Cl - Start (mmol) mmol 0 mmol End (mmol) = = Concentration(M) 0.1mmol/100mL = 0.01 M 0.0mmol/100 ml = 0.000M Now the equilibrium: B + H O BH + + OH - K b = [BH + ][OH - ] I 0.01M M 0 [B] C -x -- +x +x K b = (0.000+x)(x) E 0.01-x x x 0.01-x M After some algebra, 0 = x + (0.000+K b )x 0.01K b Solving for x, we get x = [OH - ] = M, [H + ] = K w /[OH - ] = x10-1 M ph =

7 Possibly Useful Information K a K b = K W = 1.0 x Don t eat the yellow snow. 0.51z 1 log c i z i i G = H -T S = -RTlnK x b b 4ac a H O H + + OH - K w = 1.00x10-14 ph = -log[h + ] 7

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