796 Chapter 16 Aqueous lonic Equlium EXAMPLE 16.15 Complex Ion Equilibria You mi

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796 Chapter 16 Aqueous lonic Equlium EXAMPLE 16.15 Complex Ion Equilibria You mix a 200.0 mL sample of a solution that is 1.5 × 10-3 M in Cu(Nole with a 250.0 mL sample of a solution that is 0.20 M in NH, After the solution reaches equilib- rium, what concentration of Cu (ag) remains? SOLUTION balanced equation for the complex on Cu (ag) + 4 NHg) Cu(NH)2 (a) Write the equilibrium that occurs and look up the value of Ky in Table 16.3. Since this is an equilibrium problem, you have to create an ICE table, which requires the K, 1.7×1013 1.5 × 10-3mol 0.200 E × 0200 Ex 1,5x10-3mol initial concentrations of Cu and NH,. Calculate those concentrations from the given values. 0.250= 6.7 × 10-4 M 0.200 0250 LX-TE-0.11 0.20 mol 0.250 L X.20mo [NH,I-= 0.200 L + 0.250 L 0.20L+0.250 L =0.11M Construct an ICE table for the reaction and write Cu (ag)+4 NHylag)Cu(NH) (aq) down the initial concentrations of each species Initial 67X10 Change Equil 0.11 ince the equilibrium constant is large and the Cu' (ag) +4 NH (ag)Cu(NH)(aq) oncentration of ammonia is much larger than the oncentration of Cu", you can assume that the action will be driven to the right so that most of Initial67X10 e Cu2+ is consumed. Unlike previous ICE tables, Change 6.7X1 24-67 × 10 4) here you let x represent the change in concentra Equil n in going to equilibrium, here you let x resent the small amount of Cu that remains en equilibrium is reached. INH3l 0.11 [Cu(NH3),21 0.0 (+67 × 10 ) 0.11 6.7 × 10-4 [Cu(NH)).2.1 ostitute the expressions for the equilibrium centrations into the expression for K and e for x 6.7 X 10 6.7 × 10-4 K40.11)1 10-4 13 rm that x is in concentration c 2 is valid.

Explanation / Answer

You need to know by hard the common complexes in analytical chemistry.

i.e.

when we add Cu+2 and NH3 in solution, if we add plenty of NH3, we do not form only

Cu(NH3)2 ...

but

Cu(NH3)2 + 2NH3 --> Cu(NH3)4+2

tip: if only "Kb" is adde din the description, they are talking about simply Cu+2 and NH3... if they talk about Kf, they probably are talking about a formation coefficient, of a complex ion

the complex ions of Cu+e and NH3 are comonly Cu(NH4)4+2

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