I Chapter 17 I Metal lons: Colorful and Essential SAMPLE EXERCISE 17.2 Calculati
ID: 525672 • Letter: I
Question
I Chapter 17 I Metal lons: Colorful and Essential SAMPLE EXERCISE 17.2 Calculating the Concentration Loe of a Free Metal lon in Equilibrium with a Complex Ammonia gas is dissolved in a 1.00 x 10 M solution of cuso. so that initi INH, "200 x 10 the concentration of cu ions in the the reaction mixture has come to chemical Collect and organize The concentration of Cuso, means that (cu before is 100 x The of the (NH, is We know from the text preceding this Sample that the reaction mation of Cu(NH) Also, the values UNHJ, cu "l, and [Cu(NH) are related by the formation of 5.0 x 1013 Analyze The stoichiometric ratio of NH, no Cu is 4:1 but initially[NHJMcu so there is more than enough NH,to convert all the cu into Cu(NHN *.Becau kis that nearly all the cu ions are converted to complexions and tiny concentration of Cu ions, which we will call x, remains at equilibrium: The change in Cu lis equal to the equilibrium value (x)less the initial value (100 x 10 x-100 x 104 The decrease in concentration described by this cquation is equal in magnitude bur oppo- site in sign to the increase in the concentration of complex ions A 1.00 x 10- The balanced equation tells us that 4 moles of NHs are consumed for every mole of cu so the change in ammonia concentration is 4 [NH 44[Cu 4 (x-1.00 x 10 4) We can list these initial values and changes in concentration in a RICE table: Reaction (R 4 NH, Cu(NH,he NH, (M) 1.00 x 10 2.00 x 10 Change (C r-1.00 x 10" 4 (e-100 x 10 a 00 x 10 Given the large walue of we should obtain alcu hvalue at equilibrium tha is much less than 1.00 x 10 M. solve Fint we complete the row of equilibrium concentratians ofNH and culNH, Reaction (R 4 NH NH, (M) Change (c) x 1.00 x 10 4 (x-100 x 10 a 00 x 10 Equilibrium (E) (1.60 x 10 4x 1.00 x 17.4 The nan oxidati (in the Comp 1. S 2. IExplanation / Answer
You have initial [Cu2+] = 1.0 x 10-4
after the complexation remaining [Cu2+] = x = This we need to find out
The change of Delta [Cu2+] = x-1.0 x 10-4
Simillarly, initial there is no complex formation, so [Cu(NH3)42+] = 0
The change will be = -delta[Cu2+] = -(x-1.0 x 10-4) = 1.0 x 10-4 -x
From the equation, we know that 4 moles of NH3 are consumed by each mole of Cu2+. the the change in
ammonia concentration
delta[NH3] = 4 delta[Cu2+] = 4 (x-1.00x10-4)
Please go through the steps in the RICE table: you just need to add the initial and change to get the equilibrium
value. Here we have ammonia initial (2.00 x 10-3) and change 4 x (x-1.00x10-4). So, the sum will be the equilibrium
value of ammonia = 4x + 1.6 x 10-3
reaction : Cu 2+ + 4NH3 <----> Cu(NH3)42+
Initial 1.0 x 10-4 2.00 x 10-3 0
Change x-1.00x10-4 4(x-1.00x10-4) (1.00x10-4-x)
for simplicity we can write it as
(x-1.00x10-4) 4(x-0.100x10-3) (1.00x10-4-x) + 0
Equilibrium 1.0 x 10-4 2.00 x 10-3 = (1.00x10-4-x)
(total) + x-1.00x10-4 + 4x-0.40x10-3
= x = 4x + 1.6 x 10-3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.