Please help with these. Show work :) thanks Chem 160 - Spring 2019 Worked Proble
ID: 1042232 • Letter: P
Question
Please help with these. Show work :) thanks Chem 160 - Spring 2019 Worked Problems for Chapter 17(part 3) complex ion formation, coordination-compound nomenclature and structure Kr is mixed with 150. mL of 1.0 MNHs(aq). The Ke for Cu(NH) is 5.0x10s (medium) Determine the concentration of ammonia at equilibrium when 350. mL of 1.0 M Cu(NO,)(aq) 2, (medium) Calculate the concentration of free copper(I) ion in an aqueous sol ution that is 0.17 CuNO and 3.2 M NHs given that the Ke of Cu(NH)2 is 5.0x10 K and K (medium) What is the molar solubility of silver chloride in 2.50 M NaCN aq) given that Kap of AgCl 1.6*10-10 and Kr of AgCN) is 1.0x1017 Chard) What is the molar solubility of aluminum hydroxide in a buffered solution with a pH 11.35 given that the Kp of Al(OE),- 1.8x103 and Krof AI(OE)4 is 1.1x10
Explanation / Answer
2) Cu(NO3)2 ionizes completely to produce Cu2+ which reacts with NH3 to form the complex.
Cu(NO3)2 (aq) -------> Cu2+ (aq) + 2 NO3- (aq)
Cu2+ (aq) + 4 NH3 (aq) -------> [Cu(NH3)4]2+ (aq)
Since Kf is large, we will assume that all the added Cu2+ complexes with NH3 and the equilibrium concentration of [Cu(NH3)4]2+ is 0.17 M. We can see that
1 mole Cu2+ = 4 moles NH3
Hence, 0.17 M Cu2+ = 4*0.17 M = 0.68 M NH3
Set up the ICE chart as below.
Cu2+ (aq) + 4 NH3 (aq) -------> [Cu(NH3)4]2+ (aq)
initial 0.17 3.2 0.0
change -0.17 -0.68 +0.17
intermediate 0.0 (3.2 – 0.68) = 2.52 0.17
equilibrium x (2.52 + 4x) (0.17 – x)
The equilibrium constant is given as
Kf = [Cu(NH3)4]2+/[Cu2+][NH3]4 = (0.17 – x)/(x)(2.52 – 4x)4
Since Kf is large, we can assume x << 0.17 M and write,
5.0*1013 = (0.17)/(x)(2.52)4
====> x = (0.17)/(2.52)4(5.0*1013) = 8.43*10-17
The equilibrium concentration of Cu2+ is 8.43*10-17 M (ans).