CHEM 104 - Chemistry II Exam 3.
Question 1 10 / 10 pts
le.pdf)
Click this link to access the Periodic Table.
(https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab
This may be helpful throughout the exam.
Show the calculation of the molar solubility (mol/L) of Cu(OH)2, Ksp of Cu(OH)2 = 1.6 x 10 -19 .
Your Answer:
Cu (OH) (s) --> Cu ^+2 (aq) + 2OH- (aq) Ksp = [Cu^+2] [OH-] ^2 Ksp = (x) (2x) ^2 = 4x^3
when x = molar solubility of Cu(OH)2 Ksp = 1.6 x10^-19 = 4x^3 solving for x = ((1.6x10^-19)/4) ^(1/3)
3.42 x 10^ (-7) mol/L
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Question 2 10 / 10 pts
le.pdf)
Click this link to access the Periodic Table.
(https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_Tab
This may be helpful throughout the exam.
Show the calculation of the Ksp of AgCl if the solubility of AgCl is 0.0001921 g/100 ml.MW of AgCl = 143.35
AgCl (s) Ag + (aq) + Cl - (aq)
Your Answer:
AgCL(s) --> Ag + (aq) + Cl- (aq) Ksp = [Ag+] [ Cl-] = (x) (x) = x^2 where x = molar solubility of AgCl MW of AgCl = 143.35
Cu(OH)2 (s) (s) Cu +2 (aq) + 2 OH -1 (aq) (s) (2s) 1.6 x 10 -19 = [Cu +2 ] x [OH -1 ] 2
1.6 x 10 -19 = [s] x [2s] 2
1.6 x 10 -19 = 4 s 3
s = 3.42 x 10 -7 mol/L
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