3. Solid sodium iodide is slowly added to a solution that is 0.020 M Pb2+ and 0.
ID: 503795 • Letter: 3
Question
3. Solid sodium iodide is slowly added to a solution that is 0.020 M Pb2+ and 0.020 M Ag+. Ksp (PbI2) = 1.4 ×10–8; Ksp (AgI) = 8.3 ×10–17. (a) What compound will precipitate first? (Calculations are required to support your answer.) (b) Calculate the [Ag+] concentration when PbI2 just begins to precipitate. 3. Solid sodium iodide is slowly added to a solution that is 0.020 M Pb2+ and 0.020 M Ag+. Ksp (PbI2) = 1.4 ×10–8; Ksp (AgI) = 8.3 ×10–17. (a) What compound will precipitate first? (Calculations are required to support your answer.) (b) Calculate the [Ag+] concentration when PbI2 just begins to precipitate. 3. Solid sodium iodide is slowly added to a solution that is 0.020 M Pb2+ and 0.020 M Ag+. Ksp (PbI2) = 1.4 ×10–8; Ksp (AgI) = 8.3 ×10–17. (a) What compound will precipitate first? (Calculations are required to support your answer.) (b) Calculate the [Ag+] concentration when PbI2 just begins to precipitate.Explanation / Answer
a) pbI2 - pb++ + 2I-
ksp=(pb++) / (I-)2
1.48*10^-8 = 0.020 /(I-)2
(I-)2 =sqrt (1.48*10^-8/0.020
= 0.027M
AgI - Ag+ + I-
Ksp=(Ag+) (I-)
8,3*10^-17 =( 0.020 ) I-
(I-) = 4.15*10-14 M required to precipitate AgI.
You need a lot less I- for AgI than pbI2. So AgI will precipitate first.
b) (Ag+) concentration- ksp= (Ag+) (I-)
8.3*10^-17 /0.027
= 3.07 *10-16M
pbI2 = (pb++) (I-)
= 1.4*10^-8 /0.027
= 5.185*10^-7 M
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.