1. In an unsaturated solution, what is the maximum concentration of Pb2+, if we
ID: 984987 • Letter: 1
Question
1. In an unsaturated solution, what is the maximum concentration of Pb2+, if we managed to fixed the I- concentration to 1 x 10-3M?
Pbl2->Pb2+ + 2I- Ksp=7.9 x 10-9
2. 100mL Of 5mM CoCl2 reacted with 100mL of 4mL K2CO3 to form CoCO3 and KCl. 1) Identify the ions and precipitates present in the solution and 2) what concentration will you observe precipitates forming? 3) What is the concentration of the ions in solution, accounting for activity, assuming rCo2+ =0.749 rCO32- = 0.742?
Calculation process and the answer? Thank you!
Explanation / Answer
Pbl2->Pb2+ + 2I- Ksp=7.9 x 10-9
Ksp = [Pb+2][I- ]2
[I] = 1*10-3 M
7.9*10-9 = [Pb+2](1*10-3 )2
[Pb+2] = 7.9*10-9 /1*10-6
[Pb+2] = 7.9*10-3 M >>>answer
2. CoCl2 + K2CO3 -----> CoCO3 +2 KCl
Co+2 and CO32- ions in the precipitates
no of moles of CoCl2 = molarity * volume in L
= 5*10-3 *0.1 = 5*10-4 moles
no of moles of K2CO3 = 4*1`0-3 * 0.1 = 4*10-4 moles
limiting reagent is K2CO3
CoCl2 + K2CO3 -----> CoCO3 +2 KCl
1 moles of K2CO3 react with CoCl2 to form1 mole CoCO3
4*10-4 moles of K2CO3 react with CoCl2 to form 4*10-4 moles of CoCO3
precipitate concentraction = 4*10-3 M
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