Complete the calculation summary below Report ALL answers to the correct signifi

Question

Complete the calculation summary below Report ALL answers to the correct significant digits. For the reaction: PblaPb2+21-1 the Kp [Pb.1- 2(0) Test Tube 0.012 M Pb(NO)2 0.030 M KI 0.20 M KNO, Total Volume Initial Molarity Pb+2 Initial Molarity 1-1 5.0 mL 3.0 ml 2.0 mL mL 5.0 mL 5.0 mL 0.0 mL mL Saturated solution prepared by dissolving 2(4) x 10-M × 10-3M × 10-3M ?× 10-3M Calculate the SOLUBILITY PRODUCT CONSTANT OF LEADD IODIDE for each trial below Test Tube 2 Absorbance 8x10-3M 6.8 x 10-3M Equilibrium Molarity 1-1 × 10-3M x 10-3M Equilibrium Molarity Pb2 of PBI

Explanation / Answer

For Part 1, I'm sure how to answer, but for part 2, I'm not pretty sure how to calculate the equilibrium molarities of Pb2+. So, I'm answering part 1 thoroughly and part 2 partially.

Part 1: Test tube 1

Total volume = 10 mL

Initial Molarity Pb2+ = 0.012 mmol/mL * 5 mL/10 mL = 0.006 M

Initial Molarity I- = 0.03 mmol/mL * 3 mL/10 mL = 0.009 M

Part 1: Test tube 2

Total volume = 10 mL

Initial Molarity Pb2+ = 0.012 mmol/mL * 5 mL/10 mL = 0.006 M

Initial Molarity I- = 0.03 mmol/mL * 5 mL/10 mL = 0.015 M

Part 2

Ksp of PbI2 = [Pb2+]equilibrium * [I-]2?equilibrium

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