Lead (II) Bromide is a slightly soluble solute in water. The following equation represents the equilibrium of PB12 with its ions in solution. PbBr2(s) →PB²*(aq) + 2 Br(aq) Ksp= 8x105 Where Ksp represents the solubility product constant. a) Calculate the solubility of PbBr2. b) 0.3 ml of 0.5 M solution of potassium bromide KBr is added to 2 L of 0.02M solution of lead nitrate Pb(NO3)2. Will a precipitate of lead bromide (PbBr2) be formed?

Lead (II) Bromide is a slightly soluble solute in water. The following equation represents the equilibrium of PB12 with its ions in solution. PbBr2(s) →PB²*(aq) + 2 Br(aq) Ksp= 8x105 Where Ksp represents the solubility product constant. a) Calculate the solubility of PbBr2. b) 0.3 ml of 0.5 M solution of potassium bromide KBr is added to 2 L of 0.02M solution of lead nitrate Pb(NO3)2. Will a precipitate of lead bromide (PbBr2) be formed?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Question

Lead (II) Bromide is a slightly soluble solute in water. The following equation
represents the equilibrium of PB12 with its ions in solution.
PbBr2(s)
→PB²*(aq) + 2 Br(aq)
Ksp= 8x105
Where Ksp represents the solubility product constant.
a) Calculate the solubility of PÜB.2.
b) 0.3 ml of 0.5 M solution of potassium bromide KBr is added to 2 L of 0.02M
solution of lead nitrate Pb(NO3)2. Will a precipitate of lead bromide (PbBr2) be
formed?

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