Calculate the solubility of Pb(BrO3)2 in 0.100 M KBrO3.
Pb(BrO3)2 ⇌ Pb2+ + 2BrO3- Ksp: 7.9 × 10-6
Calculate the solubility of Pb(BrO3)2 in 0.100 M KBrO3. Pb(BrO3)2 ⇌ Pb2+ + 2BrO3- Ksp:...
Calculate the solubility of Pb(BrO3)2 in: a) Water Pb(BrO3)2 ⇌ Pb2+ + 2BrO3- Ksp: 7.9 × 10-6 b) 0.100 M KBrO3. Pb(BrO3)2 ⇌ Pb2+ + 2BrO3- Ksp: 7.9 × 10-6
Explain why solubility of Pb(BrO3)2 in water is different from its solubility in 0.100 M KBrO3.
Explain why solubility of Pb(BrO3)2 in water is different from its solubility in 0.100 M KBrO3.
A solution contains Pb(BrO3)2 and PbS. Propose a way in which you could separate Pb(BrO3)2 (Ksp: 7.9 × 10-6) from PbS (Ksp: 3 × 10-28) contained in the solution.
What would the solubility of [Pb(OH)3]-(aq) be in 0.100 M Ba(OH)2 in the presence of excess Pb(OH)2(s)? Ksp Ba(OH)2 = 5.0 x 10-3 Ksp Pb(OH)2 = 1.43 x 10-20 Kf [Pb(OH)3]- = 3.8 x 1014
What would the solubility of [Pb(OH)3]-(aq) be in 0.100 M Ba(OH)2 in the presence of excess Pb(OH)2(s)? Ksp Ba(OH)2 = 5.0 x 10-3 Ksp Pb(OH)2 = 1.43 x 10-20 Kf [Pb(OH)3]- = 3.8 x 1014
What would the solubility of [Pb(OH)3]-(aq) be in 0.100 M Ba(OH)2 in the presence of excess Pb(OH)2(s)? Ksp Ba(OH)2 = 5.0 x 10-3 Ksp Pb(OH)2 = 1.43 x 10-20 Kf [Pb(OH)3]- = 3.8 x 1014
Given that the solubility product of PbI2 is expressed as Ksp = [Pb2+][I–]2, calculate the Ksp of PbI2 from the concentration of Pb2+ found in Step 6 of the Data Analysis (which was 2x10^-6). (Hint: Think about the stoichiometry involved when PbI2 was initially formed… how much I– was left unreacted?)
Calculate the concentration of 103 in a 9.27 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(IO3)2 is 2.5 x 10-13. Assume that Pb(IO3), is a negligible source of Pb2+ compared to Pb(NO3)2. [103] = M A different solution contains dissolved NalO3. What is the concentration of Naloz if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.20 x 10-6 M? concentration: M
Calculate the concentration of IO3 in a 1.25 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(IO3)2 is 2.5 x 10-13. Assume that Pb(IO3), is a negligible source of Pb2+ compared to Pb(NO3)2. [103] = 2 x10-10 M A different solution contains dissolved NalO3. What is the concentration of NaIO, if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 7.00 x 10-6 M? concentration: -1.396 x10-5 M