Question

The cations Pb2+(aq) and Ba2+(aq) can be precipitated as almost insoluble sulfates. The Ksp values for...

The cations Pb2+(aq) and Ba2+(aq) can be precipitated as almost insoluble sulfates. The Ksp values for PbSO4(s) and BaSO4(s) are 2.5 × 10−8 and 1.1 × 10−10, respectively.

If you have a solution that is 0.010 mol L-1 in both Pb2+(aq) and Ba2+(aq) ions, and the concentration of SO42-(aq) ions is gradually increased, the less soluble salt will precipitate first.

What is the concentration of the cation (Pb2+ or Ba2+) that precipitates first remains in solution just before the second sulfate salt begins to precipitate?

The answer is 4.4x10^-5
Could you please show working out :)

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Can Ba2+ be separated from Pb2+ by the addition of Na2SO4? Assume the starting solution is...
Can Ba2+ be separated from Pb2+ by the addition of Na2SO4? Assume the starting solution is 0.10 M in both cations and use whatever concentration of sulfate you like. (Ksp of PbSO4 = 1.3 x 10-8), while that of( BaSO4 = 1.5 x 10-9). Calculate how much of the first cation to precipitate is left in solution when the second starts to fall out of solution. Is it less than 1% of the initial present?
The cations Hg2 2+, Ag+, and Pb2+ can be precipitated as insoluble chlorides: Hg2Cl2 (Ksp=1.2x10^-18), AgCl...
The cations Hg2 2+, Ag+, and Pb2+ can be precipitated as insoluble chlorides: Hg2Cl2 (Ksp=1.2x10^-18), AgCl (Ksp=1.8x10^-10), PbCl2 (Ksp=1.7x10^-5). If you have a solution containing these three cations, each with a concentration of 0.10 Mol/L, what is the order in which they precipitate as hydroxides?
A 1.00 L solution at 25˚C containing both 0.0020 M Ba2+ and 0.0020 M Pb2+. A...
A 1.00 L solution at 25˚C containing both 0.0020 M Ba2+ and 0.0020 M Pb2+. A solution containing dissolved SO42– is slowly added to the solution containing 0.0020 M Ba2+ and 0.0020 M Pb2+. a) Which salt will precipitate first, BaSO4 or PbSO4? Briefly explain your answer. b) Calculate the concentration of SO42– that must be added to create a saturated solution with 0.0020 M Pb2+. (Ignore Ba2+.) c) T or F The Ksp value of PbSO4 will increase as...
Ba2+ is added to a solution to precipitate sulfate as BaSO4 (Ksp = 1.1 x 10-10)....
Ba2+ is added to a solution to precipitate sulfate as BaSO4 (Ksp = 1.1 x 10-10). a. The concentration of [Ba2+] is raised to 0.0020 M and solid BaSO4 is present in the system; calculate the equilibrium concentration of SO42-. b. Suppose KCl is added to the system to increase the ionic strength to 0.100 M; calculate the activity of the Ba2+ in this solution. c. Is equilibrium [SO42-] increased, decreased or unchanged by the addition of KCl ?
The concentration of SO42–(aq) in a sample of river water can be determined using a precipitation...
The concentration of SO42–(aq) in a sample of river water can be determined using a precipitation titration in which a salt of Ba2+(aq) is the standard solution and BaSO4(s) is the precipitate formed. What is the concentration of SO42– (aq) in a 42.6 mL sample if 5.00 mL of a 0.00100 M Ba2+(aq) solution is needed to precipitate all the SO42–(aq) in the sample?
An aqueous solution of barium nitrate is added dropwise to a solution containing 0.10 M sulfate...
An aqueous solution of barium nitrate is added dropwise to a solution containing 0.10 M sulfate ions and 0.10 M fluoride ions. The Ksp value for barium sulfate = 1.1 x 10-10 and the Ksp value for barium fluoride = 1.7 x 10-6 a.) Which salt will precipitate from solution first? b.) What is the minimum [Ba2+] concentration necessary to precipitate the first salt? c.) What is the minimum [Ba2+] concentration necessary to precipitate the second salt? c.) What is...
Selective Precipitation Given: Ksp(Ag2CrO4) = 1.2×10-12 Ksp(BaCrO4) = 2.1×10-10 For a solution which is initially 0.003...
Selective Precipitation Given: Ksp(Ag2CrO4) = 1.2×10-12 Ksp(BaCrO4) = 2.1×10-10 For a solution which is initially 0.003 M in Ba2+ and 0.003 M in Ag+, it is desired to precipitate one of these ions as its chromate (CrO42−) salt to the maximum possible degree without precipitating any of the other cation. From this solution, which ion can be selectively first-precipitated by controlled addition of CrO42− and, ideally, what is the maximum possible percentage of that ion which can be exclusively so...
Given: Ksp(Ag2CrO4) = 1.2×10^-12 Ksp(BaCrO4) = 2.1×10^-10 For a solution which is initially 0.004 M in...
Given: Ksp(Ag2CrO4) = 1.2×10^-12 Ksp(BaCrO4) = 2.1×10^-10 For a solution which is initially 0.004 M in Ba2+ and 0.004 M in Ag+, it is desired to precipitate one of these ions as its chromate (CrO42−) salt to the maximum possible degree without precipitating any of the other cation. From this solution, which ion can be selectively first-precipitated by controlled addition of CrO42− and, ideally, what is the maximum possible percentage of that ion which can be exclusively so precipitated?
Lead ions can be precipitated from solution with KClaccording to the following reaction: Pb2+(aq)+2KCl(aq)→PbCl2(s)+2K+(aq) When 28.8...
Lead ions can be precipitated from solution with KClaccording to the following reaction: Pb2+(aq)+2KCl(aq)→PbCl2(s)+2K+(aq) When 28.8 g KCl is added to a solution containing 25.7 gPb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.1 g . 1. Determine the limiting reactant (Pb2+ or KCL) 2. Determine theoretical yield of PbCl2 3. Determine percent yield for the reaction
Ksp = 1.2 x 10-5 for silver sulfate, so the equilibrium: Ag2SO4(s) ⟺2Ag+(aq) + SO42-(aq) Note:...
Ksp = 1.2 x 10-5 for silver sulfate, so the equilibrium: Ag2SO4(s) ⟺2Ag+(aq) + SO42-(aq) Note: large concentration of ions in solution is > 1 small concentration of ions in solution is < 1 Group of answer choices lies far to the left and the concentration of ions in solution is small lies far to the left and the concentration of ions in solution is large lies far to the right and the concentration of ions in solution is small...