A solution contains 1.0 × 10-3 M Pb2*. Drops of a 5.0 x 10-1M solution of NaBr were added through a microburet until the [Br ] in the solution mix reached 3.0 x 10-5 M. Will a precipitate form? (Ksp (PbBr2) = 4.67 x 10-6 Neglect the slight change in volume resulting from the added NaBr solution.) No, a precipitate will not form. O Yes, a precipitate will form.

A solution contains 1.0 × 10-3 M Pb2*. Drops of a 5.0 x 10-1M solution of NaBr were added through a microburet until the [Br ] in the solution mix reached 3.0 x 10-5 M. Will a precipitate form? (Ksp (PbBr2) = 4.67 x 10-6 Neglect the slight change in volume resulting from the added NaBr solution.) No, a precipitate will not form. O Yes, a precipitate will form.

Chemistry by OpenStax (2015-05-04)

1st Edition

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Chapter15: Equilibria Of Other Reaction Classes

Section: Chapter Questions

Problem 11E: The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities...

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The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSeO4, 0.0118 g/100 mL. (b) Ba(BrO3)2H2O, 0.30 g/100 mL. (c) NH4MgAsO46H2O, 0.033 g/100 mL. (d) La2(MoO4)3, 0.00179 g/100 mL
The following solutions are mixed: 1.0 L of 0.00010 M NaOH and 1.0 L of 0.0014 M MgSO4. Is a precipitate expected? Explain.
Equal volumes of a 5.28 x 10-4 M nickel (II) nitrate solution and a 3.18 x 10-³ M sodium carbonate solution are mixed. The Ksp of nickel (II) carbonate is 1.42 x 10-7. Qsp = 4.20 x 10-7. A precipitate will not form Qsp = 1.68 x 10-6. A precipitate will form Qsp = 4.20 x 10-7. A precipitate will form Qsp = 1.68 x 10-6. A precipitate will not form Qsp cannot be calculated without knowing the specific volumes of the two solutions that were mixed
In order to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp= 1.57x10^-9What is the concentration of the excess reagent?What is the molar solubility of Ba(IO3)2 in this solution? 2.04x10^-5 M 4.07x10^-5 M 6.15x10^-5 M 8.30x10^-5 M
When 200.0 mL of 5.15 x 10-4 M Ba(NO3)2 is added to 150.0 mL of 8.25 x 10-4 M Na2SO4. Ksp(BaSO4) = 1.08 x 10-10. A precipitate choose your answer.... Ksp because Qsp choose your answer... choose your answer... SIGMA = form
4. Cuci ksp - 1.9x 10 Is it possible to precipitate 99.0 % of 0.010 M Ce3 by adding CrO,? without precipitating 0.010 M Ca*2 ? For CaC204 Kgo = 1.3 x 10% and for Cez(C2O4)3 Ksp = 3.0 x 10. 2.
You prepare a solution by mixing 400.0 mL of 1 x 10-4 M Mg(NO3)2 and 500.0 mL of 1.00 x 10-4 M NaF. What is the observed outcome? For MgF2, Ksp = 6.4 x 10-9. No precipitate forms because Qsp < Ksp. No precipitate forms because Qsp = Ksp. No precipitate forms because Qsp > Ksp. A precipitate forms because Qsp > Ksp. A precipitate forms because Qsp < Ksp.
will there be a precipitate in the mixture of 50 ml of Pb(NO3)2 5.0 x 10-3 mol/L with 100 ml of potassium thiocyanate 5.0 x 10-3 mol / l? kps = 2.0 x 10-5.
4 2+ 2- 4 100.0mL of 0.0010MCa(NO3)2 and 100.0mL of 0.0010MNa 2 HPOд are mixed. Determine if CaHPO will precipitate. Use Solubility Products as needed. First determine the concentrations of Ca and HPO present when the solutions are mixed. [Ca2+] = [M [HPO2*]=,]M Incorrect. One Ca2 ion is produced from the dissociation of one formula unit of 2+ Ca(NO3)2 Therefore the Ca 2+ 4 in the Ca(NO3)2 solution is 0.0010M. When the solutions are added together the volume doubles, so the concentration of Ca 2+ is halved, to 0.00050M. Calculate Q. Q= @
A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba+2 ions. At what F- concentration will BaF2 start to precipitate? Neglect volume changes. For BaF2, Ksp = 1.7 x 10-6 2.7 × 10-3 M 1.1 × 10-2 M 2.4 × 10-8 M 5.9 × 10-5 M 1.2 x 10-4 M
5) 3 L contaminated air 50 mL 0.0116 M in an air pollution analysis Carbon dioxide (CO2) BaCO3 is passed through Ba (OH) 2 solution. as precipitated. Excess of base, next to phenol phthalate (f.f.) indicator It is titrated with 23.6 mL of 0.0108 M HCl. CO2 in this air sample Calculate its concentration in ppm. (Density of CO2 Take it as 1.98 g / L. C = 12, O = 16 g / mol).
5. A mixture contains 1.0 x 10-³ M Cu²+ and 1.0 x 103 M Mn²+. We want to separate these ions by selective precipitation. Ksp(Cu(OH)2) = 1.6 x 10-19, Ksp(Mn(OH)2) = 2 x 10-13. a. What is the minimum concentration of OH required to precipitate Cu(OH)2 in this solution? \- b. What is the minimum concentration of OH required to precipitate Mn(OH)2 in this solution? If we were to add solid NaOH to the mixture, which cation would precipitate first and at which pH? d. At which pH would the second precipitate form? c.

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