Calculate the value of Q and determine if a precipitate form when 01.00 mg of (NH4)2CO3 ismixed with 2.50 mg of CaBr2 in 1.0 L of solution? Ksp of CACO3 = 4.5 X 109CACO3 5 Ca²+ + CO?+02.1x10-902.1x10-601.3x10-1004.6x10-9ONo because Q < KspOYes because Q > KspOYes because Q < KspONo because Q > Ksp

The solubility reaction of CaCO3 is given as, => CaCO3 (s) --------> Ca2+ (aq) + CO32- (aq) Ksp = 4.5 X 10-9 Given : Mass of (NH4)2CO3 = 1.00 mg = 0.001 g (since 1 g = 1000 mg) Mass of CaBr2 = 2.50 mg = 0.0025 g And volume of solution = 1.0 L Molar mass of (NH4)2CO3 = Atomic mass of N X 2 + Atomic mass of H X 8 + Atomic mass of C + Atomic mass of O X 3 => Molar mass of (NH4)2CO3 = 14 X 2 + 1 X 8 + 12 + 16 X 3 = 96 g/mol. And molar mass of CaBr2 = Atomic mass of Ca + Atomic mass of Br X 2 = 40 + 79.9 X 2 = 199.8 g/mol. Since mass = moles X molar mass => 0.001 = moles of (NH4)2CO3 X 96 => Moles of (NH4)2CO3 = 1.0417 X 10-5 mol approx. And 0.0025 = moles of CaBr2 X 199.8 => Moles of CaBr2 = 1.2513 X 10-5 mol approx. Since moles = concentration X volume of solution in L And volume of solution is given 1.0 L. Hence moles = concentration. => Concentration of CaBr2 = 1.2513 X 10-5 M And concentration of (NH4)2CO3 = 1.0417 X 10-5 MSince each molecule of (NH4)2CO3 contains 1 CO32- ion and each molecule of CaBr2 contains 1 Ca2+ ion. Hence concentration of CO32- ions, [CO32- ] = concentration of (NH4)2CO3 = 1.0417 X 10-5 M And concentration of Ca2+ ions, [Ca2+ ] = concentration of CaBr2 = 1.2513 X 10-5 M The reaction quotient of the solubility reaction is given by, => Q = [CO32- ] X [Ca2+ ] = 1.2513 X 10-5 X 1.0417 X 10-5 = 1.30 X 10-10 . Since the value of Q < Ksp. Hence the reaction will proceed in forward direction producing only ions and not precipitate. Hence there will be no precipitation because Q < Ksp.

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Chemistry

Calculate the value of Q and determine if a precipitate form when 01.00 mg of (NH4)2CO3 is mixed with 2.50 mg of CaBr2 in 1.0 L of solution? Ksp of CaCO3 = 4.5 X 10-9 CaCO3 = Ca²+ + Co?+ 02.1x10-9 02.1x10-6 01.3x10-10 04.6x10-9 ONo because Q < Ksp O Yes because Q > Ksp O Yes because Q < Ksp ONo because Q > Ksp

Calculate the value of Q and determine if a precipitate form when 01.00 mg of (NH4)2CO3 is mixed with 2.50 mg of CaBr2 in 1.0 L of solution? Ksp of CaCO3 = 4.5 X 10-9 CaCO3 = Ca²+ + Co?+ 02.1x10-9 02.1x10-6 01.3x10-10 04.6x10-9 ONo because Q < Ksp O Yes because Q > Ksp O Yes because Q < Ksp ONo because Q > Ksp

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

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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12. Please write clearly
26
Ksp and Precipitation
A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of the metal ions from solution. At what concentration of sulfide ion will a precipitate begin to form? What is the identity of the precipitate? Ksp(CuS) = 1.3 × 10-36, Ksp(FES) = 6.3 × 10-18 == 3.6 × 10-35 M, FeS 3.6 x 10-35 M, CuS 1.4 x 10-16 M, CUS 1.4 × 10-16 M. FeS No precipitate will form at any concentration of sulfide ion.