A solution is 5 mM in each of the following ions: number ion K sp of M(OH) 2 1 M
ID: 681224 • Letter: A
Question
A solution is 5 mM in each of the following ions:number ion Ksp of M(OH)2 1 Mg2+ 1.8e-11 2 Cd2+ 2.5e-14 3 Co2+ 1.6e-15 4 Zn2+ 4.4e-17 5 Cu2+ 2.2e-20Indicate which of the metal ions would precipitate (or start toprecipitate) at each of the following pH values. Indicate youranswer with the number of the ion. Use 0 to indicate noprecipitate. If more than one precipitate is expected, list thenumbers in increasing order and separate them with commas. Forexample, 3,4,5 is ok but 5,4,3 is not.
pH = 6.00:
pH = 9.00:
What is the pH to the nearest 0.1 pH unit at whichZn(OH)2 begins to precipitate?
pH =number ion Ksp of M(OH)2 1 Mg2+ 1.8e-11 2 Cd2+ 2.5e-14 3 Co2+ 1.6e-15 4 Zn2+ 4.4e-17 5 Cu2+ 2.2e-20
Explanation / Answer
1 ) For Mg(OH)2 Ksp = [Mg+2][HO-]2 1.8*10-11 = (5*10-3 )[HO-]2 Therfore [HO-] = 6.0*10-5 M This is the minium concentration to form theprecipitate. 2) For Cd(OH)2 Ksp = [Cd+2][HO-]2 2.5*10-14 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.23*10-6 M 3) For Co(OH)2 Ksp = [Co+2][HO-]2 1.6*10-15 = (5*10-3 )[HO-]2 Therfore [HO-] = 5.65*10-7 M 4) For Zn(OH)2 Ksp = [Zn+2][HO-]2 4.4*10-17 = (5*10-3 )[HO-]2 Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M 2.5*10-14 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.23*10-6 M 3) For Co(OH)2 Ksp = [Co+2][HO-]2 1.6*10-15 = (5*10-3 )[HO-]2 Therfore [HO-] = 5.65*10-7 M 4) For Zn(OH)2 Ksp = [Zn+2][HO-]2 4.4*10-17 = (5*10-3 )[HO-]2 Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M Therfore [HO-] = 2.23*10-6 M 3) For Co(OH)2 Ksp = [Co+2][HO-]2 1.6*10-15 = (5*10-3 )[HO-]2 Therfore [HO-] = 5.65*10-7 M 4) For Zn(OH)2 Ksp = [Zn+2][HO-]2 4.4*10-17 = (5*10-3 )[HO-]2 Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M 1.6*10-15 = (5*10-3 )[HO-]2 Therfore [HO-] = 5.65*10-7 M 4) For Zn(OH)2 Ksp = [Zn+2][HO-]2 4.4*10-17 = (5*10-3 )[HO-]2 Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M Therfore [HO-] = 5.65*10-7 M 4) For Zn(OH)2 Ksp = [Zn+2][HO-]2 4.4*10-17 = (5*10-3 )[HO-]2 Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M 4.4*10-17 = (5*10-3 )[HO-]2 Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M Therfore [HO-] = 9.38*10-8 M 5) For Cu(OH)2 Ksp = [Cu+2][HO-]2 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M 2.2*10-20 = (5*10-3 )[HO-]2 Therfore [HO-] = 2.09*10-9 M Therfore [HO-] = 2.09*10-9 M At pH = 6 , we getthe pOH = 8 [HO-] = 10-8 Therfore in the given ions Cu+2 does not forms theprecipitate.But the remaining ions forms the precipitation at thispH At pH= 9 , pOH = 5 Therefore [HO-] = 10-5M At this pH Any ion can not form the precipitate.Related Questions
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