macmillan learning Calculate the concentration of IO3 in a 5.89 mM Pb(NO3)2 solu
ID: 540360 • Letter: M
Question
macmillan learning Calculate the concentration of IO3 in a 5.89 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(l03)2 is 2.5 x 10-3. Assume that Pb(IO3)2 is a negligible source of Pb2 compared to Pb(NO3)2 Number different solution contains dissolved NalO3. What is the concentration of NalO3 if adding excess Pb(I03)2(s) produces [Pb2-8.00 x106 M? Incorrect. You have found [103 1. 1.8x10 M, which is the sum of IO3 from Pb(IO3)2 and NalO3. You know that for every mole of Pb2 there are two moles of l03 in solution, and for every mole of NalO3 dissolved, one mole of IO3 is produced Number los l=Iloil from Nalo 3 + 110, ] from Pb(103) You know [IOS] and [Pb21. Solve for [NaIOI Previous Give Up & View Solution # Try Again Next ExitExplanation / Answer
Pb(NO3)2 suppliments Pb+2, given [Pb+2] from Pb(NO3)2= 5.89mM= 5.89*10-3 M
Pb(IO3)2 ---------->Pb+2+ 2[IO3-1]
KSp = [Pb+2] [IO3-1]2
2.5*10-13= (5.89*10-3)[IO3-1]2, [IO3-2] = 6.51*10-6M
2. NaIO3 suppliments [IO3-]
given [Pb+2] = 8*10-6
KSp = 2.5*10-13= 8*10-6 [IO3-]2, [IO3-]= 1.8*10-4 ,this consits if [IO3-] from NaIO3 and 2 times [IO3-] from previous calculations
1.8*10-4 = [NaIO3] +2*6.51*10-6
[NaIO3]= 0.000164 M
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