A different solution contains dissolved NaIO3. What is the concentration of NaIO

Question

A different solution contains dissolved NaIO3. What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces [Pb2 ] = 4.60 × 10-6 M?

Calculate the concentration of IO3 in a 1.72 mM Pb(NO3)2 solution saturated with Pb(IO3)2. The Ksp of Pb(IO3)2 is 2.5 x1018. Assume that Pb (IO3)2 is a negligible source of Pb2* compared to Pb(NO3)2. Number A different solution contains dissolved NalO3. What is the concentration of NalO3 if adding excess Pb(IO3)2(s) produces [Pb2+] = 4.60 x 10-6 M? Number Nalo,I=

Explanation / Answer

1) Pb(IO3)2Pb2+ + 2 IO3-

ksp(Pb(IO3)2) =[Pb2+][IO3-]^2

given ksp(Pb(IO3)2)=2.5*10^-13 M^3

[Pb2+]=1.72 mM=1.72*10^-3M

[IO3-]^2= ksp(Pb(IO3)2)/[Pb2+]=2.5*10^-13/1.72*10^-3=1.45*10^-10

[IO3-]= square root (1.45*10^-10)=1.2*10^-5 M(answer)

2) ksp(Pb(IO3)2) =[Pb2+][IO3-]^2

Given

[Pb2+]=4.60*10^-6M

[IO3-]^2= ksp(Pb(IO3)2)/[Pb2+]=2.5*10^-13/4.6*10^-6=0.54*10^-6

[IO3-]= square root (0.54*10^-6)=0.74*10^-3 =7.4*10^-4M=[IO3-] in the solution

[IO3-] in the solution=[IO3-] produced by partial dissociation of Pb(IO3)2 + [IO3-] produced by partial dissociation of NaIO3

But [IO3-] produced by partial dissociation of Pb(IO3)2=2*[Pb2+]=2*4.60*10^-6M=9.2*10^-6M

So, 7.4*10^-4M=9.2*10^-6M+[IO3-] produced by partial dissociation of NaIO3

[IO3-] produced by partial dissociation of NaIO3=7.4*10^-4M-9.2*10^-6M=7.4*10^-4M-0.092 *10^-4=7.31*10^-4M

[NaIO3]= [IO3-] produced by partial dissociation of NaIO3=7.31*10^-4M

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