Question

Calculate the concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution saturated with Pb(IO3)2 . The Ksp of Pb(IO3)2 is 2.5×10−13 . Assume that Pb(IO3)2 is a negligible source of Pb2+ compared to Pb(NO3)2 . [IO−3]= M A different solution contains dissolved NaIO3 . What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.60×10−6 M ?

Answer 1

Answer:

See explaination

Explanation:

1) Pb(NO3)2 => Pb2+ + 2 NO3-

[Pb2+] = [Pb(NO3)2] = 7.56 mM = 7.56 x 10-3 M

Pb(IO3)2 <=> Pb2+ + 2 IO3-

Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13

7.56 x 10-3 x [IO3-]2 = 2.5 x 10-13

[IO3-] = 5.75 x 10-6 M ≈ 5.8 x 10-6 M

(2) [Pb2+] = 1.7 x 10-6 M

Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13

1.7 x 10-6 x [IO3-]2 = 2.5 x 10-13

[IO3-] = 3.83 x 10-4 M

[IO3-]from Pb(IO3)2 = 2 x [Pb2+]

= 2 x 1.7 x 10-6 = 3.4 x 10-6 M

[IO3-]from NaIO3 = [IO3-] - [IO3-]from Pb(IO3)2

= 3.83 x 10-4 - 3.4 x 10-6

= 3.80 x 10-4 M

NaIO3 => Na+ + IO3-

[NaIO3] = [IO3-]from NaIO3

= 3.80 x 10-4 M ≈ 3.8 x 10-4 M