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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 ?

Respuesta :

kendrich

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

zainsubhani

Answer:

1) Concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution:

[tex].[IO^{-3}]=1.228*10^{-5} M[/tex]

2) Concentration of NaIO3:

[tex].[NAIO3][/tex]=[tex]2*10^{-4} M[/tex]

Explanation:

1) Concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution:

The reaction will be:

[tex]Pb(NO3)2[/tex] ⇒ [tex]Pb^{+2} +2NO^{-3}[/tex]

[tex]Concentration\ of\ Pb^{+2}=Pb(NO3)2=1.65*10^{-3}\ M[/tex]

Now,

[tex]Pb(IO3)2[/tex] ⇄ [tex]Pb^{+2}+2IO^{-3}[/tex]

Ksp=Concentration of [tex]Pb^{+2}[/tex] * (Concentration of [tex]2IO^{-3}[/tex])^2

[tex]Ksp=[Pb^{+2}][/tex]*[tex][2IO^{-3}]^2[/tex]

[tex]2.5*10^{-13}=1.65*10^{-3}*[IO^{-3}]^2\\.[IO^{-3}]^2=\frac{2.5*10^{-13}}{1.65*10^{-3}} \\.[IO^{-3}]^2=1.51*10^{-10} M\\.[IO^{-3}]=1.228*10^{-5} M[/tex]

2) Concentration of NaIO3:

Now,

[[tex]Pb^{+2}[/tex]]=[tex]5.60*10^{-6} M[/tex]

[tex]Ksp=[Pb^{+2}]*[2IO^{-3}]^2[/tex]

[tex]2.5*10^{-13}=5.60*10^{-6} *[IO^{-3}]^2\\.[IO^{-3}]^2=\frac{2.5*10^{-13}}{5.60*10^{-6}} \\.[IO^{-3}]^2=4.46*10^{-8} M\\.[IO^{-3}]=2.112*10^{-4} M[/tex]

Again:

[tex]Concentration\ of\ IO^{-3} from\ Pb(IO3)2 = 2* Concentration\ of\ Pb^{+2}\\.[IO^{-3}]_{From\ Pb(IO3)2}=2*5.60*10^{-6}\\.[IO^{-3}]_{From\ Pb(IO3)2}=1.12*10^{-5} M[/tex]

Calculating the concentration of  [tex]IO^{-3}[/tex] from NaIO3:

[tex]Concentration\ of\ IO^{-3}_{[From\ NaIO3}]=[IO^{-3}]-Concentration\ of\ IO^{-3}_{[From\ Pb(IO3)2}]\\Concentration\ of\ IO^{-3}_{[From\ NaIO3}]=2.112*10^{-4}-1.12*10^{-5}\\Concentration\ of\ IO^{-3}_{[From\ NaIO3}]=2*10^{-4} M[/tex]

Reaction:

[tex]NaIO3=>Na^{+}+IO^{-3}[/tex]

Concentation of NaIO3= Concentration of [tex]IO^{-3}[/tex]

[tex].[NAIO3][/tex]=[tex]2*10^{-4} M[/tex]

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