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

See explaination

Step-by-step 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

Answer 2

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

`.[IO^(-3)]=1.228*10^(-5) M`

2) Concentration of NaIO3:

`.[NAIO3]`=
`2*10^(-4) M`

Step-by-step explanation:

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

The reaction will be:

`Pb(NO3)2` ⇒
`Pb^(+2) +2NO^(-3)`

`Concentration\ of\ Pb^(+2)=Pb(NO3)2=1.65*10^(-3)\ M`

Now,

`Pb(IO3)2` ⇄
`Pb^(+2)+2IO^(-3)`

Ksp=Concentration of
`Pb^(+2)` * (Concentration of
`2IO^(-3)`)^2

`Ksp=[Pb^(+2)]`*
`[2IO^(-3)]^2`

`2.5*10^(-13)=1.65*10^(-3)*[IO^(-3)]^2\\.[IO^(-3)]^2=(2.5*10^(-13))/(1.65*10^(-3)) \\.[IO^(-3)]^2=1.51*10^(-10) M\\.[IO^(-3)]=1.228*10^(-5) M`

2) Concentration of NaIO3:

Now,

[
`Pb^(+2)`]=
`5.60*10^(-6) M`

`Ksp=[Pb^(+2)]*[2IO^(-3)]^2`

`2.5*10^(-13)=5.60*10^(-6) *[IO^(-3)]^2\\.[IO^(-3)]^2=(2.5*10^(-13))/(5.60*10^(-6)) \\.[IO^(-3)]^2=4.46*10^(-8) M\\.[IO^(-3)]=2.112*10^(-4) M`

Again:

`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`

Calculating the concentration of
`IO^(-3)` from NaIO3:

`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`

Reaction:

`NaIO3=>Na^(+)+IO^(-3)`

Concentation of NaIO3= Concentration of
`IO^(-3)`

`.[NAIO3]`=
`2*10^(-4) M`