Suppose you are performing a precipitation titration to study the K sp of zinc(II) iodate, Z n ( I O 3 ) 2 . The literature value of K sp is 3.9 × 10 − 6 . If you use 0.200 M K I O 3 as the titrant, what concentration of I O 3 − (in M) do you predict will be needed to start precipitation in a 0.229 M solution of Z n ( N O 3 ) 2 ?

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2021-11-30T11:10:52+01:00

The predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M

The dissociation of zinc (II) iodate can be expressed as:

[tex]\mathbf{Zn (IO_3)_2 \to Zn^{2+} + 2IO_3^-}[/tex]

Given that the solubility product constant Ksp value = 3.9 × 10⁻⁶

For the above dissociation,

[tex]\mathbf{Ksp = [Zn^{2+ }] [IO_3^-]^2}[/tex]

Since [tex]\mathbf{ [Zn^{2+ }] = [Zn(NO_3)_2] = 0.229 \ M}[/tex]

∴

[tex]\mathbf{3.9\times 10^{-8} =(0.229) \times [IO_3^-]^2}[/tex]

[tex]\mathbf{ [IO_3^-]^2 = \dfrac{3.9\times 10^{-8} }{(0.229)}}[/tex]

[tex]\mathbf{ [IO_3^-]^2 = 1.703 \times 10^{-7} \ M}[/tex]

Therefore, we can conclude that the initial predicted concentration of IO₃⁻ needed to start the precipitation titration is 1.703 × 10⁻⁷ M

Learn more about solubility product constant here:

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