Question

The mercury(I) ion Hg22+, (also called the mercurous ion) is a diatomic ion with a total...

The mercury(I) ion Hg22+, (also called the mercurous ion) is a diatomic ion with a total charge of +2. Mercury(I) iodate (MW=750.99) dissociates into three ions:

Hg2(IO3)2(s) = Hg22+(aq) + 2IO3-(aq) Ksp = 1.3*10-18

A) what is the solubility of Hg2(IO3)2 in pure water?

B) what is the solubility of Hg2(IO3)2 in 0.010 M NaIO3?

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Answer #1

A)The solubility equilibrium is

Hg2(IO3)2 (s) <------------> Hg22+ (aq) +2 IO3-2(aq)

s 0 0 initial concentration

- s 2s at equilibrium

where s = solubility of salt

Then Ksp = sx(2s)2

1.3x10-18 = 4s3

Thus solubiity s = 6.875x10-7 mol/L

B) IN 0.010 M NaIO3

Hg2(IO3)2 (s) <------------> Hg22+ (aq) +2 IO3-2(aq)

x 0 0 initial concentration

- x 2x +0.01 at equilibrium

where x is the solubility in presence of sodium iodate

Since x is very small compared to 0.01 , we can take 2x +0.01 = 0.01

Thus Ksp = (x) (0.01) = 1.3x10-18

thus solubility x = 1.3 x10-16 mol/L in presence of sodium iodate

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