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Thalium (I) Iodate (TlIO3) is only slightly soluble in water. Its KSP at 250c is 3.07X10-6. Estimate the solubility of Thalium (I) iodate in water in units of grams per 100.00 ml of water.

Short Answer

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Thallium(I) Iodate and solubility

  • Thallium(I) iodate is added to mercury are lamps to improve their performance. This mainly used for underwater lighting.
  • The solubility of most solid and liquid increases with increasing temperature.

Step by step solution

01

Step(1)- [Answer of the question]

Given:-KSP at 25oc is 3.07X10-6 .

The solubility of thalium (I) iodate in water in units of grams per 100.00ml of water is 0.0664gm.

02

Step(2):- Calculation of the solubility of Thalium Iodate

The solubility reaction of Thalium(I) iodate-

TlIO3sTl+aq+IO3-aq

So, the solubility product is-

role="math" localid="1663579923815" 3.07×10-6=S2S2=3.07×10-6S=3.07×10-6S=1.75×10-3mol/liter379.29gm/molTlIO3KSP=Tl+'XIO3-'

Let, S if (TlIO3) dissolves. Then the equilibrium concentration of Tl+ and IO3- will be:-

[Tl+]' = S and [IO3-]' =S

Then, KSP = S X S

3.07×10-6=S2S2=3.07×10-6S=3.07×10-6S=1.75×10-3

The molar solubility is 1.75X10-3mol/litre .

03

Step-3: Calculation of the solubility of thalium(I) iodate

The molar mass of (TlIO3) is 379.29gm/mol

Gram solubility= molar solubility X molar mass

Gram solubility = ( 379.29gm/mol ) X ( 1.75X10-3mol/litre)

= 663.72X10-3gm/lit

Thus, the solubility in gram per 100ml in water is:-

Gram/100ml solubility= 663.75×10-31000mlX1LXgL

= 663.75×10-3100ml

= 0.0664gm/100ml

04

Step-4:- Conclusion

Hence, gram solubility represents the solubility of thalium iodate in units of gm/100ml.

Hence the solubility is 0.0664gm/100ml.

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