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Silver ion is being considered as a reagent for separating 1037 from CO3 in a solution that is 0.060 M in K2CO3 and 0.070 M i
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Answer #1

i)

Solubility equilibrium of AgIO3 is

AgIO3(s) <-------> Ag+(aq) + IO3-(aq)

Ksp = [Ag+] [IO3-] = 3.1×10-8

[Ag+] × 0.070M= 3.1 ×10-8M2

[Ag+]= 4.43 × 10-7M

Therefore,

minimum concentration of Ag+ required to start precipitate IO3- is 4.43 ×10-7M

solubility equilibrium of Ag2CO3

Ag2CO3(s) <-------> 2Ag+(aq) + CO32-(aq)

Ksp = [Ag+]2[CO32-] = 8.1×10-12

[Ag+]2× 0.060M = 8.1 ×10-12M3

[Ag+]2 = 1.33×10-10M2

[Ag+] = 1.15 × 10-5M

minimum concentration of Ag+ required to start precipitate of CO32- is 1.15 ×10-5M

IO3- ​​​​​​ion require low concentration of Ag+ ​​​​​​

Therefore,

IO3- would precipitate first

ii)

Substitute the concentration Ag+ required start precipitate CO32- in Ksp expression of AgIO3

[Ag+][IO3-] = 3.1 ×10-8

1.15 ×10-5 M × [IO3-] = 3.1 ×10-8M2

[IO3-] = 2.70× 10-3M

Therefore,

Concentration of IO3- when CO32- start to precipitate =

2.70 × 10-3M

iii)

℅ of IO3- remaining when CO32- start to precipitate = (0.00270M/0.070M)× 100 = 3.86%

Therefore

these two ions can not be seperated with 99.99% efficiency

  

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