Question

The Kₛₚ of Ag₂CrO₄, AgCl, AgBr and AgI
are respectively, 1.1 × 10⁻¹², 1.8 × 10⁻¹⁰, 5.0 × 10⁻¹³, 8.3 × 10⁻
¹⁷. Which one of the following salts will precipitate last if
AgNO₃ solution is added to the solution containing equal
moles of NaCl, NaBr, NaI and Na₂CrO₄? [2015]
(a) AgCl (b) AgBr
(c) Ag₂CrO₄ (d) AgI

Answer 1

Answer:

gatsyya

Explanation:

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Answer 2

Ag2CrO4 will precipitate last because solubility product is less than ionic product.

Explanation:

Ksp of Ag2CrO4 = 1.1*10^-12

Ksp of AgCl= 1.8*10^-10

Ksp of AgBr= 5.0*10^-3

Ksp of AgI = 8.3*10^-17

From the data given of Ksp value we can know the ionic product.

The solubility product of Ag2Cr04

s=[2Ag]^+ [CrO4^2-]          where s is solubility

1-s=  2s    s

Ksp=2s^2*s/1-s

1.1*10^-12=2s^2*s/1-s

 1.1*10^12= 4s^3

             =1.3*10^-5 is the solubility of Ag2Cr04

solubility of AgI  is 9*10^-9

                   AgBr is 7.1*10^-9

So from the data it can be seen that Ag2CrO4 has highest solubility it will precipitate last.

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