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The Solubility Product Constant of Ag2cro4 and Agbr Are 1.1 × 10–12 and 5.0 × 10–13 Respectively. Calculate the Ratio of the Molarities of Their Saturated Solutions. - CBSE (Science) Class 11 - Chemistry

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Question

The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13respectively. Calculate the ratio of the molarities of their saturated solutions.

Solution

Let s be the solubility of Ag2CrO4.

Then `Ag_2CrO_4 "↔" Ag^(2+) + 2CrO_4^(-)`

`K_(sp) = (2s)^2 s = 4s^3`

`1.1 xx 10^(-12) = 4s^3`

`s =   6.5 xx 10^(-5) M`

Let s´ be the solubility of AgBr.

`AgBr_(s) ↔ Ag^+ + Br^(-)`

`K_(sp) = s^('2) = 5.0 xx 10^(-13)`

`:. s' = 7.07 xx 10^(-7) M`

Therefore, the ratio of the molarities of their saturated solution is s/s' = `(6.5 xx 10^(-5) M)/(7.07 xx 10^(-7)M)` = 91.9

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Solution The Solubility Product Constant of Ag2cro4 and Agbr Are 1.1 × 10–12 and 5.0 × 10–13 Respectively. Calculate the Ratio of the Molarities of Their Saturated Solutions. Concept: Concept of Solubility Equilibria of Sparingly Soluble Salts.
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