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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.

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प्रश्न

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.

संख्यात्मक
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उत्तर

Let s be the solubility of Ag2CrO4.

Then \[\ce{Ag_2CrO_4 <-> Ag^{2+} + 2 CrO_4-}\]

`"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.

\[\ce{AgBr_{(s)} <=> Ag^+ + Br-}\]

`"K"_("sp") = "s"^('2) = 5.0 xx 10^(-13)`

`therefore "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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Concept of Solubility Equilibria of Sparingly Soluble Salts
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अध्याय 6: Equilibrium - EXERCISES [पृष्ठ २३८]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 11
अध्याय 6 Equilibrium
EXERCISES | Q 7.68 | पृष्ठ २३८

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