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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.
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Solution
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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