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Predicting solubility of a salt in a solvent

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The solubility product of silver chromate, Ag2CrO4, at 25 degrees Celsius is 4.05 x 10^-12 mol^3/L^3. Calculate the solubility of silver chromate in:
a) pure water
b) 0.0500 M silver nitrate solution
c) 0.0500 M potassium chromate solution

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The molar solubility product Ksp = [Ag+]^2 * [CrO4 ^2-] ----(1)
<br>
<br>for the reaction Ag2CrO4(s) ---&gt; 2Ag^+ + CrO4^2-
<br>
<br>and the molar solubility, or the molar concentration of silver chromate going in to the solution, is equal to [CrO4^2-] and is equal to [Ag+]/2
<br>
<br>Or in terms of the concentraion of the chromate ion,
<br>
<br>Ksp = {2[CrO4^2-]}^2 * [CrO4^2-] = 4 ...

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