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# What is the solubility of Ag2CrO4 in 0.10 M K2CrO4?

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- ?Lv 72 months ago
Ag2CrO4(s) = 2 Ag+(aq) + CrO4 2- (aq) Ksp = 1.2 x 10-12

We have for a solution of potassium chromate 0.1M

Ksp=[Ag+]^2*[CrO24−]

if s moles of the salt dissolves we get

(2s)2*(s+0.1) = 1.2×10−12

since s is small compared with 0,1 you can write:

Ksp = 4s^2 *.1 =0.4s^2 = 1,2×10−12

=SQRT( 1.2×10−12 / 0.4)

= 1.73*10^-6 moles/ liter

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