Strontium chromate, SrCrO4, has a Ksp value of 4.00×10−5 . What happens when strontium and chromate solutions are mixed to give 2.00×10−2 M Sr2+ and 1.50×10−3 M CrO42−?
SrCrO4(s) <-----> Sr2+(aq) + CrO42-(aq)
Ksp = [Sr2+]*[CrO42-]
Now, when [Sr2+] = 2*10-2 M & [CrO42-] = 1.5*10-3 M
Then reaction quotient, Kq = [Sr2+]*[CrO42-] = 3*10-5
Since, Kq < Ksp , therefore dissolution will occur
Strontium chromate, SrCrO4, has a Ksp value of 4.00×10−5 . What happens when strontium and chromate solutions are mixed...
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What is the value of Q when the SrCrO4 solution contains 2.00×10−2 M Sr2+ and 1.50×10−3M CrO42−? Calculate the value of Q.
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