K„ is the solubility product. It is just a normal K for a solubility reaction like the one below. Ba, (AsO,),(s) 3 Ba" (aq) + 2 AsO (aq) Km =1.1 x 10-13 What is the equilibrium concentration of Ba+ when 30.0 g of Ba, (AsO,), (s) is placed in 250 mL of water?

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How I find the equilibrium constant? 

I was able to get K to be 0.001789 by having Ba and AsO = to S. 

I got 1.1e-13= [3S]^3 [2S]^2 so S=0.001789 then I plugged that in to find my Ba but got it wrong. 

Ksp is the solubility product. It is just a normal K for a solubility reaction like the one below.
Ba, (AsO,), (s) =
3 Ba?+ (aq) + 2 AsO (ag)
Kp = 1.1 x 1013
What is the equilibrium concentration of Ba?+ when 30.0 g of Ba, (AsO,), (s) is placed in 250 mL of water?
Transcribed Image Text:Ksp is the solubility product. It is just a normal K for a solubility reaction like the one below. Ba, (AsO,), (s) = 3 Ba?+ (aq) + 2 AsO (ag) Kp = 1.1 x 1013 What is the equilibrium concentration of Ba?+ when 30.0 g of Ba, (AsO,), (s) is placed in 250 mL of water?
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