At 25 degrees celcius CaSO4(s)<--->Ca2+(aq)+CrO42-(aq) has an equilibrium consta

Question

At 25 degrees celcius CaSO4(s)<--->Ca2+(aq)+CrO42-(aq) has an equilibrium constant of 7.1*10^-4. if the reaction occurs in a 400mL solution what mass of calcium chromate dissolves?

Explanation / Answer

[Its not CaSO4...its CaCrO4.]************** Reaction : CaCrO4----->Ca2+ + CrO42-*****at time =0 (i.e, at start), let there be 1 mole of CaCrO4 and zero moles of Ca+2 and CrO42-*****after sufficient time, when equilibrium is established, then ,say, X fraction of total amount of CaCrO4 has been dissociated.i.e.,(1* X) mole of CaCrO4 has been dissociated . So then, amount of CaCrO4 remaining is (1-1*X) moles=(1-X) moles. And amount of Ca2+ and Cro42- produced is X moles for each.********** Now, we know, K(eqm. Const.)=[Ca+2][CrO42-]/[CaCrO4]********Here [Ca+2]=[CrO42-]=(X/400*1000) moles/litre******[CaCrO4]= {(1-X)/400*1000} moles/litre******so we get, K=[{(X^2)*(1000)}/{(1-X)*(400)}]*****putting K=7.1*10^-4, we get X=0.0497352 moles******Mol. Wt. of CaCrO4=156 g/mol****hence mass of Calcium Chromate dissolved=156*0.0497352 gram=7.7587 gram (Ans.).

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