Consider the reaction CaSO4(s) -->Ca2+(aq) +SO42-(aq) At 25 oC the equilibrium c

Question

Consider the reaction CaSO4(s) -->Ca2+(aq) +SO42-(aq) At 25 oC the equilibrium constant is Kc= 2.4 x 10-5 for this reaction. If the resulting solution has a volume of 1.2 L, whatis the minimum mass of CaS)4(s) needed to achieve equilibrium?

Explanation / Answer

let the minimum mass required be 'x' grams i.e. x grams of CaSO4 dissolves to give the aqueous solution. Molecular mass of CaSO4 = 136.146 gm/mole => number of moles = x/(136.146) moles => number of moles of [Ca2+] formed = x/(136.146) = number of moles of [SO42-] formed => Concentration of [Ca2+]= x/(136.146*1.2) = Concentration of [SO42-] Hence, Kc = [Ca2+][SO42-] So, 2.4*10^-5 = [x/(136.146*1.2)]^2 => x = 0.8003 grams

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