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Working out Ks from solubility

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What is the Ks for CaCO 3 (s) when s(CaCO 3 (s))= 5.01 x 10 -6 mol L -1 ? Write the equation for the solid at equilibrium with water CaCO 3 (s) Ca 2+ + CO 3 2- Work out the concentrations of the aqueous ions. If the solubility of CaCO 3 (s)= 5.01 x 10 -6 mol L -1 [Ca 2+ ] = 5.01 x 10 -6 mol L -1 [CO 3 2- ] = 5.01 x 10 -6 mol L -1

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Write the Ks expression. Remember do not include water or solids into the expression. Ks = [Ca 2+ ] [CO 3 2- ] Substitute the numerical values for [Ca 2+ ] and [CO 3 2- ] into the expression. Ks = [5.01 x 10 -6 ] [5.01 x 10 -6 ] Ks = [2.51 x 10 -11 ]

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What is the Ks for Ag 2 S(s) 1.26 x 10 -17 moles of Ag 2 S(s) dissolves in 1 litre of water? Write the equation for the solid at equilibrium with water Ag 2 S(s) 2Ag + + S 2- Work out the concentrations of the aqueous ions. If the solubility of Ag 2 S(s)= 1.26 x 10 -17 mol L -1 [Ag + ] = 2 x s(Ag 2 S) = 1.26 x 10 -17 mol L -1 [S 2- ] = s(Ag2S) = 2.52 x 10 -17 mol L -1

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Write the Ks expression. Remember do not include water or solids into the expression. Ks = [Ag + ] 2 [S 2- ] Substitute the numerical values for [Ag + ] and [S 2- ] into the expression. Ks = [2.52 x 10 -17 ] 2 [1.26 x 10 -17 ] Ks = [8.00 x 10 -51 ]

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17.1 Liquid – vapour equilibrium

17.1 Liquid – vapour equilibrium

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