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Solubility Products Calculations for CaCO3 and Ag2S in Water

Learn how to calculate the solubility product constants (Ks) for CaCO3 and Ag2S in water based on given concentrations of the solid compounds. Understand the equations for the solids at equilibrium and work out the concentrations of the aqueous ions involved. Practice calculating Ks and applying numerical values to determine the solubility product constants.

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Solubility Products Calculations for CaCO3 and Ag2S in Water

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  1. Working out Ks from solubility

  2. What is the Ks for CaCO3(s) when s(CaCO3(s))= 5.01 x 10-6 mol L-1 ? Write the equation for the solid at equilibrium with water CaCO3(s)  Ca2+ + CO32- Work out the concentrations of the aqueous ions. If the solubility of CaCO3(s)= 5.01 x 10-6 mol L-1 [Ca2+] = 5.01 x 10-6 mol L-1 [CO32-] =5.01 x 10-6 mol L-1

  3. Write the Ks expression. Remember do not include water or solids into the expression. Ks = [Ca2+][CO32-] Substitute the numerical values for [Ca2+] and [CO32-] into the expression. Ks = [5.01 x 10-6][5.01 x 10-6] Ks = [2.51 x 10-11]

  4. What is the Ks for Ag2S(s) 1.26 x 10-17 moles of Ag2S(s) dissolves in 1 litre of water? Write the equation for the solid at equilibrium with water Ag2S(s)  2Ag+ + S2- Work out the concentrations of the aqueous ions. If the solubility of Ag2S(s)= 1.26 x 10-17 mol L-1 [Ag+] = 2 x s(Ag2S) = 1.26 x 10-17 mol L-1 [S2-] =s(Ag2S)= 2.52 x 10-17 mol L-1

  5. Write the Ks expression. Remember do not include water or solids into the expression. Ks = [Ag+]2 [S2-] Substitute the numerical values for [Ag+] and [S2-] 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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