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Working out Ks from solubility

Working out Ks from solubility. 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.

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Working out Ks from solubility

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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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