working out ks from solubility
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
Working out Ks from solubility
• 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 [CO3
2-] = 5.01 x 10-6 mol L-1
Write the Ks expression. Remember do not include water or solids into the expression.
Ks = [Ca2+] [CO32-]
Substitute the numerical values for [Ca2+] and [CO3
2-] into the expression.
Ks = [5.01 x 10-6] [5.01 x 10-6]
Ks = [2.51 x 10-11]
• 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
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]