Ionic product and predicting precipitates. What is the solubility of Pb(OH) 2 in 0.15 mol L –1 KOH? K s (Pb(OH) 2 ) = 6 × 10 –16. 1 Write the equilibrium.

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Presentation transcript:

Ionic product and predicting precipitates

What is the solubility of Pb(OH) 2 in 0.15 mol L –1 KOH? K s (Pb(OH) 2 ) = 6 × 10 –16. 1 Write the equilibrium equation and the expression for K s. 2 Rearrange the expression for K s to solve for [Pb 2+ ].

3 In 0.15 mol L –1 KOH, [OH – ] = 0.15 mol L –1. Substitute this value into the equation to calculate [Pb 2+ ]. Since [Pb 2+ ] = [Pb(OH) 2 ], the solubility of Pb(OH) 2 in 0.15 mol L –1 KOH is 2.7  10 –14 mol L –1.

Will a precipitate of MgCO 3 form when 20.0 mL of mol L –1 Na 2 CO 3 is added to 40.0 mL of 0.1 mol L –1 MgSO 4. K s (MgCO 3 ) = 1 × 10 –5. 1 Write the equilibrium expression and K s. 2 Work out the individual concentrations of each ion in the mixed solution.

3 Work out the ionic product by putting these values into the K s expression. 4 Compare the calculated IP to the K s to determine whether a precipitate will form. IP of 4.45 × 10 –4 is greater than K s of 1 × 10 –5 therefore a precipitate will form.