To Precipitate or not 6-6

Steps 1. Balanced chemical equation. 2. Find the molarity of the reactants CaVa = CbVb Va is the total volume ( add the volume of the reactants). Ca = (CbVb) /Va 3. Dissociate each reactant, find the concentration of each ion. 4. Write the Ksp expression of the solid. 5. Find the trial Ksp or Q. 6. Is Q > Ksp, if so a precipitate forms.

Will a precipitate of silver chloride form if one drop of (0 Will a precipitate of silver chloride form if one drop of (0.050 ml) of 5.0 M silver nitrate solution is added to 1.5 L of 0.20 M sodium chloride, if the Ksp is 1.8 x 10 -8 ? Balance the equation AgNO3 (aq) + NaCl (aq)→ AgCl (s) +NaNO3 (aq)

2. New concentration AgNO3 NaCl 0.05 ml of 5.0 M 1.5 l of 0.20 M Ca = (CbVb) /Va Ca = (CbVb) /Va =(0.00005 l)(5.0 mol/l) =(1.5 l)(0.20mol/l) 1.5 l 1.5 l ( add the two volumes together, the 0.00005 l is insignificant to the 1.5 l)

Molarity of reactants AgNO3 NaCl 0.00025 mol 0.30 mol 1.5 L 1.5 L = 1.67 x 10-4 mol/l = 0.20 mol/l

3. dissociation →[Ag+] + [NO3-] 1.67 x 10-4 M 1.67 x 10-4 M → [Na+] + [Cl-] 0.20 M 0.20 M

4. Write Ksp expression for silver chloride. AgCl (s) ------> Ag+ (aq) + Cl- (aq) 5. Solve for Q Ksp = [Ag + (aq) ] [Cl - (aq)] =( 1.67 x 10-4)(0.20) = 3.34 x 10-5

6. Compare Q to Ksp If Q > ksp 3.34 x 10-5 > 1.8 x 10-8 A ppt will form.

Determine if a precipitate of Ca(OH)2 will form is 20 ml of 0 Determine if a precipitate of Ca(OH)2 will form is 20 ml of 0.30 mol/l CaCl2 solution is mixed with 10 ml of 0.90 mol/l NaOH, if the Ksp for Ca(OH)2 is 5.1 X 10-6.

solution CaCl2 (aq) + 2 NaOH (aq) 20 ml of 0.30M 10 ml of 0.90M → Ca(OH)2 (s) + 2 NaCl (aq)

CaCl2 (aq) + 2 NaOH (aq) Ca = (CbVb) /Va Ca = (CbVb) /Va =(20 ml)( 0.030 mol/l) =(10 ml)(0.90mol/l) 30 ml 30 ml = 0.2 M = 0.30 M →Ca2+ + 2Cl- → 2 Na+ + 2 Cl- 0.2 0.4 0.30 0.30

Ksp expression for Ca(OH)2 is: Ca(OH)2 ----> Ca 2+ (aq) + 2 OH- (aq) Ksp = [Ca 2+ ] [OH-] 2 Q = ( 0.20)(0.30)2 Q = 0.018 Ksp = 5.1 x 10 -6 Q > Ksp therefore a ppt forms