1 What weight of sulfur (FW = ) ore which should be taken so that the weight of BaSO 4 (FW = ) precipitate will be equal to half of the percentage sulfur in the sample. Solution S  BaSO 4 mmol S = mmol BaSO 4 mg S/at wt S = mg BaSO 4 /FW BaSO 4 mg S = at wt S x ( mg BaSO 4 / FW BaSO 4 ) mg S = x (1/2 %S/233.40) mg S = %S %S = (mg S/mg sample) x 100 %S = %S/mg sample) x 100 mg sample = %S x 100 /%S = mg

2 A mixture containing only FeCl 3 (FW = 162.2) and AlCl 3 (FW = ) weighs 5.95 g. The chlorides are converted to hydroxides and ignited to Fe 2 O 3 (FW = 159.7) and Al 2 O 3 (FW = ). The oxide mixture weighs 2.62 g. Calculate the percentage Fe (at wt = 55.85) and Al (at wt = 26.98) in the sample. Solution Fe  FeCl 3 1 mol Fe = 1 mol FeCl 3 g Fe/at wt Fe = g FeCl 3 / FW FeCl 3 Rearrangement gives g FeCl 3 = g Fe (FW FeCl 3 /at wt Fe)

3 In the same manner g AlCl 3 = g Al ( FW AlCl 3 /at wt Al) g FeCl 3 + g AlCl 3 = 5.95 g Fe (FW FeCl 3 /at wt Fe) + g Al ( FW AlCl 3 /at wt Al) = 5.95 assume g Fe = x, g Al = y then: x (FW FeCl 3 /at wt Fe) + y ( FW AlCl 3 /at wt Al) = 5.95 x (162.2/55.85) + y (133.34/26.98) = x y = 5.95 (1)

4 The same treatment with the oxides gives 2 Fe  Fe 2 O 3 mol Fe = 2 mol Fe 2 O 3 g Fe/at wt Fe = 2 (g Fe 2 O 3 /FW Fe 2 O 3 ) g Fe 2 O 3 = 1/2 g Fe (FW Fe 2 O 3 /at wt Fe) In the same manner g Al 2 O 3 = 1/2 g Al (FW Al 2 O 3 /at wt Al) g Fe 2 O 3 + g Al 2 O 3 = /2 g Fe (FW Fe 2 O 3 /at wt Fe) + 1/2 g Al (FW Al 2 O 3 /at wt Al) = /2 x (159.7/55.85) + 1/2 y (101.96/26.98) = x y = 2.62 (2)

x y = 5.95 (1) 1.43 x y = 2.62 (2) from (1) and (2) we get x = 1.07 y = 0.58 % Fe = (1.07/5.95) x 100 = 18.0% % Al = (0.58/5.95) x 100 = 9.8%

6 Problem Consider a g sample containing 75% potassium sulfate (FW ) and 25% MSO 4. The sample is dissolved and the sulfate is precipated as BaSO 4 (FW ). If the BaSO 4 ppt weighs g, what is the atomic weight of M 2+ in MSO 4 ? K 2 SO 4  BaSO 4 MSO 4  BaSO 4

7 mmol BaSO 4 = mmol K 2 SO 4 + mmol MSO 4 Rearranging and solving:

8 A mixture of mercurous chloride (FW ) and mercurous bromide (FW ) weighs 2.00 g. The mixture is quantitatively reduced to mercury metal (At wt ) which weighs 1.50 g. Calculate the % mercurous chloride and mercurous bromide in the original mixture. Hg 2 Cl 2  2Hg Hg 2 Br 2  2Hg

9 Answer mmol Hg = 2 mmol Hg 2 Cl mmol Hg 2 Br 2 Rearranging and solving:

10 A 10.0 mL solution containing Cl - was treated with excess AgNO 3 to precipitate g of AgCl. What was the concentration of Cl - in the unknown? (AgCl = g/mol) mmoles of Cl - = mmoles of AgCl Concentration of Cl -

11 Precipitation Equilibria Inorganic solids which have limited water solubility show an equilibrium in solution represented by the so called solubility product. For example, AgCl slightly dissolve in water giving Cl - and Ag + where AgCl (s)  Ag + + Cl - K = [Ag + ][Cl - ]/[AgCl (s) ] However, the concentration of a solid is constant and the equilibrium constant can include the concentration of the solid and thus is referred to, in this case, as the solubility product, k sp where K sp = [Ag + ][Cl - ]

12 It should be clear that the product of the ions raised to appropriate power as the number of moles will fit in one of three cases: 1. When the product is less than K sp : No precipitate is formed and we have a clear solution 2. When the product is equal to k sp : We have a saturated solution 3. When the product exceeds the k sp : A precipitate will form It should also be clear that at equilibrium of the solid with its ions, the concentration of each ion is constant and the precipitation of ions in solution does occur but at the same rate as the solubility of precipitate in solution. Therefore, the concentration of the ions remains constant at equilibrium.

13 The molar solubility depends on the stoichiometry of the salt. A 1:1 salt is less soluble than a nonsymmetric salt with the same Ksp

14 Three situations will be studied: a. Solubility in pure water b. Solubility in presence of a common ion c. Solubility in presence of diverse ions Two other situations will be discussed in later chapters a.Solubility in acidic solutions b.solubility in presence of a complexing agent

15 Solubility in Pure Water Example Calculate the concentration of Ag + and Cl - in pure water containing solid AgCl if the solubility product is 1.0x Solution First, we set the stoichiometric equation and assume that the molar solubility of AgCl is s. AgCl (s)  Ag + + Cl -

16 K sp = [Ag + ][Cl - ] 1.0x = s x s = s 2 s = 1.0 x M [Ag + ] = [Cl - ] = 1.0 x M

17 Example Calculate the molar solubility of PbSO 4 in pure water if the solubility product is 1.6 x Solution K sp = [Pb 2+ ][SO 4 2- ] 1.6 x = s x s = s 2 s = 1.3x10 -4 M

18 Calculate the molar solubility of PbI 2 in pure water if the solubility product is 7.1 x Solution K sp = [Pb 2+ ][I - ] 2 7.1x10 -9 = s x (2s) 2 7.1x10 -9 = 4s 3 s = 1.2x10 -3 M

19 If we compare the molar solubilities of PbSO 4 and PbI 2 we find that the solubility of lead iodide is larger than that of lead sulfate although lead iodide has smaller k sp. You should calculate solubilities rather than comparing solubility products to check which substance gives a higher solubility. In presence of a precipitating agent, the substance with the least solubility will be precipitated first.

20 Example What must be the concentration of Ag + to just start precipitation of AgCl in a 1.0x10 -3 M NaCl solution. Solution AgCl just starts to precipitate when the ion product just exceeds k sp K sp = [Ag + ][Cl - ] 1.0x = [Ag + ] x 1.0 x [Ag + ] = 1.0x10 -7 M

21 Example What pH is required to just start precipitation of Fe(III) hydroxide from a 0.1 M FeCl 3 solution. K sp = 4x Solution Fe(OH) 3  Fe OH - K sp = [Fe 3+ ][OH - ] 3 4x10 -3 = 0.1 x [OH - ] 3 [OH - ] = 7x M pH = 14 – pOH pH = 14 – 12.2 = 1.8

22 Solubility in Presence of a Common Ion Example 10 mL of 0.2 M AgNO 3 is added to 10 mL of 0.1 M NaCl. Find the concentration of all ions in solution and the solubility of AgCl formed. First we find mmol AgNO 3 and mmol NaCl then determine the excess concentration mmol Ag + = 0.2 x 10 = 2 mmol Cl - = 0.1 x 10 = 1 mmol AgCl formed = 1 mmol mmol Ag + excess = 2 – 1 = 1 [Ag + ] excess = mmol/mL = 1/20 = 0.05 M

23 We can find the concentrations of NO 3 - = 0.2*10/20 = 0.1 M [Na + ] = 0.1*10/20 = 0.05 M Chloride ions react to form AgCl, therefore the only source for Cl - is the solubility of AgCl; but now in presence of 0.05 M excess Ag +

24 K sp = [Ag + ][Cl - ] 1.0x = ( s) (s) Since the solubility product is very small, we can assume 0.05>>s 1.0x = 0.05 (s) s = 2.0x10 -9 M Relative error = ( 2.0x10 -9 / 0.05) x 100 = 4x10 -6 % which is extremely small, therefore: [Cl - ] = 2.0x10 -9 M [Ag + ] = x10 -9 = 0.05 M Look at how the solubility is decreased in presence of the common ion.