CHEM1909 2004-N-7 November 2004

Marks  Calculate the molar solubility of Fe(OH)3 in a pH = 5.0 buffer solution. –38 4 2 The solubility product constant of Fe(OH)3 is 4  10 M .

- Using pOH = -log10([OH (aq)]) and pH + pOH = 14.0:

pOH = (14.0 – 5.0) = 9.0 and [OH-(aq)] = 1× 10-9 M

The solubility equilibrium and product are :

3+ - 3+ - 3 Fe(OH)3(s) Fe (aq) + 3OH (aq) Ksp = [Fe (aq)][OH (aq)]

Hence,

K (4×10-38 ) [Fe3+(aq)] = sp = =4×10-11 M [OH- (aq)] 3 (1.0×10 -9 ) 3

3+ As Fe(OH)3(s) dissolves to give 1 Fe (aq), this is also the molar solubility.

Answer: 4 × 10-11 M

CHEM1909 2006-N-9 November 2006

• –3 ° Marks The molar solubility of lead(II) fluoride, PbF 2, is found to be 2.6 × 10 M at 25 C. 2 Calculate the value of Ksp for this compound at this temperature.

The solubility equilibrium and constant for PbF 2(s) are,

2+ - 2+ - 2 PbF 2(s) Pb (aq) + 2F (aq) Ksp = [Pb (aq)][F (aq)]

As one moles of Pb 2+ (aq) and two moles of F -(aq) are produced for every mole of 2+ -3 - -3 PbF 2(s) which dissolves, [Pb (aq)] = 2.6 × 10 M and [F (aq)] = (2 × 2.6 × 10 ) = 5.2 × 10 -3 M. Hence,

-3 -3 2 -8 Ksp = (2.6 × 10 ) × (5.2 × 10 ) = 7.0 × 10

-8 Ksp = 7.0 × 10

CHEM1109 2007-N-11 November 2007

• Marks The pH of a solution can be controlled by adding small amounts of gaseous HCl. 5 Assuming no change in volume, calculate what the pH of the solution must be to just dissolve 1.00 g of NiS suspended in 1.0 L of water. + 2– –20 2 Data: H2S(aq) 2H (aq) + S (aq) K = 1.1 × 10 M –22 2 Ksp (NiS) = 1.0 × 10 M

For the dissolution NiS(s) Ni 2+ (aq) + S 2-(aq),

2+ 2- Ksp = [Ni (aq)][S (aq)].

+ 2- For the acid dissociation, H 2S(aq) 2H (aq) + S (aq),

+ − [H (aq)]2 [S 2 (aq)] K = [H2 S(aq)]

Combining these two equilibria gives, for the overall reaction,

+ 2+ NiS(s) + 2H (aq) Ni (aq) + H 2S(aq),

+++ [Ni2 (aq)][H S(aq)] K ××× -22 2 =sp =1.0 10 = -3 Keq = +++ 9.1 × 10 [H (aq)]2K 1.1 ××× 10 -20

The formula mass of NiS is (58.69 (Ni) + 32.07 (S)) g mol -1 = 90.76 g mol -1. The amount in 1.00 g is therefore,

mass(g)=== 1.00g number of moles = − − = 0.0110 mol molarmass(gmol1 ) 90.76gmol 1

In the dissolution NiS(s) Ni 2+ (aq) + S 2-(aq), so if 0.0110 mol of NiS(s) completely dissolves in 1.0 L, [Ni 2+ (aq)] = 0.011 M. From the overall reaction, each mole of NiS which dissolves produces one mole of H 2S(aq) so [H 2S(aq)] = 0.011 M. Hence,

+ 2 (0.011M)(0.011M) + [H (aq)] = −−− or [H (aq)] = 0.12 M 9.1××× 10 3

+ pH = -log 10 [H (aq)] = -log 10 (0.12) = 0.94

pH = 0.94

CHEM1109 2009-N-4 November 2009 Marks • A champagne bottle is filled with 750 mL of wine, leaving 10.0 mL of air at 6 atmospheric pressure when it is sealed with a cork. After fermentation, the pressure o inside the bottle is 6.0 atm at 20 C. Assume that the gas produced is entirely CO2 and that its solubility in the wine is the same as in water. What mass of CO2 has been produced by the fermentation? –4 Data: The mole fraction solubility of CO2 in water is 7.1 × 10 at 293 K and 1.0 atm.

-1 -1 The molar mass of H2O is (16.00 (O) + 2 × 1.008 (H)) g mol = 18.016 g mol . Assuming that the wine is entirely water with a density of 1.0 g mL-1, the bottle contains 750 g of water or: number of moles of water = = 41.67 mol .

The mole fraction of CO2 in water, , is given by: = = 7.1 × 10-4 .

Hence, the number of moles of CO2 in the wine before fermentation(1.0 atm) is given by: -4 = (7.1 × 10 )( + 41.67) -4 -4 = 7.1 × 10 + (7.1 × 10 × 41.67) -4 -4 (1.0 - 7.1 × 10 ) = (7.1 × 10 × 41.67)

= 0.0296 mol

After fermentation, the pressure is 6.0 atm so = 6.0 × 0.0296 mol. The number of moles of CO2 produced by the fermentation and dissolved in the wine is therefore:

= (6.0 -1.0) × 0.0296 mol = 0.148 mol

The increase in air pressure of 5.0 atm is due to extra CO2(g). As 1 atm = 101.3 kPa, P = (5.0 × 101.3) kPa = 506.5 kPa. The volume of air = 10.0 mL = 0.0100 L = = 1.00 × 10-5 m3. Using the ideal gas equation, PV = nRT, the number of moles of CO2(g) is: . . = = 0.00208 mol . Overall:

= = (0.148 + 0.002) mol = 0.150 mol -1 -1 The molar mass of CO2 is (12.01 (C) + 2 × 16.00 (O)) g mol = 44.01 g mol . Hence, the mass of CO2 produced by fermentation is given by: mass = number of moles × molar mass = (0.150 mol) × (44.01 g mol-1) = 6.6 g

Answer: 6.6 g ANSWER CONTINUES ON THE NEXT PAGE CHEM1109 2009-N-4 November 2009

After the bottle has been opened and all of the bubbles have been released, what volume of CO2 has escaped? Assume all the CO2 produced escapes.

When the cork is released, the pressure returns to 1.0 atm. The amount of CO2

that will remain dissolved is therefore, from above, = 0.0296 mol.

The amount of CO2 which escapes is therefore:

= (0.150 – 0.0296) mol = 0.120 mol At 1.0 atm = 101.3 kPa, this will occupy a volume: . . = . = 2.9 × 10-3 m3 = 2.9 L Answer: 2.9 × 10-3 m3 = 2.9 L

CHEM1109 2009-N-13 November 2009

• –1 Marks The solubility of BaF 2 in water is 1.30 g L . Calculate the solubility product for 2 BaF 2.

-1 -1 The molar mass of BaF 2 is (137.34 (Ba) + 2 × 19.00 (F)) g mol = 175.34 g mol . As 1.30 g dissolves in one litre, this corresponds to: . number of moles = = 0.00741 mol . 2+ - As BaF 2 dissolves to give Ba (aq) + 2F (aq), dissolution of 0.00741 mol in one litre will produce [Ba 2+ (aq)] = 0.00741 M and [F -(aq)] = 2 × 0.00741 M = 0.0148 M. Hence: 2+ - 2 2 -6 Ksp = [Ba (aq)][F (aq)] = (0.00741)(0.0148) = 1.6 × 10

Answer: 1.6 × 10 -6

• A mixture of NaCl (5.0 g) and AgNO 3 (5.0 g) was added to 1.0 L of water. What are 3 the concentrations of Ag +(aq), Cl –(aq) and Na +(aq) ions in solution after equilibrium –10 has been established? Ksp (AgCl) = 1.8 × 10 .

The molar mass of NaCl is (22.99 (Na) + 35.45 (Cl)) g mol -1 = 58.44 g mol -1. Hence: . number of moles = = 0.0856 mol .

As NaCl dissolves to give Na +(aq) + Cl -(aq), dissolution of this amount in one litre will give [Na +(aq)] = 0.0856 M and [Cl -(aq)] = 0.0856 M. -1 The molar mass of AgNO 3 is (107.87 (Ag) + 14.01 (N) + 3 × 16.00 (O)) g mol = 169.88 g mol -1. Hence: . number of moles = = 0.0294 mol .

+ - As AgNO 3 dissolves to give Ag (aq) + NO 3 (aq), dissolution of this amount in one litre will give [Ag +(aq)] = 0.0294 M. Precipitation of AgCl(s) follows Ag +(aq) + Cl -(aq)


AgCl(s). As 0.0294 mol of Ag + ions and 0.0856 mol of Cl - ions are present, the former is limiting and so 0.0294 mol of AgCl(s) will form leaving (0.0856 – 0.0294) mol = 0.0562 mol of Cl - ions. Hence, after precipitation, [Cl -(aq)] = 0.0562 M. AgCl(s) has a very low solubility and dissolves to give Ag +(aq) and Cl -(aq) with + - -10 Ksp = [Ag (aq)][Cl (aq)] = 1.8 × 10 Hence: + - -10 -9 [Ag (aq)] = Ksp / [Cl (aq)] = (1.8 × 10 ) / (0.0562) M = 3.2 × 10 M

[Ag +(aq)] = 3.2 × 10 -9 M [Cl –(aq)] = 0.0562 M [Na +(aq)] = 0.0856 M