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Calculate the molar solubility of Fe(OH) 3 in a pH = 5.0 buffer solution. The solubility product constant of Fe(OH) 3 is 4 10 –^38 M^4.
Marks 2
Using pOH = - log 10 ([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 :
Fe(OH) 3 (s) Fe3+(aq) + 3OH-(aq) K sp = [Fe3+(aq)][OH-(aq)]^3
Hence,
[Fe3+(aq)] =
- sp (^) **-
(4×10 )
= =4×10 M
[OH (aq)] (1.0×10 )
K
As Fe(OH) 3 (s) dissolves to give 1 Fe3+(aq), this is also the molar solubility.
Answer: 4 × 10-^11 M
- The molar solubility of lead(II) fluoride, PbF 2 , is found to be 2.6 × 10–3^ M at 25 °C. Calculate the value of K sp for this compound at this temperature.
Marks 2
The solubility equilibrium and constant for PbF 2 (s) are,
PbF 2 (s) Pb2+(aq) + 2F-(aq) K sp = [Pb2+(aq)][F-(aq)]^2
As one moles of Pb2+(aq) and two moles of F-(aq) are produced for every mole of PbF 2 (s) which dissolves, [Pb2+(aq)] = 2.6 × 10-3^ M and [F-(aq)] = (2 × 2.6 × 10-3) = 5.2 × 10-3^ M. Hence,
K sp = (2.6 × 10-3) × (5.2 × 10-3)^2 = 7.0 × 10-
K sp = 7.0 × 10-
- A champagne bottle is filled with 750 mL of wine, leaving 10.0 mL of air at atmospheric pressure when it is sealed with a cork. After fermentation, the pressure inside the bottle is 6.0 atm at 20 o^ C. Assume that the gas produced is entirely CO 2 and that its solubility in the wine is the same as in water. What mass of CO 2 has been produced by the fermentation? Data: The mole fraction solubility of CO 2 in water is 7.1 × 10 –4^ at 293 K and 1.0 atm.
Marks 6
The molar mass of H 2 O is (16.00 (O) + 2 × 1.008 (H)) g mol-1^ = 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 =
ܛܛ܉ܕ ܛܛ܉ܕ ܚ܉ܔܗܕ
ૠ ܔܗܕ ૡ. ష^
= 41.67 mol
The mole fraction of CO 2 in water, ࢄ۽۱ (^) , is given by:
ࢄ۽۱ (^) ൌ
(^) ۽۱ሻܙ܉ሺ (^) ۽۱ሻܙ܉ሺ ା (^) ۶ሻܔሺ۽
(^) ۽۱ሻܙ܉ሺ (^) ۽۱ሻܙ܉ሺ ାሺ.ૠ ܔܗܕሻ = 7.1 × 10 -
Hence, the number of moles of CO 2 in the wine before fermentation(1.0 atm) is given by: (^) ۽۱ (^) ሻܙ܉ሺ = (7.1 × 10-4^ )( (^) ۽۱ (^) ሻܙ܉ሺ + 41.67) = 7.1 × 10-4 (^) ۽۱ (^) ሻܙ܉ሺ + (7.1 × 10-4^ × 41.67) (^) ۽۱ (^) ሻܙ܉ሺ (1.0 - 7.1 × 10-4^ ) = (7.1 × 10-4^ × 41.67) (^) ۽۱ (^) ሻܙ܉ሺ = 0.0296 mol After fermentation, the pressure is 6.0 atm so (^) ۽۱ (^) ሻܙ܉ሺ = 6.0 × 0.0296 mol. The number of moles of CO 2 produced by the fermentation and dissolved in the wine is therefore: (^) ۽۱ (^) ሻܙ܉ሺ = (6.0 -1.0) × 0.0296 mol = 0.148 mol The increase in air pressure of 5.0 atm is due to extra CO 2 (g). As 1 atm = 101. kPa, P = (5.0 × 101.3) kPa = 506.5 kPa. The volume of air = 10.0 mL = 0.0100 L = = 1.00 × 10-5^ m 3. Using the ideal gas equation, PV = nRT , the number of moles of CO 2 (g) is:
(^) ۽۱ (^) ሺሻ =
ࢂࡼ ࢀࡾ
൫. ൈ^ ܉۾൯ሺ. ൈ ష^ ܕ^ ሻ ൫ૡ. ܕ^ Ԝ۹܉۾ ష^ ܔܗܕԜష^ ൯ሺሺାૠሻ۹ሻ
= 0.00208 mol
Overall: (^) ۽۱ (^) = (^) ۽۱ ሻܙ܉ሺ (^) ۽۱ (^) ሺሻ = (0.148 + 0.002) mol = 0.150 mol The molar mass of CO 2 is (12.01 (C) + 2 × 16.00 (O)) g mol-1^ = 44.01 g mol-1. Hence, the mass of CO 2 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
After the bottle has been opened and all of the bubbles have been released, what volume of CO 2 has escaped? Assume all the CO 2 produced escapes.
When the cork is released, the pressure returns to 1.0 atm. The amount of CO 2 that will remain dissolved is therefore, from above, (^) ۽۱ (^) ሻܙ܉ሺ = 0.0296 mol.
The amount of CO 2 which escapes is therefore:
(^) ۽۱ (^) ሺሻሻ = (0.150 – 0.0296) mol = 0.120 mol
At 1.0 atm = 101.3 kPa, this will occupy a volume:
ࢂ =
ࢀࡾ ࡼ
ሺ. ܔܗܕሻ൫ૡ. ܕ ^ Ԝ۹܉۾ ష^ ܔܗܕԜష^ ൯ሺሺାૠሻ۹ሻ ൫. ൈ ^ ൯܉۾ = 2.9 × 10-3^ m 3 = 2.9 L Answer: 2.9 × 10-3^ m 3 = 2.9 L
