Solubility Calculations This question is based on Problem 10-17 (p. 441) in our textbook; where you are to use all the reactions in Figure 10-7 to calculate the solubility of CaCO3 in cold water (T= 5° C). Please put your work in part 6a; and the solubility of CaCO3 in part 6b. The equilibrium constants at 5° C are: KH=0.065; K, for H₂CO3-3.0x10-7; Ksp=8.1x10-⁹; Ka=2.8x10-11 for HCO3; Kw=0.2x10-14.
Solubility Calculations This question is based on Problem 10-17 (p. 441) in our textbook; where you are to use all the reactions in Figure 10-7 to calculate the solubility of CaCO3 in cold water (T= 5° C). Please put your work in part 6a; and the solubility of CaCO3 in part 6b. The equilibrium constants at 5° C are: KH=0.065; K, for H₂CO3-3.0x10-7; Ksp=8.1x10-⁹; Ka=2.8x10-11 for HCO3; Kw=0.2x10-14.
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Transcribed Image Text:Solubility Calculations
This question is based on Problem 10-17 (p.
441) in our textbook; where you are to use all
the reactions in Figure 10-7 to calculate the
solubility of CaCO3 in cold water (T= 5° C).
Please put your work in part 6a; and the
solubility of CaCO3 in part 6b. The
equilibrium constants at 5° C are: KH=0.065;
K₁ for H₂CO3-3.0x10-7; Ksp=8.1x10-⁹;
Ka=2.8x10-11 for HCO3; Kw=0.2x10-14.
![PROBLEM 10-17
Repeat the above calculation for the solubility of CaCO3 in water that is also
in equilibrium with atmospheric CO₂ for a water temperature of 5°C. At this
temperature, K₁ = 0.065 for CO₂ and K₁ for H₂CO, is 3.0 x 10-7; see Prob-
lem 10-15 for other necessary data.
Finally, the residual concentrations of CO32-, of H+, and of OH- in the
system can be deduced from equilibrium constants for reactions (4), (5), and
(7), since equilibria in these processes are in effect notwithstanding the over-
all reaction (8). Thus from reaction (4),
[CO32-] = K₂p/[Ca²+] = 4.6 × 10-%/5.3 × 10-4 = 8.7 × 10-6 M
From reaction (5),
[OH-] = K₂ [CO3²-]/[HCO,-]
and finally from reaction (7)
=
= (2.1 x 10-4) × (8.7 × 10-6)/1.05 x 10-³ = 1.7 x 10-6
[H+] = K/[OH-] = 1.0 × 10-¹4/1.7 x 10-6 = 5.7 × 10-9
From this value for the hydrogen ion concentration, we conclude that,
according to this calculation, river and lake water at 25°C whose pH is
determined by saturation with CO, and CaCO, should be slightly alkaline,
with a pH of about 8.2.
Typically, the measured pH values of calcareous waters lie in the range
Air
Water
Rock, soil,
or sediments
CO₂(g)
H₂CO3
CO32- + H₂O
+
Ca2+
CaCO3(s)
→→→→-H* + HCO3¯
>
OH + HCO3
H₂O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F374ac7e4-7cbe-4278-ac11-c2618fc77dd4%2F51f53c5c-cb3b-4483-a2f2-b1b9a0032e7a%2Fpxe7cp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:PROBLEM 10-17
Repeat the above calculation for the solubility of CaCO3 in water that is also
in equilibrium with atmospheric CO₂ for a water temperature of 5°C. At this
temperature, K₁ = 0.065 for CO₂ and K₁ for H₂CO, is 3.0 x 10-7; see Prob-
lem 10-15 for other necessary data.
Finally, the residual concentrations of CO32-, of H+, and of OH- in the
system can be deduced from equilibrium constants for reactions (4), (5), and
(7), since equilibria in these processes are in effect notwithstanding the over-
all reaction (8). Thus from reaction (4),
[CO32-] = K₂p/[Ca²+] = 4.6 × 10-%/5.3 × 10-4 = 8.7 × 10-6 M
From reaction (5),
[OH-] = K₂ [CO3²-]/[HCO,-]
and finally from reaction (7)
=
= (2.1 x 10-4) × (8.7 × 10-6)/1.05 x 10-³ = 1.7 x 10-6
[H+] = K/[OH-] = 1.0 × 10-¹4/1.7 x 10-6 = 5.7 × 10-9
From this value for the hydrogen ion concentration, we conclude that,
according to this calculation, river and lake water at 25°C whose pH is
determined by saturation with CO, and CaCO, should be slightly alkaline,
with a pH of about 8.2.
Typically, the measured pH values of calcareous waters lie in the range
Air
Water
Rock, soil,
or sediments
CO₂(g)
H₂CO3
CO32- + H₂O
+
Ca2+
CaCO3(s)
→→→→-H* + HCO3¯
>
OH + HCO3
H₂O
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