Solubility CalculationsThis 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= 5⁰ C). The equilibrium constants at 5⁰ 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. What is the solubility of CaCO3 at T= 5.0 oC?
Solubility CalculationsThis 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= 5⁰ C). The equilibrium constants at 5⁰ 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. What is the solubility of CaCO3 at T= 5.0 oC?
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Solubility CalculationsThis 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= 5⁰ C). The equilibrium constants at 5⁰ 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. What is the solubility of CaCO3 at T= 5.0 oC?
![Air
Water
Rock, soil,
or sediments
CO₂(g)
H₂CO3
CO32- + H₂O
+
Ca2+
AL
CaCO3(s)
H + HCO3
OH + HCO3
H₂O
FIGURE 10-7 Reactions
among the three phases
(air, water, rocks) of the
carbon dioxide-carbonate
system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F374ac7e4-7cbe-4278-ac11-c2618fc77dd4%2Feae87a93-1a7e-4dd0-afd6-0cbd07629a67%2Fn6f7uer_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Air
Water
Rock, soil,
or sediments
CO₂(g)
H₂CO3
CO32- + H₂O
+
Ca2+
AL
CaCO3(s)
H + HCO3
OH + HCO3
H₂O
FIGURE 10-7 Reactions
among the three phases
(air, water, rocks) of the
carbon dioxide-carbonate
system.
![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, KH 0.065 for CO₂ and K₁ for H₂CO3 is 3.0 x 10-7; see Prob-
lem 10-15 for other necessary data.
-
Finally, the residual concentrations of CO2, of Ht, 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),
[CO₂²-] = K₂p/[Ca²+] = 4.6 × 10-%/5.3 × 10-4 = 8.7 × 10-6 M
From reaction (5),
[OH-] = K[CO,¹-]/[HCO,1]
=
(2.1 × 10-4) × (8.7 × 10-6)/1.05 × 10-³ = 1.7 × 10-6
and finally from reaction (7)
[H+] = K₁/[OH-] = 1.0 × 10-¹4/1.7 x 10-6 = 5.7 x 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 CO₂ 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
from 7 to 9, in reasonable agreement with our calculations. Due to the
smaller amount of bicarbonate in noncalcareous waters, their pH values are
usually close to 7. Of course if such natural waters are subject to acid rain,
the pHs can become substantially lower since there is little HCO, or CO²-
readily available with which to neutralize the acid.
About 80% of natural surface waters in the United States have pH val-
ues between 6.0 and 8.4. Lakes and rivers into which acid rain falls will have
elevated levels of sulfate ion and perhaps of nitrate ion since the principal
acids in the precipitation are H₂SO, and HNO, (see Section 4.3).
Revi
base
the
TIL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F374ac7e4-7cbe-4278-ac11-c2618fc77dd4%2Feae87a93-1a7e-4dd0-afd6-0cbd07629a67%2Fu6t8pye_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, KH 0.065 for CO₂ and K₁ for H₂CO3 is 3.0 x 10-7; see Prob-
lem 10-15 for other necessary data.
-
Finally, the residual concentrations of CO2, of Ht, 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),
[CO₂²-] = K₂p/[Ca²+] = 4.6 × 10-%/5.3 × 10-4 = 8.7 × 10-6 M
From reaction (5),
[OH-] = K[CO,¹-]/[HCO,1]
=
(2.1 × 10-4) × (8.7 × 10-6)/1.05 × 10-³ = 1.7 × 10-6
and finally from reaction (7)
[H+] = K₁/[OH-] = 1.0 × 10-¹4/1.7 x 10-6 = 5.7 x 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 CO₂ 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
from 7 to 9, in reasonable agreement with our calculations. Due to the
smaller amount of bicarbonate in noncalcareous waters, their pH values are
usually close to 7. Of course if such natural waters are subject to acid rain,
the pHs can become substantially lower since there is little HCO, or CO²-
readily available with which to neutralize the acid.
About 80% of natural surface waters in the United States have pH val-
ues between 6.0 and 8.4. Lakes and rivers into which acid rain falls will have
elevated levels of sulfate ion and perhaps of nitrate ion since the principal
acids in the precipitation are H₂SO, and HNO, (see Section 4.3).
Revi
base
the
TIL
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