Water is in equilibrium with the atmosphere at pH of 6.5. If the partial pressure of carbon dioxide in the atmosphere is 10-3.5 atm. What would be the water alkalinity? (Henry's Constant for CO2 in Water = 101.5 mole/L atm, use K1 = 10-6.3 K2 = 10-10.3). The carbonate equilibrium is defined by the following equations: CO2(aq) +H2O S H¿CO3 H2CO3 5 H* + HCO; HCO; 5 H + CO; Keg = 3.50 X 10² K = 4.20 X 10-7 K2 = 4.69 X 10*"
Water is in equilibrium with the atmosphere at pH of 6.5. If the partial pressure of carbon dioxide in the atmosphere is 10-3.5 atm. What would be the water alkalinity? (Henry's Constant for CO2 in Water = 101.5 mole/L atm, use K1 = 10-6.3 K2 = 10-10.3). The carbonate equilibrium is defined by the following equations: CO2(aq) +H2O S H¿CO3 H2CO3 5 H* + HCO; HCO; 5 H + CO; Keg = 3.50 X 10² K = 4.20 X 10-7 K2 = 4.69 X 10*"
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![Chemistry
Water is in equilibrium with the atmosphere
at pH of 6.5. If the partial pressure of
carbon dioxide in the atmosphere is 10-3.5
atm. What would be the water alkalinity?
(Henry's Constant for CO2 in Water = 10-1.5
mole/L atm, use K1 = 10-6.3 K2 = 10-10.3).
The carbonate equilibrium is defined by the following equations:
CO2(aq) +H2O 5 H¿CO3
H2CO3 5 H* + HCO;
HCO; 5 H* + CO;
Keq = 3.50 X 10²
K = 4.20 X 107
K2 = 4.69 X 10"
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F120ef40c-ba61-489a-8f0e-beaff5d51a86%2Fb789c5be-949a-4374-bab2-209a21c14ec0%2Fujcaaai_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Chemistry
Water is in equilibrium with the atmosphere
at pH of 6.5. If the partial pressure of
carbon dioxide in the atmosphere is 10-3.5
atm. What would be the water alkalinity?
(Henry's Constant for CO2 in Water = 10-1.5
mole/L atm, use K1 = 10-6.3 K2 = 10-10.3).
The carbonate equilibrium is defined by the following equations:
CO2(aq) +H2O 5 H¿CO3
H2CO3 5 H* + HCO;
HCO; 5 H* + CO;
Keq = 3.50 X 10²
K = 4.20 X 107
K2 = 4.69 X 10"
%3D
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