1. Calculate pH of pure water open to the atmosphere: CO2(dissolved) + H₂O CO2(air) со 2(dissolved) + H₂O H₂CO3 H+ + HCO3* HCO3* →H+ +CO3™ -1.47 H₂CO3 KH = [H₂CO3]/PCO2 K3 = [H][ HCO3 ]/[ H₂CO3] K4 = [H][ CO3 ]/[HCO3] Given: KH=10 M/atm Pco2 = 3x10 atm -6.35 (1) (2) (3) (4) K3 = 10 Assumption: when Pco2 is not too low, [H] = [HCO3]
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![1. Calculate pH of pure water open to the atmosphere
CO2(dissolved) + H₂O
+ H₂O → H₂CO3
CO2(air)
CO
2(dissolved)
H₂CO3H+ + HCO3¯
H+ + CO3™
HCO3
KH = [H₂CO3]/Pc02
K3 = [H][ HCO3 ]/[ H₂CO3]
K4 = [H ][ CO3 ]/[HCO3]
Given:
-1.47
KH=10 M/atm
-4
Pco2 = 3x10 atm
-6.35
K3 = 10
Assumption: when Pco2 is not too low, [H*] = [HCO3]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4cbf5201-f3bc-4285-8b65-775827ff8c37%2Fb2e61a09-1e69-467a-ad4c-39f57e7814ca%2F7g8zx5p_processed.png&w=3840&q=75)
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