A bottled carbonated mineral water (i.e. closed system) has a pH of 4.9 and an alkalinity of 4×10-³ Considering the species H₂O, H*, OH, H₂CO³, HCO³ and CO² and the reactions: H₂O = OH + H* Kw = 10-14 H₂CO3 = HCO3 + H* K₁ = 10-6.35 HCO3 = CO3+H* K₂ = 10-10.33 Calculate the total C₁ = [H₂CO3 ]+[HCO3]+[CO3²] (Hint: you can use a Tableau). Verify any assumptions that you may have made. Now the bottle is opened to the atmosphere and CO₂ is allowed to degas to reach a new equilibrium at pH = 8.1 (Hint: consider what is the dominant carbonate specie and conservative entitie/s that de not change). a. b. C. Calculate how much CO₂ is degassed (or total carbon lost).

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Chapter13: Acids And Bases
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A bottled carbonated mineral water (i.e. closed system) has a pH of 4.9 and an alkalinity of 4x10-³M
Considering the species H₂O, H, OH, H₂CO3, HCO3 and CO² and the reactions:
H₂O = OH + H+
Kw = 10-14
K₁ = 10-6.35
K₂ = 10-10.33
H₂CO3 = HCO3 + H+
HCO3 = CO3² +H*
Calculate the total C₁ = [H₂CO3 ]+[HCO3¯]+[CO3²¯] (Hint: you can use a Tableau).
Verify any assumptions that you may have made.
Now the bottle is opened to the atmosphere and CO₂ is allowed to degas to reach a new equilibrium
at pH = 8.1 (Hint: consider what is the dominant carbonate specie and conservative entitie/s that do
not change).
a.
b.
C.
Calculate how much CO₂ is degassed (or total carbon lost).
Transcribed Image Text:A bottled carbonated mineral water (i.e. closed system) has a pH of 4.9 and an alkalinity of 4x10-³M Considering the species H₂O, H, OH, H₂CO3, HCO3 and CO² and the reactions: H₂O = OH + H+ Kw = 10-14 K₁ = 10-6.35 K₂ = 10-10.33 H₂CO3 = HCO3 + H+ HCO3 = CO3² +H* Calculate the total C₁ = [H₂CO3 ]+[HCO3¯]+[CO3²¯] (Hint: you can use a Tableau). Verify any assumptions that you may have made. Now the bottle is opened to the atmosphere and CO₂ is allowed to degas to reach a new equilibrium at pH = 8.1 (Hint: consider what is the dominant carbonate specie and conservative entitie/s that do not change). a. b. C. Calculate how much CO₂ is degassed (or total carbon lost).
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