For the following problems use the following: KH CO2 = 0.034 M/atm = 10-1.47 M/a
ID: 490750 • Letter: F
Question
For the following problems use the following:
KH CO2 = 0.034 M/atm = 10-1.47 M/atm
pK1 = 5.85 (where K1 = [H+] x [HCO3-] / [CO2]aq
pK2 = 8.97 (where K2 = [H+] x [CO32-] / [HCO3-]
1. What is the pH of pure water in equilibrium with 400 ppm CO2 ( = 4.0 x 10-4 atm = 10-3.4 atm)?
a. Why can you assume charge balance only involves H+ and HCO3-?
b. What is the pH of the pure water?
2. What is the pH of seawater with [CO32-] = 1.02 x 10-4 M in equilibrium with 316 ppm CO2?
3. What happens to the pH when CO2 levels increase to 400 ppm?
4. What is the percent increase in acidity? The key to answering this is understanding the nature of the pH scale (it is logarithmic, not linear)
Explanation / Answer
1.Henry law is
C = KH × P
= 0.034(M/atm) × 4 × 10^-4 atm
= 1.36 × 10^-3M
CO2 cconcentration in solution = 1.36 × 10^-3M
Concentration of H2CO3 = 1.36 × 10^-3M
H2CO3(aq)< ---------> HCO3- (aq) + H+(aq)
Ka= [HCO3-] [ H+]/ [ H2CO3]
1.413 ×10^-6 = X^2/1.36 × 10^-3
X = 4.385 × 10^-5
[H+] = 4.385 × 10^-5M
pH = 4.35
a) pka2 = 8.97
Ka2 = 1.072 × 10^-9
Ka2 = [H+] [ CO3^2-]/ [HCO3-]
1.072 × 10^-9 = X^2/4.385× 10^-5
X = 2.17 × 10-7
[H+ ] = 2.17 × 10^-7M
Adding this hydrogen ion concentration to hydrogen ion concentration from ka1 equilibrium is not changing considerably , so pH change is not considerable. Therefore, we can assume charge balance involve H+ and HCO3- only.
b) pH of pure water = 7.0
2. Ka2 = 1.072 × 10^-9M
1.072 × 10^-9 = (1.02× 10^-4)^2/[HCO3-]
[HCO3-] = 1.04×10^-8/1.072 × 10^-9
= 9.70M
[ H+ ] = 9.70 M
pH = 0.99
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