You add extra-pure water (without any CO2 dissolved in it) in a jar and, subsequ

Question

You add extra-pure water (without any CO2 dissolved in it) in a jar and, subsequently, you let ambient air, which contains 400ppm of CO2 (gas), to go into the jar and close the jar so that no air can come in or go out anymore. How much will the pH of the water change compared to original pH after equilibrium is reached if 75% of the CO2 remains in the gas (air) phase at equilibrium? If you add hydroxide solution (i.e., OH-) so that the pH is adjusted to 8, will the partial pressure of CO2 in the gas phase be higher, lower, or the same when the new equilibrium is reached? Clearly describe your assumptions and provide the appropriate justification for them. Henry’s law: KH = Pgas/Xaq, where KH = 1,640 atm @ 25 0C, 1atm; K1(bicarbonate) = 10-6; K2(carbonate) = 10-11. Atomic weights: C-12, O-16, H-1. Tip: it is a closed carbonate system.

Explanation / Answer

The chemical equation for CO2-water rxn:

CO2 + H2O <---> H2CO3

H2CO3 +OH- <---->HCO3-

HCO3- +OH- <---->CO32-

1) before addition of OH-

Pgas/Patm=Ygas [Ygas=mole fraction of CO2 in gas phase.(=400*10^-6 =0.0004,)but only 75% of the CO2 gas in gas phase]

Ygas=0.75*0.0004=0.0003

Pgas=0.003*1atm=0.003atm [partial pressure of CO2 in the gas phase]

2) After adding OH-

(Xaq=mole fraction of CO2 in water)

pH=pka(2) +log [CO32-]/[H2CO3] [Henderson-hasselbach equation]

pH=8

K2(carbonate) = 10-11

pk2=-log(10^-11)=11

So, log [CO32-]/[H2CO3] =pH-pka=8-11=-3

[CO32-]/[H2CO3] =0.001

KH=Pgas/Xaq=1,640 atm

X(aq)=concentration of CO2 in water=[H2CO3]=[CO2(aq)]

Also,[CO32-]=[OH-]

After addition of base OH- ,some H2CO3 dissociated so [H2CO3] =Xaq <<< Xaq in part( 1),thus pgas is also lower than part(1)

Pgas=KH*X(aq)

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