ALWAYS include units!!!

let's start with pressure

P @10m = P CO

_{2} guage + P hydrostatic

= Patm + PCO

_{2} + ρgh

Patm = 101325 Pa

PCO

_{2} = 0.1 bar * (10

^{5} Pa / bar) = 10000 Pa

Phydro = (1008 kg/m

^{3}) * (9.80m/s

^{2}) * (10m) * (1Pa / 1kg/ms

^{2}) = 98784 Pa

Ptotal = 101325Pa + 10000 Pa + 98784 Pa = 210109 Pa = 2.10x10

^{5} Pa

then let's find kH...

*I'm going to use a reference value first to verify the 6.1g/L #* kH* for CO

_{2} in H

_{2}O @ 25°C = 29 atm/M * (101325 Pa / atm) = 2.94x10

^{6} Pa/M @25°C

* note: that is from the wikipedia henry's law page.

then at 2°C,

kH @2°C = kH@25°C * exp((-dHsol/R)(1/T

_{2} - 1/T

_{1}))

= 2.94x10

^{6} Pa/M * exp(-2400K)*(1/275K - 1/278K))

= 1.50x10

^{6 } Pa/M

and finally

[CO

_{2}] = 2.10x10

^{5} Pa * (1M / 1.50x10

^{6} Pa) * (44.01g/mol) = 6.2 g/L

meaning.. we

**verified** the answer of 6.1g/L using an alternate kH

*********

now let's go back to your solution

(1) ALWAYS INCLUDE UNITS !!!

(2) your atmospheric pressure is 10

^{5} Pa not 105 Pa. 101325 Pa

(3) take a look at your kH CO

_{2} @ T = 2° value.. that's NOT Pa / M

that value is Pa per MOLE FRACTION of CO

_{2} in the liquid phase <

sigh>

**********

ok.. so let's see if we can solve this with YOUR kH value

We're going to simplify this by making 2 assumptions

(1) that beer is essentially water

(2) The mole fraction of CO

_{2} in H

_{2}O is going to be very small

(I can easily show - and you should be able to also - that 6g/L is about 0.14 mol CO

_{2} per 55.5 mol H

_{2}O. So that is a reasonable assumption).

making this equation approximately true.

mol CO

_{2} mol CO

_{2} χ CO

_{2} = ----------------------- ≈ ------------

mol CO

_{2} + mol H

_{2}O mol H

_{2}O

and now I can rewrite that kH value as

84.1x10

^{6} Pa 84.1x10

^{6}Pa * (mol H

_{2}O)

kH = -------------- ≈ ----------------------------

χ CO

_{2} (mol CO

_{2})

and now we can finish up.

2.10x10

^{5} Pa mol CO

_{2 } 44.01g CO

_{2 } 55.5 mol H

_{2}O

[CO

_{2}] = -------------- * ------------------------------ * -------------- * --------------- = 6.1 g / L

1 (84.1x10

^{6} Pa) * (mol H2O) mol CO

_{2 } 1 L H

_{2}O