# Dissolution of CO2 in water

The following question has troubled me for a while.

For gases that are slightly soluble in water, there is a proportional relationship between the partial pressure, $$P$$, and the mole fraction, $$x$$, of the gas molecules dissolved in water (Henry's law):

$$\ce{P = k_H*x}$$

A container ($$V$$ = $$570$$ $$mL$$) is filled with water ($$V_l$$ = $$500$$ $$mL$$) and pressurized with $$CO_2$$ gas ($$P_0$$ = $$50$$ $$atm$$), before it is allowed to stand at $$10^\circ$$C until the vapor–liquid equilibrium is established.

Calculate the pressure of $$CO_2$$ $$[atm]$$ in the container and the amount of $$CO_2$$ $$[mol]$$ dissolved in the water. The Henry coefficient of $$CO_2$$ for water at $$10^\circ$$C is $$k_H$$ = $$0.104 × 10^4$$ $$atm$$, and we will consider that the reaction of $$CO_2$$ in water can be ignored.

I did the following calculations but I am unsure with the outcome.

$$\ce{n_{total}\approx n(H_2O) = \frac{\rho V}{M(H_2O)} = 27.753 mol}$$

$$\ce{x = \frac{P_0}{k_H} = 0.048077}$$

$$\ce{n(CO_2)_{dissolved} = x*n_{total} \approx 1.3343 mol}$$

$$\ce{P_1 = \frac{nRT}{V} = \frac {1.3343*0.0831451*1.01325*283}{0.570} = 55.8 atm}$$

Shouldn't the final pressure be smaller than the initial pressure ($$50$$ $$atm$$)?

• yes, the problem is wanting you to split the $\ce{CO2}$ initially introduced into the container between the gas and liquid phase.
– MaxW
May 4 at 20:55

Assumptions:

• Reaction of carbon dioxide with water is neglected.
• Vapour pressure of water is negligible.
• Volume of solution does not change on dissolution of carbon dioxide.
• Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $$X_{\ce{CO2{(aq)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$$

Initially no $$\ce{CO2}$$ is dissolved in water and so initial moles of $$\ce{H2O{(l)}}$$ and $$\ce{CO2{(g)}}$$ can be calculated as follows:

$$n_{{\ce{H2O{(l)}}}_i} = \frac{\rho V}{M} =\pu{\frac{500 \times 0.9975}{18} mol}= \pu{27.71 mol}\\ n_{{\ce{CO2{(g)}}}_i} = \frac{P_iV}{RT} = \pu{ \frac{50 \times 0.07}{0.0821 \times 283} mol}= \pu{0.15 mol}\\$$

Now after the equilibrium is settled some amount of carbon dioxide will dissolve in water according to Henry's law.

Final pressure in the container can be written as:

$$P_f= \frac{n_{{\ce{CO2{(g)}}}_f}RT}{V}\\$$

And by Henry's law we can say,

$$P_f= K_H \times X_{\ce{CO2{(aq)}}}$$

Thus from both these equations we can conclude:

$$K_H \times \frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}} = \frac{n_{{\ce{CO2{(g)}}}_f}RT}{V}\\$$

We also known that,

$$n_{{\ce{CO2{(g)}}}_f} = n_{{\ce{CO2{(g)}}}_i} - n_{{\ce{CO2{(aq)}}}_f} \\$$

So,

$$K_H \times \frac{n_{\ce{CO2(aq)}_f}}{n_{\ce{H2O(l)}}} = \frac{(n_{{\ce{CO2{(g)}}}_i} - n_{{\ce{CO2{(aq)}}}_f})RT}{V}\\$$

Solving this we get,

$$n_{\ce{CO2(aq)}_f} = \pu{0.135 mol}\\$$

This correspondingly means:

$$n_{{\ce{CO2{(g)}}}_f}= \pu{0.015 mol}\\ P_f= \pu{5 atm}\\$$

The final pressure is less than initial pressure as expected. The mistake you were making was implicitly considering $$P_f=P_i$$ when finding the mole fraction of carbon dioxide in water, which clearly isn't the case.