# Henry’s law constant for

Question:

Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

Solution:

It is given that:

KH = 1.67 × 108 Pa

$p_{\mathrm{CO}_{2}}=2.5 \mathrm{~atm}=2.5 \times 1.01325 \times 10^{5} \mathrm{~Pa}$

= 2.533125 × 105 Pa

According to Henry’s law:

$p_{\mathrm{CO}_{2}}=\mathrm{K}_{\mathrm{H}} x$

$\Rightarrow x=\frac{p_{\mathrm{CO}_{2}}}{\mathrm{~K}_{\mathrm{H}}}$

$=\frac{2.533125 \times 10^{5}}{1.67 \times 10^{8}}$

= 0.00152

We can write, $x=\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{CO}_{2}}+n_{\mathrm{H}_{2} \mathrm{O}}} \approx \frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}}$

[Since, $n_{\mathrm{CO}_{2}}$ is negligible as compared to $n_{\mathrm{H}_{2} \mathrm{O}}$ ]

In 500 mL of soda water, the volume of water = 500 mL

[Neglecting the amount of soda present]

We can write:

500 mL of water = 500 g of water

$=\frac{500}{18} \mathrm{~mol}$ of water

= 27.78 mol of water

Now, $\frac{n_{\mathrm{CO}_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}}=x$

$\frac{n_{\mathrm{CO}_{2}}}{27.78}=0.00152$

$n_{\mathrm{CO}_{2}}=0.042 \mathrm{~mol}$

Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g

= 1.848 g