A soft drink was bottled with a partial pressure of $$\text{CO}_2$$ of $$3\,\text{bar}$$ over the liquid at room temperature. The partial pressure of $$\text{CO}_2$$ over the solution approaches a value of $$30\,\text{bar}$$ when $$44\,\text{g}$$ of $$\text{CO}_2$$ is dissolved in $$1\,\text{kg}$$ of water at room temperature. The approximate pH of the soft drink is __________ $$\times 10^{-1}$$. (First dissociation constant of $$\text{H}_2\text{CO}_3 = 4.0 \times 10^{-7}$$; $$\log 2 = 0.3$$; density of the soft drink $$= 1\,\text{g mL}^{-1}$$)

The gas-liquid equilibrium of carbon dioxide in water obeys Henry’s law, which in its simplest form is written as

$$P = k_{\mathrm H}\,x_{\mathrm{CO_2}}$$

where $$P$$ is the partial pressure of the gas in bar, $$x_{\mathrm{CO_2}}$$ is its mole-fraction in the liquid phase and $$k_{\mathrm H}$$ is Henry’s constant (in the same pressure unit).

First we evaluate $$k_{\mathrm H}$$ from the data at $$30\;\text{bar}$$. The problem states that $$44\;\text{g}$$ of $$\mathrm{CO_2}$$ (whose molar mass is $$44\;\text{g mol}^{-1}$$) are dissolved in $$1\;\text{kg}$$ of water:

$$n_{\mathrm{CO_2}}=\frac{44\ \text{g}}{44\ \text{g mol}^{-1}}=1\ \text{mol}$$

The number of moles of water in $$1\;\text{kg}$$ is

$$n_{\mathrm H_2O}=\frac{1000\ \text{g}}{18\ \text{g mol}^{-1}}=55.5\ \text{mol}$$

Hence the mole-fraction of dissolved $$\mathrm{CO_2}$$ is

$$x_{\mathrm{CO_2}}=\frac{1}{1+55.5}=\frac{1}{56.5}\approx 1.77\times10^{-2}$$

Using the equilibrium value $$P = 30\;\text{bar}$$ we calculate Henry’s constant:

$$k_{\mathrm H}=\frac{P}{x_{\mathrm{CO_2}}}=\frac{30}{1.77\times10^{-2}}\approx1.70\times10^{3}\ \text{bar}$$

Now the soft drink is bottled at a lower pressure of $$3\;\text{bar}$$. Applying the same Henry’s law constant, the new mole-fraction of dissolved $$\mathrm{CO_2}$$ is

$$x'_{\mathrm{CO_2}}=\frac{P'}{k_{\mathrm H}}=\frac{3}{1.70\times10^{3}}\approx1.77\times10^{-3}$$

The corresponding number of moles of $$\mathrm{CO_2}$$ present per $$55.5$$ moles of water is obtained through

$$n'_{\mathrm{CO_2}} \approx x'_{\mathrm{CO_2}}\times n_{\mathrm H_2O}=1.77\times10^{-3}\times55.5\approx9.8\times10^{-2}\ \text{mol}$$

Because the density of the drink is given as $$1\ \text{g mL}^{-1}$$, $$1\;\text{kg}$$ of solution very nearly occupies $$1\;\text{L}$$. Hence the molarity of $$\mathrm{CO_2}$$ (and therefore of the hydrated form $$\mathrm{H_2CO_3}$$) is

$$C_0\;(\mathrm{H_2CO_3})\approx0.098\ \text{mol L}^{-1}\;(\text{≈}0.10\ \text{M})$$

Only the first dissociation of carbonic acid is significant for the pH, and its equilibrium is

$$\mathrm{H_2CO_3}\rightleftharpoons\mathrm{H^+}+\mathrm{HCO_3^-},\qquad K_a=4.0\times10^{-7}$$

Let $$[\,\mathrm{H^+}\,]=x\;\text{M}$$ at equilibrium. Then $$[\,\mathrm{HCO_3^-}\,]=x\;\text{M}$$ and $$[\,\mathrm{H_2CO_3}\,]=C_0-x$$. The acid-dissociation expression is

$$K_a=\frac{x^2}{C_0-x}$$

Because carbonic acid is weak, $$x\ll C_0$$, so $$C_0-x\approx C_0$$. Hence

$$x\approx\sqrt{K_aC_0}=\sqrt{(4.0\times10^{-7})(1.0\times10^{-1})}=\sqrt{4.0\times10^{-8}}=2.0\times10^{-4}\ \text{M}$$

The hydrogen-ion concentration therefore is $$[\,\mathrm{H^+}\,]=2.0\times10^{-4}\ \text{M}$$.

The definition of pH is $$\mathrm{pH}=-\log[\,\mathrm{H^+}\,]$$. Using $$\log2=0.3$$ we find

$$\mathrm{pH}=-\log(2.0\times10^{-4})=-(\log2 + \log10^{-4})=-(0.3-4)=3.7$$

Finally, the statement in the question asks for the numerical value that multiplies $$10^{-1}$$, i.e. we write

$$3.7 = 37\times10^{-1}$$

So, the answer is $$37$$.