1. Use Henry’s law to find dissolved $CO_2$ in the soft drink

At room temperature, we are told that when $44\,\text{g}$ of $CO_2$ is dissolved in $1\,\text{kg}$ of water, the equilibrium partial pressure is $30\,\text{bar}$.

• Moles of $CO_2$ in $44\,\text{g}$: $n = \frac{44}{44} = 1\,\text{mol}$

So, for this reference condition: $\frac{n}{\text{kg water}} = 1\,\text{mol kg}^{-1}$ corresponds to $p_{CO_2} = 30\,\text{bar}$

By Henry’s law, dissolved concentration is proportional to partial pressure.

For the soft drink bottled at $3\,\text{bar}$: $\text{dissolved } CO_2 = 1\times \frac{3}{30} = 0.1\,\text{mol kg}^{-1}$

Since density of soft drink is $1\,\text{g mL}^{-1}$, $1\,\text{kg}$ solution is approximately $1\,\text{L}$. Hence we take $C \approx 0.1\,\text{M}$ for dissolved carbonic acid species.

2. Treat dissolved $CO_2$ as weak acid $H_2CO_3$

First dissociation: $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$

Given: $K_a = 4.0\times 10^{-7}$

Initial concentration: $C = 0.1\,\text{M}$

Let dissociation be $x$:

• $[H^+] = x$
• $[HCO_3^-] = x$
• $[H_2CO_3] = 0.1 - x \approx 0.1$

Then $K_a = \frac{x^2}{0.1}$

So, $x^2 = 4.0\times 10^{-7}\times 0.1 = 4.0\times 10^{-8}$

$x = 2\times 10^{-4}\,\text{M}$

Thus, $[H^+] = 2\times 10^{-4}$

3. Calculate pH

$\text{pH} = -\log(2\times 10^{-4})$ $= 4 - \log 2$ Given $\log 2 = 0.3$, $\text{pH} = 4 - 0.3 = 3.7$

4. Match with required format

The question asks for pH in the form: $\underline{\hspace{2cm}}\times 10^{-1}$

Since $3.7 = 37\times 10^{-1}$

the required integer is: $\boxed{37}$

5. Comparison with stored answer

Stored correct answer = $37$

Our derived answer also = $37$.

So the answer agrees.