A soft drink was bottled with a partial pressure of CO2 of 3 bar over the liquid at room temperature

Q: Why do we assume the volume of the solution is 1 L?

The density of the soft drink is given as $1 \text{ g/mL}$. Therefore, 1 kg (or 1000 g) of the soft drink has a volume of exactly 1000 mL, which is 1 L. The dissolved gas doesn't significantly change the overall volume.

Q: Why did we ignore the second dissociation constant of $\text{H}_2\text{CO}_3$?

The first dissociation constant ($K_{a1}$) is much larger than the second dissociation constant ($K_{a2}$). Therefore, almost all the $\text{H}^+$ ions in the solution come from the first dissociation step, making the second step negligible for pH calculation.

Q: Can we use the exact quadratic equation instead of the approximation $1-\alpha \approx 1$?

You can, but it's unnecessary. Since $K_a$ is very small ($4.0 \times 10^{-7}$), the degree of dissociation $\alpha$ is less than 5% (it's actually 0.2%). This makes the approximation highly accurate and saves a lot of calculation time.

Q: What is the role of Henry's Law here?

Henry's Law allows us to relate the solubility of the gas at different pressures. Since we know the mass dissolved at 30 bar, we use the direct proportionality ($m \propto P$) to find the mass dissolved at the new pressure of 3 bar.

Q: Why is the answer 37 and not 3.7?

The question specifically asks for the value to fill in the blank for the expression $... \times 10^{-1}$. Since the calculated pH is 3.7, we rewrite it as $37 \times 10^{-1}$, making the integer answer 37.

Question

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

Sigma Percentile · Accepted Answer

The Fizz Physics: Calculating the pH of a Soft Drink

Have you ever wondered what gives your favorite soft drink that sharp, tangy bite? It's not just the flavorings; it's the physics and chemistry of dissolved carbon dioxide! In this problem, we dive into the fascinating world of Henry's Law and ionic equilibrium to calculate the exact pH of a freshly opened bottle of soda.

### The Setup: Henry's Law in Action

Imagine a sealed bottle of soft drink. Inside the factory, it is pressurized to a whopping 30 bar, forcing 44 grams of $	ext{CO}_2$ to dissolve into just 1 kg of water. But when you open it at room temperature, the pressure drops to a mere 3 bar. What happens to all that dissolved gas?

Enter **Henry's Law**, which states that the mass of a dissolved gas is directly proportional to its partial pressure ($m \propto P$).

Since the pressure drops by a factor of 10 (from 30 bar to 3 bar), the mass of the dissolved $	ext{CO}_2$ must also drop by a factor of 10. 
$$m_2 = \frac{44}{30} 	imes 3 = 4.4 	ext{ g}$$
So, at room temperature, only 4.4 grams of $	ext{CO}_2$ remain dissolved in the water.

### Finding the Concentration

To find the pH, we first need the molar concentration of the dissolved $	ext{CO}_2$. 
The molar mass of $	ext{CO}_2$ is 44 g/mol. Therefore, 4.4 grams corresponds to:
$$n = \frac{4.4}{44} = 0.1 	ext{ mol}$$

We are given that the density of the soft drink is $1 	ext{ g/mL}$. This means 1 kg of water has a volume of exactly 1 Liter. 
Thus, the molarity ($C$) of the carbonic acid ($	ext{H}_2	ext{CO}_3$) formed by the dissolved $	ext{CO}_2$ is:
$$C = \frac{0.1 	ext{ mol}}{1 	ext{ L}} = 0.1 	ext{ M}$$

### The Weak Acid Equilibrium

When $	ext{CO}_2$ dissolves in water, it forms carbonic acid, a weak acid that partially dissociates:
$$	ext{H}_2	ext{CO}_3 ightleftharpoons 	ext{H}^+ + 	ext{HCO}_3^-$$

Let $\alpha$ be the degree of dissociation. At equilibrium, the concentrations are:
- $[	ext{H}_2	ext{CO}_3] = C(1-\alpha)$
- $[	ext{H}^+] = C\alpha$
- $[	ext{HCO}_3^-] = C\alpha$

The equilibrium constant $K_a$ is given by:
$$K_a = \frac{[	ext{H}^+][	ext{HCO}_3^-]}{[	ext{H}_2	ext{CO}_3]} = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)}$$

Because carbonic acid is a very weak acid ($K_a = 4.0 	imes 10^{-7}$), $\alpha$ is extremely small compared to 1. We can safely approximate $(1-\alpha) \approx 1$, simplifying our equation to:
$$K_a \approx C\alpha^2$$

Now, we plug in our known values to find $\alpha$:
$$4.0 	imes 10^{-7} = 0.1 	imes \alpha^2$$
$$\alpha^2 = 40 	imes 10^{-7} = 4 	imes 10^{-6}$$
$$\alpha = 2 	imes 10^{-3}$$

### The Final pH Calculation

With $\alpha$ in hand, finding the hydrogen ion concentration is straightforward:
$$[	ext{H}^+] = C\alpha = 0.1 	imes 2 	imes 10^{-3} = 2 	imes 10^{-4} 	ext{ M}$$

Finally, we calculate the pH using the standard logarithmic formula:
$$	ext{pH} = -\log[	ext{H}^+]$$
$$	ext{pH} = -\log(2 	imes 10^{-4}) = 4 - \log 2$$

Given that $\log 2 = 0.3$:
$$	ext{pH} = 4 - 0.3 = 3.7$$

The question asks for the answer in the format of $... 	imes 10^{-1}$. 
$$3.7 = 37 	imes 10^{-1}$$

Therefore, the integer value to fill in the blank is **37**. The next time you take a sip of a fizzy drink, remember the elegant math that dictates its acidic bite!

Enter Numerical Value:

Visualized Solution

$P_{1} = 30 bar, P_{2} = 3 bar$

$m \propto P$

$m_{2} = 4.4 g$

$C = 0.1 M$

$H_{2} CO_{3} ⇌ H^{+} + HCO_{3}^{-}$

$K_{a} \approx C α^{2}$

$α = 2 \times 1 0^{- 3}$

$[H^{+}] = 2 \times 1 0^{- 4} M$

$pH = 3.7$

$Answer = 37$

The Sigma Insight: pH, Buffer and Indicator

The Setup

Finding the Concentration

The Weak Acid Equilibrium

The Final pH Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​=30 bar,P2​=3 bar

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m∝P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​=4.4 g

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C=0.1 M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​CO3​⇌H++HCO3−​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ka​≈Cα2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=2×10−3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [H+]=2×10−4 M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pH=3.7

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Answer=37