© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understanding the Problem
The question describes a soft drink containing dissolved CO2 at a partial pressure of 3 bar. When 44 g of CO2 is dissolved in 1 kg of water at 30 bar, we want to find how much CO2 dissolves at 3 bar and then use the first dissociation of carbonic acid (H2CO3) to determine the pH. The key steps involve:
Relating dissolved CO2 to the partial pressure.
Using the dissociation equilibrium of carbonic acid to find [H+].
Computing the pH from [H+].
Step 2: Determining the Mass of CO2 Dissolved at 3 bar
It is given that at 30 bar, 44 g of CO2 dissolves in 1 kg of water. Hence, if the pressure is reduced from 30 bar to 3 bar (i.e., one-tenth of 30 bar), the mass of CO2 dissolved becomes one-tenth of 44 g (assuming approximate proportionality):
Mass of CO2 dissolved at 3 bar = 4.4 g.
Step 3: Calculating the Number of Moles of Dissolved CO2
The molar mass of CO2 is 44 g mol–1. Hence, the moles of CO2 dissolved in 1 kg of water at 3 bar is:
\text{Moles of CO}_{2} = \dfrac{4.4 \text{ g}}{44 \text{ g mol}^{-1}} = 0.1 \text{ mol}.
Step 4: Writing the Dissociation of Carbonic Acid
When CO2 dissolves in water, it forms carbonic acid (H2CO3):
\text{CO}_{2} + \text{H}_{2}\text{O} \to \text{H}_{2}\text{CO}_{3}.
Carbonic acid (H2CO3) then dissociates as follows:
\text{H}_{2}\text{CO}_{3} \rightleftharpoons \text{H}^{+} + \text{HCO}_{3}^{-}.
Let the initial concentration of H2CO3 be 0.1 M (from the 0.1 mol in 1 kg of water, approximating the volume as 1 L because density is 1 g/mL). Let \alpha be the degree of dissociation. Then at equilibrium:
H2CO3
H+
HCO3–
t = 0
0.1
0
0
t = teq
0.1(1 – \alpha )
0.1 \alpha
0.1 \alpha
Step 5: Setting Up the Equilibrium Expression
The first dissociation constant of carbonic acid is given as K_{a} = 4.0 \times 10^{-7} . Therefore:
K_{a} = \dfrac{[\text{H}^{+}][\text{HCO}_{3}^{-}]}{[\text{H}_{2}\text{CO}_{3}]}
= \dfrac{(0.1 \alpha) (0.1 \alpha)}{0.1 (1 - \alpha)} = \dfrac{0.1 \alpha^{2}}{1 - \alpha}.
Since K_{a} = 4.0 \times 10^{-7} , we have
4.0 \times 10^{-7} = \dfrac{0.1 \alpha^{2}}{1 - \alpha}.
Step 6: Approximating the Degree of Dissociation
Because K_{a} is small, \alpha is typically very small, so 1 - \alpha \approx 1 . Hence:
4.0 \times 10^{-7} = 0.1 \alpha^{2}.
Solving for \alpha ,
\alpha^{2} = \dfrac{4.0 \times 10^{-7}}{0.1} = 4.0 \times 10^{-6},
\alpha = 2.0 \times 10^{-3}.
Step 7: Calculating [H+] and pH
The concentration of H+ ions at equilibrium is
[\text{H}^{+}] = 0.1 \times \alpha = 0.1 \times 2.0 \times 10^{-3} = 2.0 \times 10^{-4}.
Hence, the pH is given by
\text{pH} = -\log [\text{H}^{+}] = -\log\left(2.0 \times 10^{-4}\right).
Using \log(2) \approx 0.3 ,
\text{pH} = -[\log(2) + \log(10^{-4})] = -[0.3 + (-4)] = -(-3.7) = 3.7.
Therefore, the pH is approximately 3.7 , which can be written as 37 \times 10^{-1} .
Final Answer
The pH of the soft drink is approximately 37 \times 10^{-1} .
