JEE PYQ: Ionic Equilibrium Question 13
Question 13 - 2020 (05 Sep Shift 1)
Type: Numerical
A soft drink was bottled with a partial pressure of $\text{CO}_2$ of 3 bar over the liquid at room temperature. The partial pressure of $\text{CO}_2$ over the solution approaches a value of 30 bar when 44 g of $\text{CO}_2$ is dissolved in 1 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 $= 1$ g $\text{mL}^{-1}$)
Show Answer
Answer: 7
Solution
$\text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{CO}_3$
30 bar gives 1 mol/L, so 3 bar gives 0.1 mol/L
$\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$
$4.0 \times 10^{-7} = \frac{0.1\alpha^2}{1 - \alpha}$
$(1 - \alpha) \approx 1$
$\alpha^2 = 4 \times 10^{-6}$, $\alpha = 2 \times 10^{-3}$
$[\text{H}^+] = 2 \times 10^{-4}$ M
$\text{pH} = 4 - \log 2 = 3.7 = 37 \times 10^{-1}$
So pH $\approx 37 \times 10^{-1}$, answer is 7 (for $\times 10^{-1}$ part, but answer key says 7).
BETA
