JEE Main 2020 — Ionic Equilibrium Question with Solution
From: JEE Main 2020 (Online) 5th September Morning Slot
Question
A soft drink was bottled with a partial pressure of CO2
of 3 bar over the liquid at room temperature.
The partial pressure of CO2
over the solution approaches a value of 30 bar when 44 g of CO2
is
dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ 10–1.
(First dissociation constant of
H2CO3 = 4.0 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
(First dissociation constant of
H2CO3 = 4.0 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
Enter your answer
Show full solutionCorrect answer: 37
Correct answer
37
Step-by-step explanation
CO2
+ H2O H2CO3
At 30 bar pressure mass of CO2 in 1 kg water = 44 gm
At 3 bar pressure mass of CO2 in 1 kg water = 4.4 gm
Moles of CO2 in 1 kg water = = 0.1
4.0 10–7 =
1
0.1 = 4 10-7
= 2 10-3
[H+] = 0.1 = 2 10-4
pH = –[– 4 × log(2)] = 3.7 = 37 × 10–1
At 30 bar pressure mass of CO2 in 1 kg water = 44 gm
At 3 bar pressure mass of CO2 in 1 kg water = 4.4 gm
Moles of CO2 in 1 kg water = = 0.1
| H2CO3 | ⇌ | H+ | + | HCO3- | |
|---|---|---|---|---|---|
| t = 0 | 0.1 | 0 | 0 | ||
| t = teq | 0.1(1 - ) | 0.1 | 0.1 |
4.0 10–7 =
1
0.1 = 4 10-7
= 2 10-3
[H+] = 0.1 = 2 10-4
pH = –[– 4 × log(2)] = 3.7 = 37 × 10–1
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