JEE MAIN - Chemistry (2020 - 5th September Morning Slot - No. 15)
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 ______ $$ \times $$ 10–1.
(First dissociation constant of
H2CO3 = 4.0 $$ \times $$ 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
(First dissociation constant of
H2CO3 = 4.0 $$ \times $$ 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1) .
Answer
37
Explanation
CO2
+ H2O $$ \to $$ 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
$$ \therefore $$ Moles of CO2 in 1 kg water = $${{4.4} \over {44}}$$ = 0.1
4.0 $$ \times $$ 10–7 = $${{0.1{\alpha ^2}} \over {1 - \alpha }}$$
$${1 - \alpha }$$ $$ \simeq $$ 1
$$ \Rightarrow $$ 0.1$${{\alpha ^2}}$$ = 4 $$ \times $$ 10-7
$$ \Rightarrow $$ $$\alpha $$ = 2 $$ \times $$ 10-3
[H+] = 0.1$$\alpha $$ = 2 $$ \times $$ 10-4
$$ \therefore $$ 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
$$ \therefore $$ Moles of CO2 in 1 kg water = $${{4.4} \over {44}}$$ = 0.1
| H2CO3 | ⇌ | H+ | + | HCO3- | |
|---|---|---|---|---|---|
| t = 0 | 0.1 | 0 | 0 | ||
| t = teq | 0.1(1 - $$\alpha $$) | 0.1$$\alpha $$ | 0.1$$\alpha $$ |
4.0 $$ \times $$ 10–7 = $${{0.1{\alpha ^2}} \over {1 - \alpha }}$$
$${1 - \alpha }$$ $$ \simeq $$ 1
$$ \Rightarrow $$ 0.1$${{\alpha ^2}}$$ = 4 $$ \times $$ 10-7
$$ \Rightarrow $$ $$\alpha $$ = 2 $$ \times $$ 10-3
[H+] = 0.1$$\alpha $$ = 2 $$ \times $$ 10-4
$$ \therefore $$ pH = –[– 4 × log(2)] = 3.7 = 37 × 10–1
Comments (0)
