A soft drink was bottled with a partial pressure of \(\mathrm{CO}_{2}\) of 3 bar over the liquid at room temperature. The partial pressure of \(\mathrm{CO}_{2}\) over the solution approaches a value of 30 bar when 44 g of \(\mathrm{CO}_{2}\) 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 \(\mathrm{H}_{2} \mathrm{CO}_{3}=4.0 \times 10^{-7} ; \log 2=0.3\); density of the soft drink \(= 1 \mathrm{~g} \mathrm{~mL}^{-1}\))

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Solutions

A soft drink was bottled with a partial pressure of

CO_{2}

of 3 bar over the liquid at room temperature. The partial pressure of

CO_{2}

over the solution approaches a value of 30 bar when 44 g of

CO_{2}

is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ____

\times 1 0^{- 1} /

(First dissociation constant of

H_{2} CO_{3} = 4.0 \times 1 0^{- 7}; lo g 2 = 0.3

; density of the soft drink

= 1 g mL^{- 1}

)

Answer: 37

By Henry's law the amount of dissolved

CO X_{2}

is proportional to its partial pressure. The data point,

44

g (one mole) dissolving per kg of water at

30

bar, fixes the constant, so at the bottling pressure of

3

bar the amount dissolved is

\frac{3}{30} \times 1 = 0.1

mol in

1

L of solution (density

1

g/mL). Treating

H X_{2} CO X_{3}

as a weak acid of concentration

0.1

M with

K_{a 1} = 4.0 \times 1 0^{- 7}

, so

p K_{a 1} = 6.4

, the pH is

\frac{1}{2} (p K_{a 1} - lo g C) = \frac{1}{2} (6.4 + 1) = 3.7

. Expressed as

\times 1 0^{- 1}

this reads

37

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Integer answer