A soft drink was bottled with partial pressure of of over the liquid at room temperature

Question

A soft drink was bottled with partial pressure of $C O_{2}$ of $6 b a r$ over the liquid at room temperature. The partial pressure of $C O_{2}$ over the solution approaches a value of $60 b a r$ when $88 g$ of $C O_{2}$ is dissolved in $1 k g$ of water at room temperature. The approximate pH of the soft drink is _________ $\times 10^{- 1}$ .

(first dissociation constant of $H_{2} C O_{3} = 4.0 \times 10^{- 7}$ , $\log 2 = 0.3$

density of the soft drink $= 1 g m l^{- 1}$ )

(first dissociation constant of $H_{2} C O_{3} = 4.0 \times 10^{- 7}$ , $\log 2 = 0.3$

density of the soft drink $= 1 g m l^{- 1}$ )

Smart Achievers · Accepted Answer

(36)

$\frac{P_{1}}{P_{2}} = \frac{m_{1}}{m_{2}} = \frac{6}{60} = \frac{m_{1}}{88}$

$m_{1} = 8.8 g$ $Mole = \frac{8.8}{44} = 0.2$

$p H = \frac{1}{2} p K_{a} - \log c = \frac{1}{2} \times 6.4 - \log (2 \times 10^{- 1})$

$K_{a} = 4 \times 10^{- 7}$ $p K_{a} = 7 - 0.6 = 6.4$ $\Rightarrow 3.2 + 0.35 = 3.55 = 35.5 \times 10^{- 1} = 36 \times 10^{- 1}$

Solution

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Solution

Ans. (36) P1P2=m1m2=660=m188 m1=8.8g Mole=8.844=0.2 pH=12pKa-logc=12×6.4-log(2×10-1) Ka=4×10-7 pKa=7-0.6=6.4⇒3.2+0.35=3.55=35.5×10-1=36×10-1

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