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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 ______ $\times 10^{-1}$ . (First dissociation constant of $H_{2}CO_{3}=4.0\times 10^{-7}; \; log\; 2=0.3;$ density of the soft drink $=1\; g\; mL^{-1}$ )

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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.

Then, $\textup{Molarity }= \frac{\textup{mole}}{\textup{Volume(L)}}$

$\textup{Molarity }= \frac{\frac{\textup{Mass}}{\textup{Molar mass}}}{\textup{Volume(L)}}$

and density = Mass/volume

so, Volume = density X Mass = 1 x 1 = 1 litre

Then, $\textup{Molarity }= \frac{\frac{\textup{44}}{\textup{44}}}{\textup{1}}=1\textup{ mole /litre}$

Now,

A soft drink was bottled with a partial pressure of $CO_{2}$ of 3 bar over the liquid at room temperature.

Reaction : $\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}$

For, 30 Bar → 1 mol/litre

Then, 3 Bar → 0.1 mol/litre

Now,

First dissociation constant :

$4 \times 10^{-7}=\frac{0.1 \alpha^{2}}{1-\alpha}$

$4 \times 10^{-7}=0.1 \alpha^{2} \quad \because {(1-\alpha\approx 1)}$

$\begin{aligned} &\alpha^{2}=4 \times 10^{-6} \\ &\alpha=2 \times 10^{-3} \end{aligned}$

So,

$\left[\mathrm{H}^{+}\right]= 0.1 \alpha = 2 \times 10^{-4} \mathrm{M}$

Now, we pH = - log([H⁺])

$\mathrm{pH}=\mathrm{-log[2\times 10^{-4}]}$

$\mathrm{pH}=-[\log(2) + (-4)\log (10)]$

$\mathrm{pH}=-[0.3 + (-4)]$

$\mathrm{pH}=3.7=37 \times 10^{-1}$

Ans = 37

Posted by

Kuldeep Maurya

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