Question

A soft drink was bottled with a partial pressure of CO₂

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)

Answer 1

Answer: The approximate pH of soft drink is $37\times {10}^{-1}$

Step 1: From the given reaction:

${\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{CO}}_3$

$30\mathrm{bar}=1\mathrm{m}/\mathrm{lit}.$

$3\mathrm{bar}=0.1\mathrm{m}/\mathrm{lit}$

Step 2:

$$\begin{gathered} {\mathrm{H}}_2{\mathrm{CO}}_3\rightleftharpoons {\mathrm{H}}^{+}+{{\mathrm{HCO}}_3}^{-} \\ \mathrm{t}=00.100 \\ \mathrm{Eq}.0.1(1-\mathrm{\alpha })0.1\mathrm{\alpha }0.1\mathrm{\alpha } \end{gathered}$$

Step 3:

From the formula,

$\mathrm{Ka}=\frac{{\mathrm{c\alpha }}^2}{1-\mathrm{\alpha }}$

$$\begin{gathered} 4\times {10}^{-7}=\frac{0.1{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }} \\ (1-\mathrm{\alpha })=1 \\ {\mathrm{\alpha }}^2=4\times {10}^{-6} \\ \mathrm{\alpha }=2\times {10}^{-3} \\ \left[{\mathrm{H}}^{+}\right]=2\times {10}^{-4}\mathrm{M} \end{gathered}$$

Step 4:

For the calculation of $\mathrm{pH}$ :

$$\begin{gathered} \mathrm{pH}=-[-4\times \log (2\left)\right] \\ =3.7 \\ \mathrm{pH}=37\times {10}^{-1} \end{gathered}$$