Question

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 \[{H_2}C{O_3} = 4.0 \times {10^{ - 7}}\] ; \[\log 2 = 0.3\] ; density of the soft drink\[ = 1\,g\,m{L^{ - 1}}\])

Answer 1

Hint: Here, in this question, we have to use henry’s law to calculate the pH. When the temperature is held constant, Henry's law, a gas law, asserts that the amount of gas dissolved in a liquid is precisely proportional to the partial pressure of that gas above the liquid. pH is the negative logarithm (base 10) of hydrogen ion (or hydronium ion).

Complete Step by Step Solution:
The pressure a particular gas exerts within a mixture of other gases is referred to as its partial pressure. For illustration, if a container contains a mixture of three gases—oxygen, nitrogen, and carbon dioxide—its partial pressure is equal to the pressure oxygen exerts on the container's walls, and its individual partial pressures are equal to those of nitrogen and carbon dioxide. The sum of the partial pressures of the gases (oxygen, nitrogen, and carbon dioxide) in the mixture exerts the total pressure on the container walls.

Here, in this question,
\[{P_{1\left( {C{O_2}} \right)}} = 3\,bar\]
\[{P_{2\left( {C{O_2}} \right)}} = 30\,bar\]
The molar mass of \[C{O_2}\] is \[44\,g\,mo{l^{ - 1}}\] .
\[44\,g\]of \[C{O_2}\] is dissolved in \[1\] kg of water at room temperature, that means \[1\] mol of \[C{O_2}\] is dissolved.
Partial pressure of \[C{O_2}\] can be calculated by the equation as follows:
\[{P_{C{O_2}}} = {K_H} \times {n_{C{O_2}}}\]
For the initial condition (before dissolution),
\[ {P_{1\left( {C{O_2}} \right)}} = {K_H} \times {n_{1\left( {C{O_2}} \right)}} \\
   \Rightarrow 3 = {K_H} \times {n_{1\left( {C{O_2}} \right)}}\,\,\,\,\,\,...(1) \\ \]
For the final condition (after dissolution),
\[ {P_{2\left( {C{O_2}} \right)}} = {K_H} \times {n_{2\left( {C{O_2}} \right)}} \\
   \Rightarrow 30 = {K_H} \times 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2) \\ \]
Now, let us divide equation \[(1)\] with equation \[(2)\] as follows:
\[ \dfrac{3}{{30}} = \dfrac{{{K_H} \times {n_{C{O_2}}}}}{{{K_H} \times 1}} \\
   \Rightarrow {n_{C{O_2}}} = \dfrac{3}{{30}} \\
   \Rightarrow {n_{C{O_2}}} = 0.1\,mol \\ \]
Now, given the value of \[{K_a} = 4.0 \times {10^{ - 7}}\] .
We can calculate pH by using the value of \[{K_a}\] by the equation as follows:
\[pH = \dfrac{1}{2}\left( {p{K_a} - \log \,c} \right)\]
First, calculate \[p{K_a}\] as follows:
\[ p{K_a} = - \log {K_a} \\
   \Rightarrow p{K_a} = - \log \left[ {4 \times {{10}^{ - 7}}} \right] \\
   \Rightarrow p{K_a} = 6.4 \\ \]
Substituting values,
\[ pH = \dfrac{1}{2}\left( {p{K_a} - \log \,c} \right) \\
   \Rightarrow pH = \dfrac{1}{2}\left( {6.4 - \log \dfrac{1}{{10}}} \right) \\
   \Rightarrow pH = \dfrac{1}{2}\left( {6.4 - 1} \right) \\ \]
Further solving,
\[ pH = \dfrac{1}{2} \times 7.4 \\
   \Rightarrow pH = 3.7 \\
   \Rightarrow pH = 37 \times {10^{ - 1}} \\ \]
Therefore, the pH of the soft drink is \[37 \times {10^{ - 1}}\].

Note: We can measure the acid's strength using its \[p{K_a}\] value. If an acid's \[p{K_a}\] value is excessively high, it is very weak, and if it is excessively low, it is a strong acid. This means strength and \[p{K_a}\] values are inversely related.