State Henry's law. Calculate the mole fraction of \(CO_2\) in water at \(298\,K\) under \(760\,mm\,Hg\).
(Given : \(K_H\) for \(CO_2\) in \(H_2O\) at \(298\,K = 1.25 \times 10^6\,mm\,Hg\))
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Solution and Explanation
Statement of Henry's Law: At constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas dissolved in the solution. Mathematically, \[ p=K_Hx \] where \[ p=\text{partial pressure of the gas} \] \[ K_H=\text{Henry's law constant} \] \[ x=\text{mole fraction of the gas in solution} \] This equation can be rearranged to calculate the mole fraction as \[ x=\frac{p}{K_H} \]
Step 1: Writing the given data. The pressure of carbon dioxide is \[ p=760\,mm\,Hg \] Henry's law constant is \[ K_H=1.25\times10^6\,mm\,Hg \] We have to calculate the mole fraction of carbon dioxide in water.
Step 2: Applying Henry's law. Using the relation \[ p=K_Hx \] we obtain \[ x=\frac{p}{K_H} \] Substituting the given values, \[ x=\frac{760}{1.25\times10^6} \]
Step 3: Performing the numerical calculation. \[ x=\frac{760}{1250000} \] \[ x=0.000608 \] Expressing the answer in scientific notation, \[ x=6.08\times10^{-4} \] Thus, the mole fraction of carbon dioxide dissolved in water is \[ \boxed{x_{CO_2}=6.08\times10^{-4}} \]
Step 4: Interpreting the result. The obtained mole fraction is very small, indicating that only a small amount of carbon dioxide dissolves in water under ordinary atmospheric pressure. This is expected because gases generally have limited solubility in liquids unless subjected to higher pressures. The result also confirms Henry's law: as pressure increases, the mole fraction of dissolved gas increases proportionally.
Final Answer: \[ \boxed{x_{CO_2}=6.08\times10^{-4}} \]
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