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Chapter 13: Problem 43

Under an \(\mathrm{O}_{2}(\mathrm{g})\) pressure of \(1.00 \mathrm{atm}, 28.31 \mathrm{mL}\) of \(\mathrm{O}_{2}(\mathrm{g})\) dissolves in \(1.00 \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C} .\) What will be the molarity of \(\mathrm{O}_{2}\) in the saturated solution at \(25^{\circ} \mathrm{C}\) when the \(\mathrm{O}_{2}\) pressure is 3.86 atm? (Assume that the solution volume remains at \(1.00 \mathrm{L}\).)

Short Answer

Expert verified
The molarity of O2 in the saturated solution at 25°C and pressure of 3.86 atm would be 3.86 times the initial molarity computed in step 2.

Step by step solution

01

Understanding Henry's Law

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the solution. It is represented by the formula: \(C= k \cdot P\) where C is the concentration of the gas in the solution, P is the pressure of the gas above the solution, and k is the Henry’s law constant.
02

Calculate initial molarity

First, we will calculate the molarity (M) of O2 under 1 atm pressure. Given that 28.31 mL of O2 gas dissolves in 1.00 L of water, remember that molarity is defined as moles of solute per liter of solution. The molar volume of a gas at STP is 22.4 L/mol, so we can calculate moles of O2 as \(moles= \frac{Volume}{Molar\,volume} =\frac{28.31\,mL }{ 22.4 \times 10^3 \,mL/mol}\). Then the molarity M = \( \frac{moles}{ Volume\, of\, solution} \) . Should be computed to know the molarity of O2 in the initial condition.
03

Calculate molarity at higher pressure

According to Henry's Law, if we increase the pressure, the solubility (or molarity) of the gas will increase proportionally. So if the pressure is 3.86 times bigger (3.86 atm), the molarity will also be 3.86 times bigger. So multiply the initial molarity (calculated in step 2) by 3.86 to find the new molarity at 3.86 atm.

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