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Fish need at least 4 ppm dissolved \(\mathrm{O}_{2}\) for survival. (a) What is this concentration in \(\mathrm{mol} / \mathrm{L}\) ? (b) What partial pressure of \(\mathrm{O}_{2}\) above the water is needed to obtain this concentration at \(10^{\circ} \mathrm{C}\) ? (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(1.71 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) -atm. \()\)

Short Answer

Expert verified
The concentration of 4 ppm dissolved O₂ is equivalent to \(1.25 \times 10^{-4} \) mol/L. To achieve this concentration at 10°C, the partial pressure of O₂ above the water should be approximately 0.0731 atm.

Step by step solution

01

Conversion of Units

To convert the concentration from ppm to mol/L, we first need to convert ppm to mg/L and then to mol/L using the molar mass of oxygen (32 g/mol). Given concentration = 4 ppm of dissolved O₂. 1 ppm is equivalent to 1 mg/L, so 4 ppm means 4 mg/L of dissolved O₂. Next, we will convert the mass of O₂ in mg to moles, and then divide by the volume in L to get the concentration in mol/L.
02

Calculation of Concentration

First, convert the mass (mg) to moles, and then divide by the volume (1 L) to calculate the concentration (mol/L). \[ \frac{4 \, \text{mg} \, \text{O}_{2}}{1 \, \text{L}} \times \frac{1 \, \text{g}}{1000 \, \text{mg}} \times \frac{1 \, \text{mol}}{32 \, \text{g}} = \frac{4}{32000}\, \text{mol/L} \] Now, we can simplify the fraction: \[ \frac{4}{32000} = \frac{1}{8000} = 1.25 \times 10^{-4} \, \text{mol/L} \] So, the concentration of dissolved oxygen in mol/L is \(1.25 \times 10^{-4}\) mol/L. #b) Calculate partial pressure using Henry's Law#
03

Henry's Law Formula

Henry's Law states that the concentration of a gas in a liquid is proportional to its partial pressure above the liquid. The formula for this relationship is: \[ C = kH \times P \] Where, C = concentration of the gas in mol/L, kH = Henry's Law constant (given as \(1.71 \times 10^{-3}\,\text{mol/L-atm} \) in the problem) P = partial pressure of the gas in atm. Our aim is to calculate the partial pressure (P), so we will rearrange the equation to solve for P: \[ P = \frac{C}{kH} \]
04

Calculation of Partial Pressure

Now we can plug in the values for 'C' and 'kH' to get the partial pressure of oxygen above the water: \[ P = \frac{1.25 \times 10^{-4}\, \text{mol/L}}{1.71 \times 10^{-3}\, \text{mol/L-atm}} \approx 0.0731 \, \text{atm} \] Therefore, the partial pressure of O₂ needed above the water to obtain the required concentration of dissolved oxygen at 10°C is approximately 0.0731 atm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ppm to mol/L conversion
Understanding how to convert parts per million (ppm) to moles per liter (mol/L) is essential in chemistry, especially when dealing with solutions. One ppm is equivalent to 1 milligram per liter (mg/L). For instance, if you have 4 ppm of dissolved oxygen (O₂), it translates to 4 mg/L.

Next, you need to convert this mass from milligrams to moles. Start by using the molar mass of oxygen, which is 32 g/mol. Since 1 g equals 1000 mg, you convert mg to g, and then use the molar mass to change it to moles.

The conversion process looks like this:
  • Convert 4 mg to grams: \ \(4 \text{ mg} = \frac{4}{1000} \text{ g} = 0.004\text{ g}\).
  • Convert grams to moles: \ \(\frac{0.004 \text{ g}}{32 \text{ g/mol}} = 0.000125 \text{ mol/L} = 1.25 \times 10^{-4} \text{ mol/L}\).

This means 4 ppm of oxygen in water equals approximately \(1.25 \times 10^{-4}\) mol/L.
oxygen solubility
Oxygen solubility is a key factor in aquatic environments because fish and other organisms rely on dissolved oxygen to survive. Solubility is determined by various factors, including temperature and the pressure of the gas above the liquid.

Henry's Law describes this relationship between the pressure of a gas and its solubility in a liquid. According to the law, the concentration of a dissolved gas is directly proportional to its partial pressure above the liquid:
  • \( C = k_H \times P \)
Where \(C\) is the concentration of the gas, \(k_H\) is the Henry's Law constant (specific to the gas and temperature), and \(P\) is the partial pressure.

For oxygen at 10°C, the Henry's Law constant \(k_H\) is given as \(1.71 \times 10^{-3} \text{ mol/L-atm}\). Knowing this constant allows you to calculate how pressure affects oxygen solubility.
partial pressure calculation
Calculating the partial pressure of a gas, like oxygen, involves using Henry’s Law. After determining the concentration of dissolved oxygen, you can use this information to find the necessary pressure.

The rearranged Henry's Law formula helps calculate the partial pressure:
  • \( P = \frac{C}{k_H} \)
Here, \(C\) is the concentration of oxygen you calculated earlier \( (1.25 \times 10^{-4} \text{ mol/L}) \) and \(k_H\) is \(1.71 \times 10^{-3} \text{ mol/L-atm}\).

Substitute these values to find the pressure:
  • \( P = \frac{1.25 \times 10^{-4} \text{ mol/L}}{1.71 \times 10^{-3} \text{ mol/L-atm}} \approx 0.0731 \text{ atm}\).

This approximation shows the partial pressure needed to maintain the required oxygen solubility at 10°C.

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Most popular questions from this chapter

A dilute aqueous solution of an organic compound soluble in water is formed by dissolving \(2.35 \mathrm{~g}\) of the compound in water to form \(0.250 \mathrm{~L}\) solution. The resulting solution has an osmotic pressure of \(0.605 \mathrm{~atm}\) at \(25^{\circ} \mathrm{C}\). Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) \(0.75 \mathrm{~L}\) of \(1.5 \times 10^{-2} M \mathrm{KBr}\), (b) \(125 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KBr}\) (c) \(1.85 \mathrm{~L}\) of a solution that is \(12.0 \% \mathrm{KBr}\) by mass (the density of the solution is \(1.10 \mathrm{~g} / \mathrm{mL}\) ), (d) a \(0.150 \mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough \(\mathrm{KBr}\) to precipitate \(16.0 \mathrm{~g}\) of \(\mathrm{AgBr}\) from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\)

(a) A sample of hydrogen gas is generated in a closed container by reacting \(2.050 \mathrm{~g}\) of zinc metal with \(15.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution is \(122 \mathrm{~mL}\). Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\), ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}-\mathrm{atm} .\) Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?

How does increasing the concentration of a nonvolatile solute in water affect the following properties: (a) vapor pressure, (b) freezing point, (c) boiling point; (d) osmotic pressure?

List four properties of a solution that depend on the total concentration but not the type of particle or particles present as solute. Write the mathematical expression that describes how each of these properties depends on concentration.

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