Problem 42 - Chemistry: Principles and Reactions

Question

Chlorine can remove the foul smell of \(\mathrm{H}_{2} \mathrm{~S}\) in water. The reaction is $$\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)+\mathrm{S}(s)$$ If the contaminated water has \(5.0\) ppm hydrogen sulfide by mass, what volume of chlorine gas at STP is required to remove all the \(\mathrm{H}_{2} \mathrm{~S}\) from \(1.00 \times 10^{3}\) gallons of water \((d=1.00 \mathrm{~g} / \mathrm{mL}) ?\) What is the \(\mathrm{pH}\) of the solution after treatment with chlorine?

Step-by-Step Solution

Answer

Answer: The volume of chlorine gas required is 12.44 L at STP, and the pH of the solution after treatment with chlorine is 3.53.

Step 1: 1. Determine the mass of H2S in the water

Since the concentration of hydrogen sulfide is given in ppm (parts per million), we have 5.0 mg H2S per 1 million mg (1 L) of water. The volume of water is given as \(1.00 \times 10^{3}\) gallons, and since 1 gallon is approximately equal to 3.785 L, we can convert this to liters: $$V_\text{water} = 1.00 \times 10^{3} \, \text{gal} \times \frac{3.785 \, \text{L}}{1 \, \text{gal}} = 3.785 \times 10^{3} \, \text{L}$$ Next, we can find the mass of H2S in the water: $$m_\text{H2S} = 3.785 \times 10^{3} \, \text{L} \times \frac{5.0 \, \text{mg}}{1 \, \text{L}} = 1.8925 \times 10^{4} \, \text{mg}$$

Step 2: 2. Convert the mass of H2S to moles

Using the molar mass of H2S (34.08 g/mol), we can convert the mass of H2S in mg to moles: $$n_\text{H2S} = \frac{1.8925 \times 10^{4} \, \text{mg}}{34.08 \, \text{g/mol} \times \frac{1 \, \text{g}}{1000 \, \text{mg}}} = 0.5555 \, \text{mol}$$

Step 3: 3. Determine the moles of Cl2 needed

From the balanced chemical equation, we can see that we need 1 mole of Cl2 for each mole of H2S, so the moles of Cl2 needed are the same as the moles of H2S: $$n_\text{Cl2} = 0.5555 \, \text{mol}$$

Step 4: 4. Find the volume of Cl2 gas at STP

We can use the ideal gas law to find the volume of Cl2 gas at STP (0°C and 1 atm). First, we need to create the equation: $$PV = nRT$$ Since the conditions are at STP, we can use the standard molar volume of 22.4 L/mol: $$V_\text{Cl2} = n_\text{Cl2} \times V_\text{molar,STP} = 0.5555 \,\text{mol} \times 22.4 \,\text{L/mol} = 12.44 \,\text{L}$$

Step 5: 5. Determine the pH of the solution after treatment with chlorine

After the reaction, the concentration of \(\mathrm{H}^{+}\) ions is doubled in comparison to the initial concentration of \(\mathrm{H}_{2}\mathrm{S}\). First, we determine the initial concentration of \(\mathrm{H}_{2}\mathrm{S}\): $$[\mathrm{H}_{2}\mathrm{S}] = \frac{0.5555 \, \text{mol}}{3785 \, \text{L}} = 1.47 \times 10^{-4} \, \text{M}$$ Now we can determine the concentration of \(\mathrm{H}^{+}\) ions after the reaction: $$[\mathrm{H}^{+}] = 2 \times [\mathrm{H}_{2}\mathrm{S}] = 2 \times 1.47 \times 10^{-4} \, \text{M} = 2.94 \times 10^{-4} \, \text{M}$$ Finally, we can calculate the pH of the solution: $$\text{pH} = -\log{[\mathrm{H}^{+}]} = -\log{(2.94 \times 10^{-4})} = 3.53$$ So, the volume of chlorine gas required to remove all the \(\mathrm{H}_{2}\mathrm{S}\) from \(1.00 \times 10^{3}\) gallons of water is 12.44 L at STP, and the pH of the solution after treatment with chlorine is 3.53.

Key Concepts

Stoichiometry

In chemistry, stoichiometry is all about understanding the relationships between reactants and products in a chemical reaction. Essentially, it allows us to determine how much of each substance is needed for a reaction to occur, as well as what will be produced. To conduct stoichiometric calculations, you have to begin with a balanced chemical equation. This equation tells you the ratio of moles of each substance involved in the reaction.
Let's consider the reaction of hydrogen sulfide (\(\mathrm{H_2S}\)) and chlorine (\(\mathrm{Cl_2}\)). The balanced chemical equation: \[\mathrm{H_2S(aq) + Cl_2(aq) \to 2H^+(aq) + 2Cl^-(aq) + S(s)}\] tells us that one mole of \(\mathrm{H_2S}\) reacts with one mole of \(\mathrm{Cl_2}\). This direct 1:1 relationship is crucial for determining the amount of \(\mathrm{Cl_2}\) needed to react fully with a given amount of \(\mathrm{H_2S}\).
When working with stoichiometry, it's common to use mole ratios as conversion factors for translating between amounts of reactants and products.

Molarity Calculations

Molarity is a way to express concentration, which refers to the amount of solute dissolved in a given volume of solution. It's usually represented by the formula \[\text{Molarity (M) = } \frac{\text{moles of solute}}{\text{liters of solution}}\] Understanding molarity is key to making precise chemical solutions and calculations.
In the original problem, we begin by determining the moles of \(\mathrm{H_2S}\) dissolved in water. Given that the water sample contains \(5.0\) parts per million (ppm) of \(\mathrm{H_2S}\), we can use this to find the initial molarity for the reaction. The concentration of \(\mathrm{H_2S}\) was calculated to be \(1.47 \times 10^{-4}\) M.
Knowing the molarity helps us calculate the amount of \(\mathrm{Cl_2}\) needed to treat the water, maintaining an effective neutralization of the contaminants.

Ideal Gas Law

The ideal gas law combines several gas laws to describe the behavior of ideal gases. The formula is expressed as \[PV = nRT\] where:

• \(P\) is the pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the gas constant (0.0821 L atm/mol K),
• \(T\) is the temperature in Kelvin.

In standard temperature and pressure (STP) conditions, which are \(0\)°C and \(1\) atm, one mole of an ideal gas occupies \(22.4\) liters.
For the chlorine gas required in the reaction, we can rearrange the ideal gas law to solve for the volume: \[V = nRT/P\] However, at STP, you can simply multiply the moles of gas by the molar volume (\(22.4\) L/mol). Thus, for \(0.5555\) moles of \(\mathrm{Cl_2}\), the volume is \(12.44\) L.

Acid-Base Chemistry

Acid-base chemistry is crucial to understand the pH changes in a solution during reactions. pH is a measure of hydrogen ion concentration, with \[pH = -\log{[H^+]}\]. It tells us how acidic or basic a solution is.
In the given reaction, the treatment of \(\mathrm{H_2S}\) with \(\mathrm{Cl_2}\) results in the production of \(\mathrm{H^+}\) ions. The initial concentration of \(\mathrm{H_2S}\) was calculated, and since the formation of \(\mathrm{H^+}\) ions is twice that of \(\mathrm{H_2S}\), we use this to determine the concentration of \(\mathrm{H^+}\).
The resulting pH of \(3.53\) indicates an acidic solution. This illustrates how stoichiometric calculations and an understanding of acid-base chemistry are interconnected in determining the effects of chemical treatments on water pH.