Question

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second \((\mathrm{L} / \mathrm{s})\) upstream of a manufacturing plant. The plant discharges \(3.50 \times\) \(10^{3} \mathrm{~L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm} \mathrm{HCl}\) into the stream. (See Exercise 113 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of \(\mathrm{HCl}\) in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{~L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What mass of \(\mathrm{CaO}\) is consumed in an \(8.00-\mathrm{h}\) work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+}\). Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Short answer

The stream's total flow rate downstream of the first plant is \(5.35 \times 10^4 L/s\), and the concentration of HCl downstream from that plant is 4.25 ppm. The second plant consumes 235,984.61 g of CaO in an 8-hour workday. After the second plant returns 90% of the water used, the concentration of Ca^2+ downstream is 15.47 ppm.

Step-by-step solution

Calculate the total flow rate downstream from the first plant

To calculate the total flow rate downstream, we add the stream's initial flow rate to the flow rate of the water discharged by the first plant: Total flow rate downstream = Initial flow rate + Discharged flow rate Total flow rate downstream = \(5.00 \times 10^4 L/s + 3.50 \times 10^3 L/s\) Total flow rate downstream = \(5.35 \times 10^4 L/s\)

Calculate the concentration of HCl downstream from the first plant

First, we'll find the mass of HCl added by the first plant: Mass of HCl = Discharged flow rate * HCl concentration Mass of HCl = \(3.50 \times 10^3 L/s * 65.0 ppm\) Note: 1 ppm = \(1 \times 10^{-6} g/L\) Mass of HCl = \(3.50 \times 10^3 L/s * 65.0 * 10^{-6} g/L\) Mass of HCl = \(2.275 \times 10^{-1} g/s\) Next, we'll calculate the concentration of HCl downstream: HCl concentration downstream = Mass of HCl / Total flow rate downstream HCl concentration downstream = \(\frac{2.275 \times 10^{-1} g/s}{5.35 \times 10^4 L/s} \) HCl concentration downstream = \(4.25 \times 10^{-6} g/L = 4.25 ppm\)

Calculate the mass of CaO needed to neutralize HCl for the second plant in 8 hours

The second plant takes \(1.80 \times 10^4 L/s\) of water at a concentration of 4.25 ppm of HCl. To neutralize it, we need to use the stoichiometry of the reaction: 1 mol CaO neutralizes 2 mol HCl: \(CaO + 2H^{+} \rightarrow Ca^{2+} + H_2O\) First, we calculate the moles of HCl to be neutralized: Moles of HCl = Flow rate * Time * HCl concentration_downstream / molar_mass_HCl Moles of HCl = \(\frac{1.80 \times 10^4 L/s * 8.00 h * 60 min/h * 60 s/min * 4.25 \times 10^{-6} g/L}{36.46 g/mol}\) Moles of HCl = \(8409.85 mol\) Next, we calculate the moles of CaO needed to neutralize it: Moles of CaO = Moles of HCl / 2 = \(4204.93 mol\) Now, we calculate the mass of CaO needed: Mass of CaO = Moles of CaO * molar_mass_CaO Mass of CaO = \(4204.93 mol * 56.08 g/mol\) Mass of CaO = \(235984.61 g\)

Calculate the concentration of Ca^2+ downstream of the second plant

The original stream water contained 10.2 ppm Ca^2+. Additionally, the waste water from the second plant contains Ca^2+ from the neutralization reaction (converting CaO to Ca^2+). The mass of Ca^2+ added by the waste water is the same as the mass of CaO used in the neutralization, so: Mass of Ca^2+ added = Mass of CaO used = \(235984.61 g\) Now, we need to determine how much water is returned to the stream after being used by the second plant: Water returned = Water used by the second plant * 0.90 Water returned = \(1.80 \times 10^4 L/s * 0.90\) Water returned = \(1.62 \times 10^4 L/s\) Now, we can find the total flow rate after the second plant: Total flow rate downstream_second_plant = Total flow rate downstream_first_plant - Water used by the second plant + Water returned Total flow rate downstream_second_plant = \(5.35 \times 10^4 L/s - 1.80 \times 10^4 L/s + 1.62 \times 10^4 L/s\) Total flow rate downstream_second_plant = \(5.17 \times 10^4 L/s\) Finally, we can calculate the Ca^2+ concentration downstream: Ca^2+ concentration downstream = (Initial Concentration * Initial Flow Rate + Ca^2+ mass added) / Total Flow Rate downstream_second_plant Ca^2+ concentration downstream = \(\frac{10.2 \times 10^{-6} g/L * 5.00 \times 10^4 L/s + 235984.61}{5.17 \times 10^4 L/s}\) Ca^2+ concentration downstream = \(15.47 \times 10^{-6} g/L = 15.47 ppm\)

Key concepts

Flow Rate Calculation

Flow rate calculation is essential for understanding how much water passes a point in the stream each second. It's like measuring the speed at which water flows in a river.
Here, we calculate the flow rate downstream from the first plant. We start by considering the initial flow rate of the stream upstream, which is given as \(5.00 \times 10^{4} \mathrm{~L/s}\). To find the new flow rate downstream, you add the flow of the water discharged by the manufacturing plant, \(3.50 \times 10^{3} \mathrm{~L/s}\).

It’s just simple addition:

• Total flow rate = Initial flow rate + Discharged flow rate

Plug the numbers in and you get:

• Total flow rate = \(5.00 \times 10^{4} \mathrm{~L/s} + 3.50 \times 10^{3} \mathrm{~L/s} = 5.35 \times 10^{4} \mathrm{~L/s}\)

This calculation helps us understand how much water is flowing downstream, an important first step to analyzing any changes in the water's properties further down the stream.

Concentration in ppm

Concentration measured in parts per million (ppm) tells us how much of a given substance is present per million parts of the solution. For example, 1 ppm is equivalent to 1 milligram of substance per liter of water.
The concentration of hydrochloric acid (HCl) downstream from the first plant is calculated by considering the mass of the acid being discharged. Since the plant releases water with 65 ppm of HCl at a rate of \(3.50 \times 10^{3} \mathrm{~L/s}\), we can calculate the mass of HCl added as:

• Mass of HCl = Discharged flow rate × HCl concentration
• Mass of HCl = \(3.50 \times 10^{3} \mathrm{~L/s} \times 65 \times 10^{-6} \mathrm{g/L}\)
• Mass of HCl = \(2.275 \times 10^{-1} \mathrm{g/s}\)

Now, using the total flow rate downstream, the concentration of HCl becomes:

• HCl concentration downstream = Mass of HCl / Total flow rate
• HCl concentration downstream = \(\frac{2.275 \times 10^{-1} \mathrm{g/s}}{5.35 \times 10^{4} \mathrm{~L/s}}\)
• HCl concentration downstream = 4.25 ppm

This shows how diluted the HCl is when mixed with the larger volume of water.

Stoichiometry

Stoichiometry is like a recipe for chemical reactions, using the balanced chemical equation to relate the amount of reactants to products.
In the reaction where lime (CaO) is used to neutralize HCl, we use the equation:
\[\mathrm{CaO} + 2\,\mathrm{H}^{+} \rightarrow \mathrm{Ca}^{2+} + \mathrm{H}_2\mathrm{O}\]
From this equation, one mole of CaO neutralizes two moles of HCl. First, determine the moles of HCl for the entire duration the plant operates (an 8-hour workday):

Use:

• Moles of HCl = \(\frac{\text{Flow rate} \times \text{Time} \times \text{Concentration}}{\text{Molar mass of HCl}}\)

Calculating gives:

• Moles of HCl = \(\frac{1.80 \times 10^4 \mathrm{L/s} \times 28800 \mathrm{~s} \times 4.25 \times 10^{-6} \mathrm{g/L}}{36.46 \mathrm{g/mol}} = 8409.85 \) mol
• Moles of CaO needed = Moles of HCl / 2 = 4204.93 mol

Lastly, convert moles of CaO to mass:

• Mass of CaO = Moles of CaO \(\times\) molar mass of CaO
• Mass of CaO = \(4204.93 \times 56.08 \mathrm{g/mol} = 235984.61 \mathrm{g}\)

This step shows how much lime is needed for neutralization.

Neutralization Reaction

Neutralization reactions are all about balancing acidic and basic substances to produce water and sometimes salts. Here, lime (CaO) neutralizes hydrochloric acid in the stream water.
The chemical reaction involved is:

\[\mathrm{CaO} + 2\,\mathrm{H}^{+} \rightarrow \mathrm{Ca}^{2+} + \mathrm{H}_2\mathrm{O}\]

In this scenario, the amount of \(\mathrm{CaO}\) used in neutralization directly affects the concentration of calcium ions \((\mathrm{Ca}^{2+})\) in the waste water.
After neutralization, 90% of the water goes back to the stream, making it crucial to calculate how much extra \(\mathrm{Ca}^{2+}\) is introduced into the downstream flow.
Calculating the new concentration of \(\mathrm{Ca}^{2+}\) is done by considering:

• Total \(\mathrm{Ca}^{2+}\) mass = \(235984.61 \mathrm{g}\)
• Water returned to stream = 90% of \(1.80 \times 10^4 \mathrm{~L/s}\)
• Total downstream flow = \(5.35 \times 10^4 \mathrm{L/s} - 1.80 \times 10^4 \mathrm{L/s} + 1.62 \times 10^4 \mathrm{L/s} = 5.17 \times 10^4 \mathrm{L/s}\)

Ca\(^{2+}\) concentration is then calculated as:

• Ca^{2+} concentration downstream = Total \(\mathrm{Ca}^{2+}\) mass / Total flow rate
• Ca^{2+} concentration downstream = \(\frac{10.2 \times 10^{-6} \mathrm{g/L} \times 5.00 \times 10^4 \mathrm{L/s} + 235984.61}{5.17 \times 10^4 \mathrm{L/s}}\)
• Ca^{2+} concentration downstream = 15.47 ppm

This calculation informs us about the increase in calcium levels due to the processing at the second plant.