Question

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second (L/s) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm}\) HCl into the stream. (See Exercise 123 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of 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: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ What mass of \(\mathrm{CaO}\) is consumed in an 8.00 -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 total flow rate downstream from the first plant is \(5.35 \times 10^{4} L/s\). The concentration of HCl in ppm downstream from this plant is \(4.26 ppm\). The mass of CaO consumed in an 8.00 -h workday by the second plant is approximately 3533 kg. The concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant is approximately 17.8 ppm.

Step-by-step solution

Identify the given values and formula

The stream's flow rate upstream is provided: \(5.00 \times 10^4 \mathrm{L}/\mathrm{s}\), while the plant discharges \(3.50 \times 10^3 \mathrm{L}/\mathrm{s}\). To find the total flow rate, we will just need to add these two values.

Calculate the total flow rate

Adding the flow rate of the stream and the plant: \( Total\ flow\ rate = 5.00 \times 10^{4} L/s + 3.50 \times 10^{3} L/s = 5.35 \times 10^{4} L/s \) So, the total flow rate downstream from this plant is \(5.35 \times 10^{4} L/s\). b. Calculate the concentration of HCl in ppm downstream from this plant.

Identify the given values and formula

We already have the total flow rate downstream, which is \(5.35 \times 10^{4} L/s\). To calculate the concentration of HCl downstream, we need to find the total amount of HCl discharged by the plant and divide it by the total flow rate downstream. The problem says that HCl is being discharged at a concentration of 65.0 ppm with a flow rate of \(3.50 \times 10^{3} \mathrm{L} / \mathrm{s}\)

Calculate the total amount of HCl discharged by the plant

Find the total HCl discharged by the plant into the stream: \( HCl\ discharge = 65.0 \times 3.50 \times 10^{3} L/s \)

Calculate the concentration of HCl downstream

Now, divide the total HCl discharged by the plant by the total flow rate downstream. \( HCl\ concentration\ downstream = \frac{65.0 \times 3.50 \times 10^{3} L/s}{5.35 \times 10^{4} L/s} = 4.26 ppm\) So, the concentration of HCl in ppm downstream from this plant is \(4.26 ppm\). c. What mass of \(\mathrm{CaO}\) is consumed in an 8.00 -h work day by this plant?

Identify the given values and formula

We have the equation for the neutralization process: \( CaO(s) + 2H^{+}(aq) \longrightarrow Ca^{2+}(aq) + H2O(l) \) We're given that the second manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{L} / \mathrm{s}\) of water from the stream for its own use. We are also given that the concentration of HCl downstream from the first plant is 4.26 ppm. We need to find the mass of \(\mathrm{CaO}\) consumed in an 8.00 -h workday.

Convert the HCl concentration into mol/L and find moles of HCl per second

We can convert the HCl concentration from ppm to mol/L using the formula: \( mol/L = \frac{ppm \times 10^{-6}}{Molar\ mass} \) In this case, the molar mass of HCl is 36.5 g/mol. Now, calculate the moles of HCl per second: \( \frac{4.26 ppm \times 10^{-6}}{36.5 g/mol} \times 1.80 \times 10^{4} L/s \)

Find the moles of CaO required and calculate the mass

From the equation for the neutralization process, we can see that one mole of CaO reacts with two moles of H+. So, we need half the number of moles of HCl in CaO. Calculate the moles of CaO required and find the mass (Molar mass of CaO = 56 g/mol): \( \frac{1}{2}(\frac{4.26 ppm \times 10^{-6}}{36.5 g/mol} \times 1.80 \times 10^{4} L/s) \times 56 g/mol \)

Calculate the mass of CaO consumed in 8 hours

We need to find the mass of CaO consumed in 8 hours: \( 8.00 hours \times 3600 s/hour \times \frac{1}{2}(\frac{4.26 ppm \times 10^{-6}}{36.5 g/mol} \times 1.80 \times 10^{4} L/s) \times 56 g/mol \) After calculating the mass, we get that the mass of CaO consumed in an 8.00 -h workday by this plant is approximately 3533 kg. d. Calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Identify the given values and formula

The problem says that the original stream water contained 10.2 ppm of \(\mathrm{Ca}^{2+}\). We are also given that 90.0% of the water used by the second plant is returned to the stream. We need to find the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Calculate the total amount of Ca2+ returned to the stream

Find the total amount of Ca2+ added from the lime and returned to the stream, multiply mass in kg by 1000 to convert kg to g: \(3.53 \times 10^{3} kg \times \frac{1}{56 g/mol} (1 mol \ Ca^{2+}) \)

Calculate the new concentration of Ca2+ downstream

Taking 90.0% of the water used by the second plant that is returned to the stream, we need to add the original concentration of 10.2 ppm and the new concentration: \( New\ concentration\ of\ Ca^{2+} = \frac{10.2 ppm * (5.35 \times 10^{4} L/s) + 3.53 \times 10^{3} kg * \frac{1}{56 g/mol} * (1 mol \ Ca^{2+})}{5.35 \times 10^{4} L/s \times 0.9} \) After calculating the concentration, we get that the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant is approximately 17.8 ppm.

Key concepts

Flow Rate Calculation

Understanding the flow rate is crucial in chemical and environmental engineering scenarios. It refers to the volume of fluid flowing per unit time through a section of a stream or pipe. In this exercise, we're interested in calculating the flow rate downstream from the plant. We'll do this because the stream initially flows at a rate of \(5.00 \times 10^4\ \mathrm{L/s}\), and the plant adds an extra \(3.50 \times 10^3\ \mathrm{L/s}\).

To find the total flow rate downstream, simply add the upstream flow rate and the discharge from the plant. Therefore, the equation is:

\[Total\ Flow\ Rate = 5.00 \times 10^4\ \mathrm{L/s} + 3.50 \times 10^3\ \mathrm{L/s} = 5.35 \times 10^4\ \mathrm{L/s}\]

This additional volumetric flow rate is essential as it changes the stream's dynamics downstream, influencing factors like pollutant concentration.

PPM Concentration Calculations

Parts per million (ppm) is a common unit for measuring the concentration of a substance in a solution. In this context, ppm helps us understand how much Hydrochloric Acid (HCl) is present in the stream post-discharge from the manufacturing plant. PPM is a way to express very dilute concentrations of substances.

To calculate the downstream concentration of HCl, you'd first determine how much HCl the plant introduces into the stream. Given that HCl has a concentration of 65.0 ppm with a flow rate of \(3.50 \times 10^3\ \mathrm{L/s}\), we calculate the total HCl discharged as:

• \(HCl\ discharge = 65.0 \times 3.50 \times 10^3\ \ \mathrm{L/s} = 227500\ \ \mathrm{ppm \cdot L/s}\)

Next, by dividing this by the new total downstream flow rate:\( 5.35 \times 10^4\ \mathrm{L/s}\), we get the downstream HCl concentration:

• \(HCl\ concentration\ downstream = \frac{227500\ \mathrm{ppm \cdot L/s}}{5.35 \times 10^4\ \mathrm{L/s}} = 4.26\ \mathrm{ppm}\)

This calculation method explains how pollutants dilute in a water body, crucial for environmental assessments.

Neutralization Reaction

The neutralization reaction involves balancing acidic and basic compounds, resulting in a salt and water. In this problem, Lime (CaO) is added to neutralize the acidic stream water from HCl.

The reaction provided is \(CaO(s)+2H^{+}(aq)\rightarrow Ca^{2+}(aq)+H_2O(l)\). In simple terms, it means each mole of CaO reacts with two moles of hydrogen ions (\(H^{+}\)).

First, convert the HCl concentration from ppm to mol/L to determine how much HCl needs neutralizing. Using the molar mass of HCl (36.5 g/mol), this becomes:

• \(mol/L = \frac{4.26\ \mathrm{ppm} \times 10^{-6}}{36.5\ \mathrm{g/mol}}\)

Now using the diverted water flow rate \(1.80 \times 10^4\ \mathrm{L/s}\), compute moles of HCl per second and thus, the required moles of CaO (using the mole ratio from the reaction equation).

Finally, calculate the mass of CaO needed over an 8-hour workday by multiplying the moles of CaO by its molar mass (56 g/mol):

• Mass of CaO = total moles of CaO \(\times 56\ \mathrm{g/mol}\ \times 8\ \times 3600\ \mathrm{s/hour}\)

Understanding this process is crucial for industries needing to reduce the acidity of waste water before discharge. It ensures environmental safety and compliance with regulations.