Problem 63 - Chemistry: An Atoms-Focused Approach

Question

High concentrations of ammonia (NH \(_{3}\) ), nitrite ion, and nitrate ion in water can kill fish. Lethal concentrations of these species for rainbow trout are \(1.1 \mathrm{mg} / \mathrm{L}, 0.40 \mathrm{mg} / \mathrm{L},\) and \(1361 \mathrm{mg} / \mathrm{L},\) respectively. Express these concentrations in molality units, assuming a solution density of \(1.00 \mathrm{g} / \mathrm{mL}\).

Step-by-Step Solution

Answer

Answer: The lethal concentrations of ammonia, nitrite ion, and nitrate ion in molality units are: 1. Ammonia: \(6.457551 × 10^{-5} \mathrm{mol} / \mathrm{kg}\) 2. Nitrite ion: \(8.695282 × 10^{-6} \mathrm{mol} / \mathrm{kg}\) 3. Nitrate ion: \(2.193228 × 10^{-2} \mathrm{mol} / \mathrm{kg}\)

Step 1: Find the molecular weights

To convert the concentrations to molality, we first need to find the molecular weights of the three species involved: 1. Ammonia (NH\(_3\)): N(14.01) + 3 * H(1.01) = 17.04 g/mol 2. Nitrite ion (NO\(_2^-\)): N(14.01) + 2 * O(16.00) = 46.01 g/mol 3. Nitrate ion (NO\(_3^-\)): N(14.01) + 3 * O(16.00) = 62.01 g/mol

Step 2: Convert mg/L to g/L

We should convert the given concentrations from mg/L to g/L: 1. Ammonia: \(1.1 \mathrm{mg} / \mathrm{L} = 1.1 \times 10^{-3} \mathrm{g} / \mathrm{L}\) 2. Nitrite ion: \(0.40 \mathrm{mg} / \mathrm{L} = 0.40 \times 10^{-3} \mathrm{g} / \mathrm{L}\) 3. Nitrate ion: \(1361 \mathrm{mg} / \mathrm{L} = 1361 \times 10^{-3} \mathrm{g} / \mathrm{L}\)

Step 3: Convert g/L to mol/L

Now, we will convert the concentrations from g/L to mol/L using the molecular weights calculated in step 1: 1. Ammonia: \(1.1 \times 10^{-3} \mathrm{g} / \mathrm{L} \times \frac{1 \mathrm{mol}}{17.04 \mathrm{g}} = 6.457551\mathrm{-}5 \mathrm{mol} / \mathrm{L}\) 2. Nitrite ion: \(0.40 \times 10^{-3} \mathrm{g} / \mathrm{L} \times \frac{1\ \mathrm{mol}}{46.01 \mathrm{g}} = 8.695282\mathrm{-}6 \mathrm{mol} / \mathrm{L}\) 3. Nitrate ion: \(1361 \times 10^{-3} \mathrm{g} / \mathrm{L} \times \frac{1 \mathrm{mol}}{62.01\mathrm{g}} = 2.193228\mathrm{-}2 \mathrm{mol} / \mathrm{L}\)

Step 4: Convert mol/L to mol/kg

Finally, we will convert the concentrations from mol/L to mol/kg using the density of the solution: 1. Ammonia: \(6.457551{-}5 \mathrm{mol} / \mathrm{L} \times \frac{1 \mathrm{L}}{1 \mathrm{kg}} = 6.457551\mathrm{-}5 \mathrm{mol} / \mathrm{kg}\) 2. Nitrite ion: \(8.695282\mathrm{-}6 \mathrm{mol} / \mathrm{L} \times \frac{1 \mathrm{L}}{1 \mathrm{kg}} = 8.695282\mathrm{-}6 \mathrm{mol} / \mathrm{kg}\) 3. Nitrate ion: \(2.193228\mathrm{-}2 \mathrm{mol} / \mathrm{L} \times \frac{1 \mathrm{L}}{1 \mathrm{kg}} = 2.193228\mathrm{-}2 \mathrm{mol} / \mathrm{kg}\)

Step 5: Final Results

The lethal concentrations of ammonia, nitrite ion, and nitrate ion in molality units are: 1. Ammonia: \(6.457551{-}5 \mathrm{mol} / \mathrm{kg}\) 2. Nitrite ion: \(8.695282\mathrm{-}6 \mathrm{mol} / \mathrm{kg}\) 3. Nitrate ion: \(2.193228\mathrm{-}2 \mathrm{mol} / \mathrm{kg}\)

Key Concepts

Molecular Weight

Understanding molecular weight is crucial when working with solutions. The molecular weight of a compound is the sum of the atomic weights of all the atoms in a molecule. This allows for the conversion of a substance's mass to moles.

**How to Calculate Molecular Weight**
For instance, to find the molecular weight of ammonia (NH extsubscript{3}), you start by identifying the atomic weights of nitrogen (N) and hydrogen (H):

• Nitrogen has an atomic weight of approximately 14.01 g/mol.
• Hydrogen has an atomic weight of about 1.01 g/mol.

Ammonia has one nitrogen atom and three hydrogen atoms; hence, its molecular weight is calculated as follows: \[14.01 \, \text{g/mol} + 3 \times 1.01 \, \text{g/mol} = 17.04 \, \text{g/mol}\] This calculation is repeated similarly for other compounds like nitrite (NO extsubscript{2} extsuperscript{-}) and nitrate (NO extsubscript{3} extsuperscript{-}). Understanding these values helps in further calculations necessary to convert concentrations into different units like molality.

Conversion Calculations

Conversion calculations are necessary to express concentrations in different units such as molality. Starting from a given concentration in mg/L, you often need to convert it to g/L first.

This involves understanding that: 1. 1 mg/L = 0.001 g/L because 1 mg is 0.001 g.
2. Multiplying the concentration in mg/L by 0.001 gives the concentration in g/L. **Example Calculation**
If you start with ammonia at 1.1 mg/L: 1. Convert 1.1 mg to grams, resulting in 0.0011 g/L. Next, convert from g/L to mol/L using molecular weights found earlier: \[\text{mol/L} = \frac{\text{g/L}}{\text{Molecular Weight (g/mol)}}\] For ammonia: \[= \frac{1.1 \times 10^{-3} \text{g/L}}{17.04 \text{g/mol}} \approx 6.46 \times 10^{-5} \text{mol/L}\] With this, you are closer to expressing the concentration in molality units.

Solution Density

Solution density is a key factor when expressing concentrations in molality. Molality ( extit{m}) is defined as the number of moles of solute per kilogram of solvent.

Often, the density of a solution is necessary to convert from molarity (mol/L) to molality (mol/kg). When a solution's density is given as 1.00 g/mL (or equivalently 1 kg/L), the conversion becomes straightforward.

**Using Density for Conversion**

• The conversion from mol/L to mol/kg is simplified by the fact that 1 L of this solution weighs 1 kg.

For our example with ammonia: \[\text{molality} = 6.457551 \times 10^{-5} \text{mol/L} \times \frac{1 \text{L}}{1 \text{kg}} = 6.457551 \times 10^{-5} \text{mol/kg}\] Thus, solution density helps directly bridge the conversion between molarity and molality, especially when the density equals 1.00 g/mL.