Problem 16 - Chemistry: An Atoms-Focused Approach

Question

In the human body, the concentrations of ultratrace essential elements are even lower than those of trace essential elements and therefore are sometimes expressed in parts per billion. Express the concentrations of each of the following elements in parts per billion: a. Bromine, \(6 \mathrm{mg} / \mathrm{L}\) b. Boron, \(0.014 \mathrm{g} / 100 \mathrm{kg}\) c. Selenium, \(5.0 \mathrm{mg} / 70 \mathrm{kg}\)

Step-by-Step Solution

Answer

Question: Convert the given concentrations of the following elements into parts per billion (ppb): a. Bromine: \(6 \mathrm{mg} / \mathrm{L}\) b. Boron: \(0.014 \mathrm{g} / 100 \mathrm{kg}\) c. Selenium: \(5.0 \mathrm{mg} / 70 \mathrm{kg}\) Answer: a. Bromine concentration: \(6,000,000 \, \mathrm{ppb}\). b. Boron concentration: \(140 \, \mathrm{ppb}\). c. Selenium concentration: ≈ \(71.43 \, \mathrm{ppb}\).

Step 1: Understanding parts per billion (ppb)

Parts per billion is a unit to express very small concentrations. 1 part per billion means 1 part in 1,000,000,000 parts, or in other words, 1 part in 10^9 parts. It can be used to express concentrations in mass/mass, volume/volume, or mass/volume units.

Step 2: Convert Bromine concentration to ppb

Given, Bromine concentration: \(6 \mathrm{mg} / \mathrm{L}\). To convert this concentration to ppb, we need to convert milligrams (mg) to nanograms (ng), since 1 ppb equals 1 ng/L. 1 mg = 1,000,000 ng So, \(6 \mathrm{mg} = 6 \times 1,000,000 \, \mathrm{ng} = 6,000,000 \, \mathrm{ng}\). Thus, Bromine concentration in ppb is \(6,000,000 \, \mathrm{ng} / \mathrm{L}\).

Step 3: Convert Boron concentration to ppb

Given, Boron concentration: \(0.014 \mathrm{g} / 100 \mathrm{kg}\). To convert this concentration to ppb, we first convert grams (g) to nanograms (ng) and kilograms (kg) to grams (g), since 1 ppb is equal to 1 ng/g. 1 g = 1,000,000,000 ng and 1 kg = 1,000 g Therefore, \(0.014 \mathrm{g} = 0.014 \times 1,000,000,000 \, \mathrm{ng} = 14,000,000 \, \mathrm{ng}\) and \(100 \mathrm{kg} = 100 \times 1,000 \, \mathrm{g} = 100,000 \, \mathrm{g}\). Thus, Boron concentration in ppb is \(14,000,000 \, \mathrm{ng} / 100,000 \, \mathrm{g}\) = \(140 \, \mathrm{ppb}\).

Step 4: Convert Selenium concentration to ppb

Given, Selenium concentration: \(5.0 \mathrm{mg} / 70 \mathrm{kg}\). To convert this concentration to ppb, we first convert milligrams (mg) to nanograms (ng) and kilograms (kg) to grams (g). 1 mg = 1,000,000 ng and 1 kg = 1,000 g Therefore, \(5.0 \mathrm{mg} = 5.0 \times 1,000,000 \mathrm{ng} = 5,000,000 \, \mathrm{ng}\) and \(70 \mathrm{kg} = 70 \times 1,000 \, \mathrm{g} = 70,000 \, \mathrm{g}\). Thus, Selenium concentration in ppb is \(5,000,000 \, \mathrm{ng} / 70,000 \, \mathrm{g}\) ≈ \(71.43 \, \mathrm{ppb}\).

Step 5: Final answers

a. Bromine concentration: \(6,000,000 \, \mathrm{ppb}\). b. Boron concentration: \(140 \, \mathrm{ppb}\). c. Selenium concentration: ≈ \(71.43 \, \mathrm{ppb}\).