Question

Silver ions can be found in some of the city water piped into homes. The average concentration of silver ions in city water is \(0.028 \mathrm{ppm} .\) (a) How many milligrams of silver ions would you ingest daily if you drank eight glasses (eight oz/glass) of city water daily? (b) How many liters of city water are required to recover \(1.00 \mathrm{~g}\) of silver chemically?

Short answer

Answer: By drinking eight glasses of city water, we ingest approximately 0.0529 mg of silver ions daily. Approximately 35,714 liters of city water are required to recover 1.00 g of silver chemically.

Step-by-step solution

Convert ounces to liters

First, we need to convert the eight glasses of city water (eight oz/glass) to liters. There are 33.814 ounces in a liter, so we can calculate the volume of eight glasses in liters: $$(8 \text{ glasses}) \times (8 \text{ oz/glass}) \times \left(\frac{1 \text{ liter}}{33.814 \text{ oz}}\right)$$

Calculate the daily intake of silver ions

Now, we can calculate the amount of silver ions we ingest daily by using the conversion factor \(0.028\mathrm{ppm}\), which means that there are \(0.028\mathrm{mg}\) of silver ions in a liter of city water. We first convert the ppm to mg/L and then multiply it by the volume of water in liters: $$0.028\text{mg silver/liter}\cdot8\cdot8\cdot\frac{1\text{L}}{33.814\text{oz}}$$

Simplify the expression

Now, we can simplify the expression above to get the amount of silver ions ingested daily in milligrams: $$0.028\cdot8\cdot\frac{8}{33.814}\approx0.0529\mathrm{mg}$$ So, you would ingest about \(0.0529\mathrm{mg}\) of silver ions daily by drinking eight glasses of city water.

Calculate how many liters of city water are needed to recover 1.00 g of silver

Finally, we need to find out how many liters of city water are required to recover \(1.00\mathrm{g}\) of silver chemically. To do this, we can use the ratio between the silver ions' concentration in city water and the mass of silver ions we want to recover: $$\frac{0.028\text{ mg silver}}{1\text{ L city water}}\:=\:\frac{1.00\cdot1000\text{ mg silver}}{x\text{ L city water}}$$ We multiplied \(1.00\mathrm{g}\) by \(1000\) to convert grams to milligrams.

Solve for the unknown x

Now, solve for the unknown x, which represents the liters of city water needed to recover \(1.00\mathrm{g}\) of silver chemically: $$x\:=\:1.00\cdot1000\cdot\frac{1\text{L city water}}{0.028\text{mg silver}}$$

Region 6: Simplify the expression

Simplify the expression to get the number of liters of city water needed: $$x\approx\frac{1000}{0.028}\approx 35714\text{L}$$ Hence, approximately \(35714\text{ liters}\) of city water are required to recover \(1.00\mathrm{g}\) of silver chemically.

Key concepts

ppm to mg/L Conversion

Understanding how to convert parts per million (ppm) to milligrams per liter (mg/L) is crucial in chemistry, especially when dealing with solutions' concentration. Essentially, ppm is a unit of concentration that denotes the mass of a chemical or contaminate per unit volume of water. It indicates that out of one million parts of a solution, one part is the solute (the substance dissolved in the solution).

In the case of water, because the density of water is approximately 1 g/mL (which is equivalent to 1000 mg/L), 1 ppm corresponds to 1 mg/L. This is due to the fact that a liter of water, which weighs 1000 g, has 1,000,000 mg in it. Thus, if a substance's concentration is 1 ppm, it would mean it contains 1 mg of that substance in every liter of water. So, to convert ppm to mg/L for water solutions, you can use a 1:1 conversion ratio without any complex calculations.

When the textbook exercise asks for the amount of silver ions you would ingest from drinking city water with a silver concentration of 0.028 ppm, we are effectively saying there are 0.028 mg of silver ions per liter of city water. To find out how much silver you ingest, we multiply the concentration in mg/L by the amount of water you drink in liters, simplifying the detailed math into an accessible concept for daily intake calculations.

Chemical Mass Recovery

Chemical mass recovery is an important concept that relates to how much of a substance can be obtained from a solution. It involves calculations that help determine the efficiency of recovery processes or the quantity of a solute present in a certain volume of solution.

For example, if someone wanted to 'recover' silver from city water, we need to consider the concentration of silver ions in ppm (or mg/L after conversion) and the total volume of water we have. The exercise provided shows a practical application of this concept: calculating how many liters of city water are needed to recover a certain mass of silver chemically.

By setting up a proportion between the concentration of silver ions and the desired mass (in mg after converting from grams), we can find the volume of water needed. This proportion also illustrates the inverse relationship between concentration and volume for a fixed mass of solute. It reveals how the dilute concentration of a solution (like city water with very low ppm of silver) requires a significant volume to accumulate a substantial amount of solute (silver ions) for recovery purposes.

Unit Conversion in Chemistry

Unit conversion is a fundamental skill in chemistry that helps students and professionals to accurately interpret and work with different measurements. The units used can significantly affect how one understands a problem and applies formulas or constants.

In the context of the textbook exercise, the crucial unit conversions included converting ounces to liters and grams to milligrams. Since scientific measurements typically use the metric system, understanding how to switch between different systems of measurement is key. For instance, the conversion rate from ounces to liters (33.814 oz to 1 L) is essential for calculating the volume of water consumed, and the conversion of grams to milligrams (1.00 g to 1000 mg) is necessary to find out how much water is needed for recovery.

It's not uncommon for students to overlook the importance of units or make simple errors during conversion, which can lead to incorrect results. Emphasizing the need to perform these conversions accurately ensures that the calculations made when solving chemistry problems reflect the correct quantities and allow for meaningful scientific interpretations.