Problem 114 - Chemistry: The Central Science

Question

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion ( \(\mathrm{ppb})\) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\), what mass of sodium arsenate would be present in a \(1.00\)-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\text { g solution }} \times 10^{9} $$

Step-by-Step Solution

Answer

The mass of sodium arsenate (\(\mathrm{Na}_{3}\mathrm{AsO}_{4}\)) present in a 1.00-L sample of drinking water that just meets the U.S. standard for arsenate in drinking water (10 ppb arsenic) is approximately \(2.78 \times 10^{-5} \: \text{g}\).

Step 1: Find the mass of arsenic in the solution.

We need to find the mass of arsenic in the 1.00-L sample of drinking water that contains 10 ppb. Using the given formula for ppb, ppb = \(\frac{\text{g solute}}{\text{g solution}} \times 10^9\) We can rearrange the formula to: g solute (arsenic) = \(\frac{\text{ppb} \times \text{g solution}}{10^9}\)

Step 2: Calculate the mass of the solution.

We know that the volume of the drinking water is 1.00 L. Because the density of water is approximately 1g/mL or 1g/cm³, the mass of 1 L (1000 mL) of water is approximately 1000 g.

Step 3: Calculate the mass of arsenic that meets the standard.

Substituting the given ppb value and the mass of the solution into the formula: g solute (arsenic) = \(\frac{10 \times 1000}{10^9} = 10^{-5} \: \text{g}\) So, 10^-5 grams of arsenic is present in the 1.00-L sample of drinking water.

Step 4: Calculate the mass of sodium arsenate.

We need to find the mass of sodium arsenate, \(\mathrm{Na}_{3}\mathrm{AsO}_{4}\), which contains 10^-5 grams of arsenic. First, we must determine the molar masses: Molar mass of \(\mathrm{As}\) = 74.92 g/mol (from periodic table) Molar mass of \(\mathrm{Na}_{3}\mathrm{AsO}_{4}\) = 3(22.99) + 74.92 + 4(16.00) = 207.98 g/mol Now, we can find the mass of sodium arsenate using the mass of arsenic and the molar masses: mass of sodium arsenate = \(\frac{\text{mass of arsenic} \times \text{molar mass of sodium arsenate}}{\text{molar mass of arsenic}}\)

Step 5: Calculate the mass of sodium arsenate that meets the standard.

Substituting the known values: mass of sodium arsenate = \(\frac{10^{-5} \times 207.98}{74.92} = 2.78 \times 10^{-5} \: \text{g}\) Therefore, 2.78 x 10^-5 grams of sodium arsenate would be present in a 1.00-L sample of drinking water that just meets the U.S. standard for arsenate in drinking water.

Key Concepts

Arsenate Analysis

Before navigating the math and chemistry behind arsenate standards in drinking water, it's vital to understand what arsenates are. Arsenates are chemical compounds containing the arsenate ion, \(\text{AsO}_4^{3-}\). This polyatomic ion is the key player in many environmental and biological contexts.
In drinking water, arsenate is often considered a contaminant due to its potential health impacts. The U.S. Environmental Protection Agency (EPA) sets a strict limit to ensure safety, capping arsenate levels at just 10 parts per billion (ppb).
This regulation ensures that any harmful effects from arsenates are minimized, keeping our drinking water safe. The analysis is centered around determining whether a given water sample surpasses these limits, often involving precise chemical calculations and measurements.

Parts Per Billion Calculation

When dealing with minute concentrations like those of arsenate in water, scientists use a scale called parts per billion (ppb). Calculating these figures is crucial for ensuring that water meets safety standards.
The general formula for parts per billion is expressed as:

• \( \text{ppb} = \frac{\text{g solute}}{\text{g solution}} \times 10^9 \)

This formula helps in determining how many grams of a substance (solute), such as arsenate, are present per billion grams of solution (water in this case).
To apply this, suppose you're working with a 1.00-liter sample of water, equivalent to 1000 grams due to water's density of 1 g/mL. Using the 10 ppb standard for arsenate:

• \( g \text{ solute (arsenic) } = \frac{10 \times 1000}{10^9} = 10^{-5} \text{ g} \)

This means in your 1-liter water sample, only 0.00001 grams of arsenic is present. Such precision underscores the delicate balance necessary for maintaining water safety.

Molar Mass Determination

Understanding molar mass is a cornerstone of chemistry, especially when calculating masses of compounds like sodium arsenate based on their constituent elements. Molar mass represents the mass of one mole of any given substance, expressed in grams per mole (g/mol).
For sodium arsenate (\(\text{Na}_3\text{AsO}_4\)), you'll find its molar mass by summing up the molar masses of its composing atoms. Here's a quick calculation to show how it's done:

• Molar mass of \(\text{Na}_3\text{AsO}_4\) = 3(22.99) + 74.92 + 4(16.00) = 207.98 \text{ g/mol}

Each sodium atom (Na) contributes 22.99 g/mol, the arsenic atom (As) adds 74.92 g/mol, and each oxygen atom (O) provides 16.00 g/mol.
This sum gives the total molar mass of sodium arsenate. Having calculated this, you can determine how much sodium arsenate corresponds to a given mass of arsenic. In our previous calculation, to find the mass equivalent to 10^-5 grams of arsenic:

• mass of sodium arsenate = \( \frac{10^{-5} \times 207.98}{74.92} = 2.78 \times 10^{-5} \text{ g} \)

Such calculations are vital for converting between different forms of the same substance, allowing scientists to ensure compliance with safety standards in various contexts.