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The maximum allowable concentration of \(\mathrm{Pb}^{2+}\) ions in drinking water is 0.05 ppm (i.e., \(0.05 \mathrm{~g}\) of \(\mathrm{Pb}^{2+}\) in 1 million grams of water). Is this guideline exceeded if an underground water supply is at equilibrium with the mineral anglesite \(\left(\mathrm{PbSO}_{4}\right)\left(K_{\mathrm{sp}}=1.6 \times 10^{-8}\right) ?\)

Short Answer

Expert verified
The guideline is exceeded; the concentration is 8.288 ppm.

Step by step solution

01

Understand the Problem

We need to determine if the concentration of \( \text{Pb}^{2+} \) ions in water at equilibrium with anglesite \( \text{PbSO}_4 \) exceeds the regulatory limit of 0.05 ppm. The mineral's solubility product \( K_{sp} \) is given as \( 1.6 \times 10^{-8} \).
02

Express Solubility Product

The solubility product expression for anglesite \( \text{PbSO}_4 \) is given as:\[K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}]\]Since anglesite dissolves as \( \text{PbSO}_4 \rightarrow \text{Pb}^{2+} + \text{SO}_4^{2-} \), both ions have the same concentration \( s \). Therefore,\[K_{sp} = s^2\]where \( s \) is the solubility of \( \text{PbSO}_4 \).
03

Calculate Molar Solubility

Set up the equation using the \( K_{sp} \):\[s^2 = 1.6 \times 10^{-8}\]Solving for \( s \):\[s = \sqrt{1.6 \times 10^{-8}} = 4.0 \times 10^{-5} \text{ M}\]This is the concentration of \( \text{Pb}^{2+} \) ion in moles per liter.
04

Convert Molar Solubility to ppm

To convert molarity to ppm, use the following:\[\text{ppm} = \frac{\text{grams of solute}}{\text{grams of solution}} \times 10^6\]Calculate the grams of \( \text{Pb}^{2+} \):\[\text{grams of Pb}^{2+} = 4.0 \times 10^{-5} \times 207.2 = 8.288 \times 10^{-3} \text{ g/L}\]Thus, ppm = \( 8.288 \) ppm.
05

Compare with the Guideline

The calculated concentration of \( \text{Pb}^{2+} \) is 8.288 ppm, which is far greater than the guideline of 0.05 ppm. Therefore, if the water is at equilibrium with anglesite, it exceeds the allowable concentration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium
In chemistry, equilibrium refers to a state in which the rate of the forward reaction equals the rate of the reverse reaction. This results in the concentrations of reactants and products remaining constant over time. In the context of solubility and the solubility product, equilibrium is crucial. When a mineral such as anglesite is in contact with water, it dissolves until a point is reached where no more can dissolve, known as saturation. At this saturation point, we have dynamic equilibrium. The dissolved ions and the undissolved solid exist together in a stable balance. For anglesite, this equilibrium can be described by its solubility product (
Drinking Water Safety
Ensuring the safety of drinking water is pivotal as contaminants like lead can pose serious health risks. Regulations such as those set by the Environmental Protection Agency in the U.S. limit the concentration of harmful ions in drinking water. For lead, the maximum allowable concentration is 0.05 parts per million (ppm). Such guidelines are designed to protect from the adverse effects of lead exposure, which can range from developmental issues in children to cardiovascular problems in adults. Regular monitoring and testing of water supplies ensure compliance with safety standards, safeguarding public health.
Lead Concentration
When we talk about lead concentration in water, we often measure it in parts per million (ppm), which indicates the number of parts of lead per million parts of water. Lead can enter water supplies through natural mineral deposits or from lead-containing materials, like pipes or fittings. Excessive lead concentration is a concern because it can cause lead poisoning. In our analysis of the underground water at equilibrium with anglesite, the lead concentration was calculated as 8.288 ppm. This is dramatically higher than the safety limit set for drinking water, highlighting the importance of understanding and managing lead contamination in water resources.
Molar Solubility
Molar solubility is a measure of how many moles of a substance can dissolve in a liter of solvent to reach saturation. For salts like anglesite

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Most popular questions from this chapter

One way to distinguish a buffer solution with an acid solution is by dilution. (a) Consider a buffer solution made of \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) and \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\) Calculate its \(\mathrm{pH}\) and the \(\mathrm{pH}\) after it has been diluted \(10-\) fold. (b) Compare the result in part (a) with the pHs of a \(0.500 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) solution before and after it has been diluted 10 -fold.

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For which of the following reactions is the equilibrium constant called a solubility product? (a) \(\mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \rightleftarrows \mathrm{Zn}(\mathrm{OH})_{4}^{2-}(a q)\) (b) \(3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) (c) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons\) \(\mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)

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