Problem 97 - Chemistry The Central Science In Si Units

Question

Lead(II) carbonate, \(\mathrm{PbCO}_{3}\), is one of the components of the passivating layer that forms inside lead pipes. (a) If the \(K_{i p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? (d) The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

Step-by-Step Solution

Answer

(a) \( 8.6 \times 10^{-8} \ \mathrm{mol/L} \); (b) 17.8 ppb; (c) Solubility increases; (d) Exceeds EPA limit.

Step 1: Write the Equilibrium Expression

For the dissolution of lead(II) carbonate in water, we can write the equilibrium reaction as follows: \[ \mathrm{PbCO}_{3} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + \mathrm{CO}_{3}^{2-} (aq) \] The solubility product, \( K_{sp} \), for this equilibrium is given by \[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{CO}_{3}^{2-}] \] where \( [\mathrm{Pb}^{2+}] \) and \( [\mathrm{CO}_{3}^{2-}] \) are the molar concentrations of lead and carbonate ions in a saturated solution.

Step 2: Determine Ion Concentrations in Saturated Solution

Since lead(II) carbonate dissociates in a 1:1 ratio to lead and carbonate ions, we can let \( [\mathrm{Pb}^{2+}] = s \) and \( [\mathrm{CO}_{3}^{2-}] = s \), where \( s \) is the solubility of \( \mathrm{PbCO}_{3} \) in mol/L. Thus, \( K_{sp} = s^2 \).The expression becomes: \[ 7.4 \times 10^{-14} = s^2 \].

Step 3: Solve for Molarity of \( \mathrm{Pb}^{2+} \)

To find \( s \), or the molarity of \( \mathrm{Pb}^{2+} \), take the square root of both sides of the equation:\[ s = \sqrt{7.4 \times 10^{-14}} \approx 8.6 \times 10^{-8} \ /mol/L \].Thus, the molarity of \( \mathrm{Pb}^{2+} \) in the saturated solution is approximately \( 8.6 \times 10^{-8} \ \mathrm{mol/L} \).

Step 4: Convert Molarity to ppb

First, convert the molarity of \( \mathrm{Pb}^{2+} \) to grams per liter using the molar mass of lead, which is 207.2 g/mol:\[ 8.6 \times 10^{-8} \ \mathrm{mol/L} \times 207.2 \ \mathrm{g/mol} = 1.78 \times 10^{-5} \ \mathrm{g/L} \].Convert grams per liter to parts per billion (ppb) as follows:\[ 1.78 \times 10^{-5} \ \mathrm{g/L} \times 10^6 = 17.8 \ \mathrm{ppb} \].The concentration of \( \mathrm{Pb}^{2+} \) ions is 17.8 ppb.

Step 5: Effect of pH on Solubility

As the \( \text{pH} \) decreases, \( [\mathrm{H}^+] \) ions increase. These \( [\mathrm{H}^+] \) ions can react with \( \mathrm{CO}_{3}^{2-} \) ions to form \( \mathrm{HCO}_{3}^{-} \) ions. \[ \mathrm{H}^+ + \mathrm{CO}_{3}^{2-} \rightleftharpoons \mathrm{HCO}_{3}^{-} \] This reduces the \( \mathrm{CO}_{3}^{2-} \) concentration, shifting the equilibrium to the right, thus increasing the solubility of \( \mathrm{PbCO}_{3} \).

Step 6: Compare to EPA Limit

The concentration of \( \mathrm{Pb}^{2+} \) ions in a saturated solution of lead(II) carbonate is 17.8 ppb. Since this exceeds the EPA permissible limit of 15 ppb, the solution's lead concentration is above the EPA limit.

Key Concepts

Lead(II) Carbonate

Lead(II) carbonate, often represented as \(\mathrm{PbCO}_{3}\), is a sparingly soluble compound. This means that it only dissolves a small amount in water. It's frequently found in environments where lead is present because it forms part of the protective layer inside lead pipes. This coating can help to reduce the release of lead into water, but understanding its solubility behavior is essential for water safety management. The compound dissolves to form lead ions \(\mathrm{Pb}^{2+}\) and carbonate ions \(\mathrm{CO}_{3}^{2-}\) in a balanced chemical reaction: \[ \mathrm{PbCO}_{3} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + \mathrm{CO}_{3}^{2-} (aq) \] This equilibrium is central to determining how much lead can leach into the water from the pipes, especially under varying conditions such as changes in pH.

Solubility Product Constant

The Solubility Product Constant, known as \(K_{sp}\), is a valuable tool for chemists when dealing with sparingly soluble compounds like lead(II) carbonate. This constant represents the maximum extent to which a compound can dissolve in water to form ions. For lead(II) carbonate, the \(K_{sp}\) value is determined by the molar concentrations of the ions at equilibrium: \[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{CO}_{3}^{2-}] \] Given the \(K_{sp}\) of \(7.4 \times 10^{-14}\), it allows us to understand how much \(\mathrm{Pb}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\) are present in equilibrium. This low \(K_{sp}\) value implies that lead(II) carbonate is not very soluble, which is critical for applications involving lead pipes as it indicates that they'll often be passivated by this layer, potentially protecting water sources.

Ion Concentration Calculations

Calculating ion concentrations in saturated solutions often begins with the equilibrium expression for the compound in question. For lead(II) carbonate, the dissolution in water is represented by \[ \mathrm{PbCO}_{3} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + \mathrm{CO}_{3}^{2-} (aq) \] This 1:1 stoichiometry means that the concentration of \(\mathrm{Pb}^{2+}\) is equal to that of \(\mathrm{CO}_{3}^{2-}\), namely \(s\), the solubility of \(\mathrm{PbCO}_{3}\). Given as \( K_{sp} = s^2 \), by solving the equation \[ s^2 = 7.4 \times 10^{-14} \] we find that \( s \approx 8.6 \times 10^{-8} \ \mathrm{mol/L} \). Converting this to a practical measure like parts per billion (ppb), as shown by the process of converting moles to grams and then to ppb, can give us a sense of how much of the ion is present in real-world quantities. For instance, the conversion could give us a value of 17.8 ppb, which needs to be compared to regulatory standards like those of the EPA.

Impact of pH on Solubility

The solubility of lead(II) carbonate is sensitive to the pH of the solution. pH can significantly affect how chemicals behave in aqueous environments. As pH decreases, indicating an increase in \([\mathrm{H}^+]\) ions, these ions react with \(\mathrm{CO}_{3}^{2-}\) ions: \[ \mathrm{H}^+ + \mathrm{CO}_{3}^{2-} \rightleftharpoons \mathrm{HCO}_{3}^{-} \] This reaction reduces the concentration of \(\mathrm{CO}_{3}^{2-}\) ions. By Le Chatelier's principle, the equilibrium shifts to the right to dissolve more \(\mathrm{PbCO}_{3}\), thereby increasing its solubility.This principle explains why acidic waters can dissolve more lead from lead carbonate layers, potentially resulting in higher lead content in drinking water. Thus, monitoring pH is crucial when assessing the risk of lead contamination in water systems.