Problem 97 - Chemistry: The Central Science in SI Units, Global Edition

Question

Lead(II) carbonate, \(\mathrm{PbCO}_{3}\), is one of the components of the passivating layer that forms inside lead pipes. (a) If the \(K_{s p}\) 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) The molarity of \(\mathrm{Pb}^{2+}\) ions in a saturated solution of lead(II) carbonate is approximately \(2.72 \times 10^{-7} \, \mathrm{M}\). (b) The concentration of \(\mathrm{Pb}^{2+}\) ions in ppb is 272 ppb. (c) The solubility of \(\mathrm{PbCO}_{3}\) will increase as the pH is lowered. (d) A saturated solution of lead(II) carbonate produces a solution with a lead ion concentration of 272 ppb, which exceeds the EPA limit of 15 ppb.

Step 1: Write the chemical equation for the dissolution of \(\mathrm{PbCO}_{3}\)

Begin by writing out the balanced chemical equation for the dissolution of \(\mathrm{PbCO}_{3}\) in water: \[\mathrm{PbCO}_3(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{CO}_3^{2-}(aq)\]

Step 2: Find the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution

Let's denote the molarity of \(\mathrm{Pb}^{2+}\) and \(\mathrm{CO}_3^{2-}\) in the saturated solution of \(\mathrm{PbCO}_{3}\) as \(x\). The solubility product constant, \(K_{sp}\), can be written as: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{CO}_3^{2-}] = x^2\] Substitute the given \(K_{sp}\) value of \(7.4 \times 10^{-14}\). and solve for \(x\): \[7.4 \times 10^{-14} = x^2\] \[x = \sqrt{7.4 \times 10^{-14}} \approx 2.72 \times 10^{-7} \, \mathrm{M}\] The molarity of \(\mathrm{Pb}^{2+}\) ions in a saturated solution of lead(II) carbonate is approximately \(2.72 \times 10^{-7} \, \mathrm{M}\).

Step 3: Convert the molarity to ppb

To convert the molarity of \(\mathrm{Pb}^{2+}\) ions to parts per billion (ppb), use the relationship: \[1 \, \mathrm{M} = 10^9 \, \mathrm{ppb}\] So, the concentration of \(\mathrm{Pb}^{2+}\) in ppb is: \[(2.72 \times 10^{-7} \, \mathrm{M}) \times \frac{10^9 \, \mathrm{ppb}}{1 \, \mathrm{M}} = 272 \, \mathrm{ppb}\]

Step 4: Determine the effect of lower pH on \(\mathrm{PbCO}_{3}\) solubility

Decreasing the pH in the solution will cause an increase in the concentration of \(\mathrm{H}^+\) ions. These ions will react with \(\mathrm{CO}_3^{2-}\) ions in the solution to produce \(\mathrm{HCO}_3^-\) (bicarbonate ions), as shown below: \[\mathrm{H}^+ + \mathrm{CO}_3^{2-} \rightleftharpoons \mathrm{HCO}_3^-\] This reaction will cause the \(\mathrm{CO}_3^{2-}\) ion concentration in the solution to decrease, which, in turn, decreases the denominator of the \(K_{sp}\) expression (as seen in Step 2). As the numerator remains constant, the decrease in the denominator will increase the solubility of \(\mathrm{PbCO}_{3}\). Therefore, the solubility of \(\mathrm{PbCO}_{3}\) will increase as the pH is lowered.

Step 5: Determine if the concentration of \(\mathrm{Pb}^{2+}\) exceeds the EPA limit

From Step 3, the concentration of \(\mathrm{Pb}^{2+}\) ions in a saturated solution of lead(II) carbonate is approximately 272 ppb. The EPA threshold for acceptable levels of lead ions in water is 15 ppb. As 272 ppb is greater than the EPA limit of 15 ppb, a saturated solution of lead(II) carbonate produces a solution in which the lead ion concentration exceeds the EPA limit.

Key Concepts

Molarity Calculation

When you're dealing with solubility equilibrium, finding the molarity of a dissolved ion is a vital skill. Molarity refers to the concentration of a solute in a solution, often expressed in moles per liter (M). In our problem with lead(II) carbonate (\(\mathrm{PbCO}_3\)), we start by identifying the solubility product constant, \(K_{sp}\), which is a measure of the dissolved ions' product of concentrations each raised to the power of their coefficients in the balanced chemical equation.

Knowing that the \(K_{sp}\) for \(\mathrm{PbCO}_3\) is \(7.4 \times 10^{-14}\), you can set up an equation where \(x\) represents the molarity of \(\mathrm{Pb}^{2+}\) and \(\mathrm{CO}_3^{2-}\) ions in the solution. It is important to understand the dissolution process: \(\mathrm{PbCO}_3(s)\rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{CO}_3^{2-}(aq)\), which shows a 1:1 molarity ratio for the ions produced.

This gives us the equation \(K_{sp} = x^2\). Solving for \(x\) will determine the molarity of \(\mathrm{Pb}^{2+}\) ions, which is found by substituting and solving \(x = \sqrt{7.4 \times 10^{-14}}\), resulting in \(2.72 \times 10^{-7} \, \mathrm{M}\). Remember, this calculation assumes the system has reached equilibrium, meaning the rate of dissolution equals the rate of precipitation.

pH Effect on Solubility

The pH of a solution can greatly influence the solubility of compounds, especially salts. Decreasing the pH means increasing the concentration of hydrogen ions (\(\mathrm{H}^+\)) in the solution. In the presence of carbonate salts like \(\mathrm{PbCO}_3\), the \(\mathrm{H}^+\) ions interact with \(\mathrm{CO}_3^{2-}\) ions to form \(\mathrm{HCO}_3^-\) ions, according to the reaction:

\[\mathrm{H}^+ + \mathrm{CO}_3^{2-} \rightleftharpoons \mathrm{HCO}_3^-\]

This transformation reduces the concentration of \(\mathrm{CO}_3^{2-}\) ions in the solution. With this decrease, the equilibrium shifts to dissolve more \(\mathrm{PbCO}_3\) to compensate and restore balance according to Le Chatelier's principle.
The lower concentration of \(\mathrm{CO}_3^{2-}\) ions effectively increases the solubility of \(\mathrm{PbCO}_3\) as more solid is dissolved to maintain the \(K_{sp}\) balance. As a result, a lower pH will indeed increase the solubility of \(\mathrm{PbCO}_3\), leading to a higher concentration of \(\mathrm{Pb}^{2+}\) ions in solution. This understanding is crucial in predicting and controlling the behavior of salts under different environmental conditions.

Solubility Product Constant (Ksp)

The solubility product constant, \(K_{sp}\), is at the core of predicting the solubility of salts in water. It represents the product of the molar concentrations of the ions involved, each raised to the power of their coefficients as described in the dissociation equation.
In the context of our exercise with \(\mathrm{PbCO}_3\), the \(K_{sp}\) value is \(7.4 \times 10^{-14}\). This low value indicates that \(\mathrm{PbCO}_3\) is not very soluble in water, as it dissociates into \(\mathrm{Pb}^{2+}\) and \(\mathrm{CO}_3^{2-}\) ions.
Understanding \(K_{sp}\) is crucial because:

• It helps in predicting whether a precipitate will form when two solutions are mixed.
• It allows the calculation of ion concentrations in saturated solutions.
• It provides insights into how changes in conditions (like pH, temperature) affect solubility.

In practical scenarios, the \(K_{sp}\) can be used to compare the solubility of different salts or to calculate the concentration of ions at equilibrium. For example, if a laboratory setting requires knowing if lead levels exceed safety thresholds, calculating \(K_{sp}\) combined with other equilibrium data can provide the necessary insights.
Thus, the \(K_{sp}\) stands as a fundamental part of solubility equilibrium, highlighting how chemical principles govern the behavior of ionic compounds in aqueous environments.