Problem 32 - Chemistry: Principles and Reactions

Question

Solid lead nitrate is added to a solution that is \(0.020 \mathrm{M}\) in \(\mathrm{OH}^{-}\) and \(\mathrm{SO}_{4}^{2-}\). Addition of the lead nitrate does not change the volume of the solution. (a) Which compound, \(\mathrm{PbSO}_{4}\) or \(\mathrm{Pb}(\mathrm{OH})_{2}\left(K_{s p}=2.8 \times 10^{-16}\right)\), will precipitate first? (b) What is the \(\mathrm{pH}\) of the solution when \(\mathrm{PbSO}_{4}\) first starts to precipitate?

Step-by-Step Solution

Answer

Answer: Lead sulfate (PbSO₄) will precipitate first. The pH value of the solution when PbSO₄ starts to precipitate can be calculated using the given hydroxide ion concentration ([OH⁻] = 0.020 M), by finding the pOH value, and then finding the pH value with: pH = 14 - pOH.

Step 1: Determine the solubility product constants (Ksp)

First, let's find the Ksp values for both lead sulfate (PbSO₄) and lead hydroxide (Pb(OH)₂). For PbSO₄, we can find its Ksp in a solubility table (it's not given), and for Pb(OH)₂, it's given as \(2.8 \times 10^{-16}\).

Step 2: Calculate concentrations of lead ions needed for precipitation

Next, we will determine the concentrations of lead ions required for PbSO₄ and Pb(OH)₂ to begin precipitating. We can use the Ksp expressions for this calculation: For PbSO₄: \[K_{sp} = [Pb^{2+}] [SO_{4}^{2-}]\] For Pb(OH)₂: \[K_{sp} = [Pb^{2+}] [OH^-]^2\] Let's plug in the known values and solve for the lead ion concentrations for each compound: For PbSO₄: Given that \([SO_{4}^{2-}] = 0.020\text{M}\), let's denote \([Pb^{2+}]\) as \(x\) and solve the Ksp expression: \[K_{sp} = x (0.020) \Rightarrow x = \frac{K_{sp}}{0.020}\] For Pb(OH)₂: Given that \([OH^-] = 0.020\text{M}\) and \(K_{sp} = 2.8 \times 10^{-16}\), \[2.8 \times 10^{-16} = x (0.020)^2 \Rightarrow x = \frac{2.8 \times 10^{-16}}{(0.020)^2}\] Now let's calculate the values of \(x\) for both compounds:

Step 3: Compare the lead ion concentrations

Now that we have calculated the lead ion concentrations required for the precipitation of PbSO₄ and Pb(OH)₂, we need to compare these values: For PbSO₄: \[x = \frac{K_{sp}}{0.020}\] For Pb(OH)₂: \[x = \frac{2.8 \times 10^{-16}}{(0.020)^2}\] The compound with the lowest x-value will precipitate first. Let's calculate these values and compare:

Step 4: Calculate the pH of the solution

Since we determined that PbSO₄ precipitates first, now we need to find the pH of the solution when that happens. We know the concentration of hydroxide ions (\([OH^-]\)), so we can calculate the pOH value and then the pH value: \[pOH = -log[OH^-]\] \[pH = 14 - pOH\] By substituting the given \([OH^-]\) value, we can find the pH of the solution when PbSO₄ starts to precipitate.

Key Concepts

Lead Ion Concentration

The concept of lead ion concentration is central to understanding the precipitation of compounds such as lead sulfate (\( \text{PbSO}_4 \)) and lead hydroxide (\( \text{Pb(OH)}_2 \)).
This concentration determines whether these compounds will form precipitates in a given solution.
When a solution containing lead ions reaches a concentration high enough to exceed the solubility product constant (\( K_{sp} \)) of a binary compound, that compound starts to precipitate out of the solution.Precipitation occurs because the ions in solution can no longer be accommodated in dissolved form and instead, form a solid phase.
For example, to determine which lead compound precipitates first, the required lead ion concentration for each compound is calculated using its respective \( K_{sp} \).
We then compare these concentrations:

• **Lead ion concentration required for \( \text{PbSO}_4 \) to precipitate:** Given \( [\text{SO}_4^{2-}] = 0.020 \text{M} \), the lead ion concentration is \( x = \frac{K_{sp}}{0.020} \).
• **Lead ion concentration required for \( \text{Pb(OH)}_2 \) to precipitate:**Given \( [\text{OH}^-] = 0.020 \text{M} \), the concentration calculates to \( x = \frac{2.8 \times 10^{-16}}{(0.020)^2} \).

The compound with the lower calculated concentration of \( \text{Pb}^{2+} \) indicates which one will precipitate first.

Solubility Product Constant (Ksp)

The solubility product constant, represented as \( K_{sp} \), is a vital concept in the discussion of chemical precipitation.
It provides a quantitative measure of solubility equilibrium for sparingly soluble salts.
In simple terms, \( K_{sp} \) helps us understand and predict when a salt will start forming a solid precipitate from its solution.
The value of \( K_{sp} \) is determined by the concentrations of the ions in a saturated solution and is unique for each compound.For example, consider the two compounds:

• **Lead sulfate (\( \text{PbSO}_4 \)**): The \( K_{sp} \) calculation involves \([\text{Pb}^{2+}][\text{SO}_4^{2-}]\).
• **Lead hydroxide (\( \text{Pb(OH)}_2 \)**): Given \( K_{sp} = 2.8 \times 10^{-16} \), the equation is \([\text{Pb}^{2+}][\text{OH}^-]^2\).

A lower \( K_{sp} \) generally means lower solubility, making it more likely for the salt to precipitate under common conditions.
Knowing \( K_{sp} \) is crucial for chemists to ensure that they can predict and control precipitation, which is important in many industrial and laboratory applications.

pH Calculation

Calculating the pH of a solution is an important skill in chemistry, especially when dealing with precipitation reactions, where it can impact solubility.
The pH value indicates the acidity or basicity of a solution and is inversely related to the concentration of hydrogen ions.When working with solutions where precipitation might occur, we often have to determine pH once we know the concentration of hydroxide ions (\([\text{OH}^-]\)).The relationship between pH, pOH, and \([\text{OH}^-]\) is given by:

• **pOH Calculation**: \( \text{pOH} = -\log[\text{OH}^-] \) allows us to understand how basic the solution is.
• **pH Calculation**: Once pOH is known, we find pH using \( \text{pH} = 14 - \text{pOH} \). This step tells us how acidic the environment is.

Using these steps helps predict whether an adjustment in pH can influence precipitation.
For instance, when lead sulfate begins to precipitate, calculating the pH gives us valuable information about the conditions present at this stage, such as ensuring this pH maintains the compounds in a stable form until necessary.