Question

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the \(\mathbf{p H}\) of a 0.020 M solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to 8.5? (b) Will \(\mathrm{AgIO}_{3}\) precipitate when 20 mL of 0.010 M \(\mathrm{AglO}_{3}\) is mixed with 10 mL of 0.015 \(M \mathrm{NaIO}_{3}\)? ( \(K_{s p}\) of \(\mathrm{AgIO}_{3}\) is \(3.1 \times 10^{8}\))?

Short answer

(a) After calculating the hydroxide ion concentration [OH⁻] and the concentration of \(\mathrm{Co^{2+}}\) ions, the ion product \(Q\) is found to be smaller than the \(K_{sp}\) for \(\mathrm{Co}(\mathrm{OH})_{2}\). Therefore, \(\mathrm{Co}(\mathrm{OH})_{2}\) will not precipitate at pH 8.5. (b) After mixing the solutions and calculating the new concentrations of \(\mathrm{Ag^{+}}\) and \(\mathrm{IO_{3}^{-}}\), the ion product \(Q\) is found to be smaller than the \(K_{sp}\) for \(\mathrm{AgIO}_{3}\). Therefore, \(\mathrm{AgIO}_{3}\) will not precipitate under these conditions.

Step-by-step solution

Calculate the hydroxide ion concentration from the given pH

The given pH of the solution is 8.5. To find the hydroxide ion concentration [OH⁻], use the relationship between pH and pOH: \(pOH = 14 - pH\) Calculate the pOH and find the [OH⁻] using: \([OH^-] = 10^{-pOH}\)

Find the concentration of \(\mathrm{Co^{2+}}\) ions

Since it is a 0.020 M solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2},\) the concentration of \(\mathrm{Co^{2+}}\) ions in the solution is 0.020 M.

Calculate the ion product \(Q\) and compare it to \(K_{s p}\)

The ion product \(Q\) is calculated as follows: \(Q = [\mathrm{Co^{2+}}][\mathrm{OH^-]}^2\) If \(Q > K_{s p}\), precipitation will occur. (b) Precipitation of \(\mathrm{AgIO}_{3}\)

Calculate the concentrations after mixing the solutions

After mixing 20 mL of 0.010 M \(\mathrm{AgIO}_{3}\) with 10 mL of 0.015 M \(\mathrm{NaIO}_{3}\), calculate the new concentrations by finding the moles of each ion and divide by the total volume.

Calculate the ion product \(Q\) and compare it to \(K_{s p}\)

Similar to the previous case, calculate the ion product \(Q\) as follows: \(Q = [\mathrm{Ag^{+}}][\mathrm{IO_{3}^{-}}]\) If \(Q > K_{s p}\), precipitation will occur.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as Ksp, is a vital concept in chemistry, particularly when discussing the solubility of ionic compounds in water. Simply put, Ksp is a measure of the solubility of a compound under equilibrium conditions. It represents the maximum product of the concentrations of the ions produced in a saturated solution.

To illustrate how Ksp works, consider a generic compound represented as AB, which dissolves in water to form A+ and B- ions. The expression for Ksp is written as an equilibrium constant:

• For the dissolution reaction: AB (s) ⇌ A⁺ (aq) + B⁻ (aq)
• The Ksp = [A⁺][B⁻]

When the ion product Q, calculated from the actual concentrations of ions in a solution, is greater than Ksp, precipitation occurs. Otherwise, the solution remains unsaturated.

Ion Product (Q)

The ion product, symbolized as Q, is a snapshot of the multiplicative concentration of ions in a solution at any moment. While Ksp is a constant under equilibrium, Q helps us decide if a solution has reached that point. It can be thought of as a real-time checkup for potential precipitation processes.

Calculating Q involves taking the product of the concentrations of the dissociated ions, each raised to the power of their stoichiometric coefficients in the balanced equation. For example:

• In the case of ionic compound AB: Q = [A⁺][B⁻]
• If Q > Ksp, the solution is supersaturated, leading to precipitation.
• If Q < Ksp, more solute can dissolve, indicating an unsaturated solution.

Thus, comparing Q and Ksp helps predict whether precipitation will occur in a given solution.

Solution pH

The pH of a solution is a measure of its acidity or basicity. It plays a significant role in many chemical reactions, including precipitation. The scale ranges from 0 to 14, with 7 being neutral. Values less than 7 denote acidic solutions, whereas those above indicate basic solutions.

Understanding the link between pH and hydroxide ion concentration is crucial for predicting precipitation events in bases. Specifically, when adjusting the pH of a solution, you affect the concentration of hydroxide ions ofollow{-OH^-}. The relationship is given by:

• 
  <[OH^-] = 10^{-pOH}>

Thus, manipulating pH can either trigger or hinder the formation of precipitates in reactions involving ions like Co(OH)₂.

Molar Concentration

Molar concentration, also called molarity, is a fundamental concept, which expresses the number of moles of a solute present per litre of solution. It provides a way to quantify how concentrated a solution is.

For example, a 1 M solution means there is 1 mole of solute per litre of solution. Here’s how to calculate it:

• Use the formula: Molarity (M) = Moles of solute / Litres of solution.
• This relationship helps in determining reactant amounts for reactions.

In precipitation reactions, knowing the molarity of the ions involved is essential, as it allows for the accurate calculation of the ion product, Q. By understanding their concentrations, you are able to predict and interpret when a precipitate will form.