Question

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

Short answer

(a) Co(OH)2 will not precipitate because the calculated ion product (IP) is smaller than the solubility product constant (Ksp) of Co(OH)2. (b) AgIO3 will precipitate since the ion product (IP) is greater than the solubility product constant (Ksp) of AgIO3.

Step-by-step solution

Find the OH- concentration

Given the pH of the solution is 8.5, we can calculate the pOH using the formula: pOH = 14 - pH pOH = 14 - 8.5 pOH = 5.5 Now, let's find the concentration of OH- ions. The formula for calculating the concentration of OH- ions is: \[OH^-] = 10^{-pOH}\] \[OH^-] = 10^{-5.5}\]

Find the Ksp of Co(OH)2

We can find the Ksp of Co(OH)2 from a reference table. The solubility product constant Ksp for Co(OH)2 is 1.6 × 10^-15.

Calculate the ion product and compare with Ksp

Now, let's calculate the ion product (IP) for Co(OH)2 using the concentrations of Co2+ and OH- ions: \[IP = [Co^{2+}][OH^-]^2\] Since the concentration of Co2+ ions is 0.020 M and we found the concentration of OH- ions as \(10^{-5.5} M\), we can plug in these values: \[IP = (0.020)(10^{-5.5})^2\] Now, compare the ion product (IP) to the solubility product constant (Ksp) for Co(OH)2: If IP > Ksp, then Co(OH)2 will precipitate. If IP < Ksp, then Co(OH)2 will not precipitate. • Part (b):

Calculate the final concentrations of Ag+ and IO3- ions

We are given that 20 mL of 0.010 M AgNO3 is mixed with 10 mL of 0.015 M NaIO3. We must find the final concentrations of Ag+ and IO3- ions. First, let's find the moles of each ion: moles of Ag+ = (20 mL)(0.010 mol/L) moles of IO3- = (10 mL)(0.015 mol/L) Next, we need to find the total volume of the mixture: total volume = 20 mL + 10 mL = 30 mL Now, let's find the final concentrations of both ions: \[Ag^+ = \frac{moles\ of\ Ag^+}{total\ volume}\] \[IO3^- = \frac{moles\ of\ IO3^-}{total\ volume}\]

Calculate the ion product for AgIO3

Now, let's calculate the ion product (IP) for AgIO3 using the final concentrations of Ag+ and IO3- ions: \[IP = [Ag^+][IO3^-]\]

Compare the ion product to Ksp of AgIO3

We are given that the solubility product constant Ksp for AgIO3 is 3.1 × 10^-8. Now, compare the calculated ion product (IP) to the solubility product constant (Ksp) for AgIO3: If IP > Ksp, then AgIO3 will precipitate. If IP < Ksp, then AgIO3 will not precipitate.

Key concepts

Precipitation Reactions

Precipitation reactions are a fascinating area of chemistry and are often used to determine whether a particular compound will remain dissolved in solution or form a solid precipitate. In simple terms, precipitation occurs when the concentration of product ions in solution exceed the solubility limit.
When two solutions containing different ions are mixed, if a product of the reaction has a lower solubility in water, that compound precipitates out. Precipitation reactions rely heavily on the solubility product constant (Ksp), which is a crucial factor in predicting whether a reaction will result in precipitation.
Some key points to understanding precipitation reactions include:

• In a reaction, if the ionic product (IP) exceeds the Ksp of the compound in question, a precipitate will form.
• If the ionic product is less than the Ksp, no precipitate will form, and the ions remain dissolved.
• The concept is essential to chemical analysis, wastewater treatment, and the formation of minerals in natural processes.

These reactions are critical for students to grasp not only for laboratory experiments but also to understand how solubility impacts substances in real-world scenarios.

Ion Product

The ion product (IP) plays a crucial role in predicting the occurrence of a precipitation reaction. It's the product of the molar concentrations of the ions involved in forming a potential precipitate, raised to the power of their coefficients in the balanced equation.
The mathematical treatment of the ion product helps chemists determine if the conditions are right for a compound to form a precipitate. The ion product is calculated in a similar manner to the equilibrium constant but applies to the initial state of a solution.
Consider the case of the compound Co(OH) ₂ as an example:

• Given the equilibrium reaction: Co(OH) ₂ ⇌ Co ²⁺ + 2 OH ^-
• The ion product formula would be: IP = [ Co ²⁺][ OH ^- ]²
• If IP > Ksp, the solution favors the formation of precipitate.
• If IP < Ksp, the ions remain in solution, and no precipitation occurs.

Understanding ion product is essential for tackling problems related to precipitation and solubility of ionic compounds in solutions.

pH Calculations

pH calculations are essential in the field of chemistry for determining the acidity or basicity of a solution. The pH scale ranges from 0 to 14, with a pH of 7 being neutral, values below 7 indicating acidity, and values above 7 indicating basicity.
For solutions with hydroxide ions, it often involves calculating the pOH first, especially in precipitation reactions involving bases. pOH is related to pH by the equation:

pOH = 14 - pH

This simple formula allows us to convert between pOH and pH quickly. For example, if a solution's pH is 8.5, the pOH can be easily found as 14 - 8.5 = 5.5.
Once pOH is known, the concentration of hydroxide ions ([OH ^-]) can be determined as follows:

• [OH ^-] = 10^-pOH

This relationship is used to predict and analyze precipitation reactions when dealing with basic compounds, such as Co(OH) ₂, and influences the calculation of the ion product.

Ksp Calculations

Ksp calculations are pivotal in understanding the solubility of slightly soluble compounds. The solubility product constant, Ksp, signifies the extent to which a compound will dissolve in water. It is defined for the equilibrium between a solid and its ions in an aqueous solution.
Ksp is a fixed value for any particular compound at a given temperature, and it helps determine whether a precipitate will form under specific conditions.
To calculate Ksp, the concentrations of the ions must be known or calculated using the given conditions:

• Start by writing the dissociation equation, e.g., for AgIO₃: AgIO ₃ ⇌ Ag ⁺ + IO ^3-
• The Ksp expression is written as: Ksp = [ Ag ⁺][ IO ^3-]
• By substituting the concentrations, one can determine if the solution is saturated (ion product equals Ksp), unsaturated (ion product is less than Ksp), or supersaturated (ion product is greater than Ksp).

Mastery of Ksp calculations allows prediction and understanding of how different conditions affect solubility, and thus, when a precipitate will form in reactions.