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 \(8.5 ?(\mathbf{b})\) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgIO}_{3}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NaIO}_{3} ?\left(K_{s p}\right.\) of \(\mathrm{AgIO}_{3}\) is \(3.1 \times 10^{-8} .\) )

Short answer

In part (a), Co(OH)₂ will precipitate from the solution when the pH is adjusted to 8.5 since the ion product (2.00 × 10^(-11)) is greater than the Ksp of Co(OH)₂ (1.6 × 10^(-15)). In part (b), AgIO₃ will precipitate after mixing the solutions, as the ion product (3.34 × 10^(-5)) is greater than the given Ksp value (3.1 × 10^(-8)).

Step-by-step solution

1. Calculate the concentration of hydroxide ions

To find the concentration of OH- ions in the solution, we will use the relationship between pH and pOH, and the relationship between pOH and the concentration of OH-: pH + pOH = 14 From the given pH value, we can calculate pOH: pOH = 14 - 8.5 = 5.5 Then, using the formula: pOH = -log[OH-] we can calculate the concentration of hydroxide ions: [OH-] = 10^(-5.5) = 3.16 × 10^(-6) M #a) Check if Co(OH)2 will precipitate#

2. Calculate the solubility product constant for Co(OH)2

The solubility product constant, Ksp, for Co(OH)2 can be expressed as: Ksp = [Co²⁺] × [OH⁻]² We are given the concentration of Co²⁺ ions as 0.020 M. Using the calculated OH- concentration, we can determine if precipitation occurs according to the Ksp value. If the ion product ([Co²⁺] × [OH⁻]²) is greater than Ksp, then Co(OH)2 will precipitate. Ion product = (0.020) × (3.16 × 10^(-6))^2 = 2.00 × 10^(-11) The Ksp of Co(OH)2 is 1.6 × 10^(-15). Since the ion product is greater than Ksp (2.00 × 10^(-11) > 1.6 × 10^(-15)), Co(OH)2 will precipitate from the solution. #b) Check if AgIO3 will precipitate after mixing the solutions#

3. Calculate the new concentrations of Ag+ and IO3- ions after mixing

After mixing the AgIO3 and NaIO3 solutions, the concentration of each ion should be calculated by applying the dilution formula C1V1=C2V2: [Ag⁺] = (0.010 M × 20 mL) / (20 mL + 10 mL) = 6.67 × 10^(-3) M [IO3⁻] = (0.015 M × 10 mL) / (20 mL + 10 mL) = 5.00 × 10^(-3) M #b) Compare the ion product to the given Ksp value#

4. Determine if AgIO3 will precipitate

Now, with the new concentrations of Ag⁺ and IO3⁻ ions, we can determine if AgIO3 will precipitate by comparing the ion product to the given Ksp value: Ion product = [Ag⁺] × [IO3⁻] = (6.67 × 10^(-3)) × (5.00 × 10^(-3)) = 3.34 × 10^(-5) The given Ksp value for AgIO3 is 3.1 × 10^(-8). Since the ion product is greater than the Ksp value (3.34 × 10^(-5) > 3.1 × 10^(-8)), AgIO3 will precipitate after mixing the solutions.

Key concepts

Precipitation Reactions

In chemistry, a precipitation reaction is a crucial type of double displacement reaction where two aqueous solutions react to form an insoluble solid, known as a precipitate. This solid emerges from the reaction mixture, a pattern often observed with ionic compounds. To predict whether a precipitate forms, understanding the concept of solubility rules and the solubility product constant ( K_{sp} ) is essential.
The solubility product constant describes the extent to which a compound dissociates into its constituent ions in solution. This constant helps us predict whether a solid will precipitate in a given solution based on the concentration of ions present. If the ion product—a calculated value from the concentrations of the ions in the solution—exceeds the K_{sp} , precipitation occurs because the solution exceeds its capacity to hold those ions in solution.
For example, in the case of Co(OH)_2 , based on the calculated ion product and the known K_{sp} , a precipitate will form if the ion product is greater. Similarly, when solutions of AgIO_3 and NaIO_3 are mixed, AgIO_3 precipitates out if the ion product surpasses its K_{sp} .

Solubility Calculations

Solubility calculations play a vital role in predicting whether a given substance will dissolve or form a precipitate under specific conditions. These calculations often involve the determination of ion concentrations in solutions before and after mixing. The relationship between pH and pOH provides insight into the concentration of hydroxide ions in a solution.
The general calculation involves converting pH to pOH using the formula:\[ \text{pH} + \text{pOH} = 14 \]From the pOH, we can find the concentration of hydroxide ions ([OH^-]) using:\[ \text{pOH} = -\log[OH^-] \]
In an example where a solution's pH is 8.5, this conversion tells us that pOH is 5.5. Using the pOH, we calculate [OH^-] as 3.16 \times 10^{-6} M. This concentration is essential in determining whether certain hydroxide compounds will precipitate, like Co(OH)_2 in this case.
When two solutions are mixed, as with AgIO_3 and NaIO_3, it's crucial to calculate the new concentrations of ions after dilution using the formula:\[ C_1V_1 = C_2V_2 \]With this, we find post-mix ion concentrations and compare their ion product to the K_{sp} to validate precipitate formation predictions.

Chemical Equilibrium

Chemical equilibrium is a state in a reversible reaction where the forward and reverse reactions occur at the same rate. This state results in constant concentrations of reactants and products. In the context of solubility and precipitation reactions, equilibrium considerations involve the dynamic balance between dissolved ions and any solid precipitate that forms.
The equilibrium equation for solubility product (K_{sp}) links the ion concentrations in equilibrium with their solid form. For a solid like Co(OH)_2 dissociating into Co²⁺ and OH⁻ ions, it can be expressed as:\[ K_{sp} = [Co^{2+}][OH^-]^2 \]
This equation helps determine whether the solid will remain dissolved or precipitate. At equilibrium, the ion product equals K_{sp}. When the ion product exceeds K_{sp}, it indicates that the solution is supersaturated, favoring precipitate formation. Conversely, if the ion product is lower, the solution can dissolve more solute.
Understanding these principles is critical for predicting and controlling reactions where equilibrium plays a role, such as in laboratory syntheses and industrial applications. Equilibrium constants, like K_{sp}, thus serve as a crucial tool in assessing the behavior of ionic compounds in solutions.