Question

If \(10.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) is added to 10.0 \(\mathrm{mL}\) of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?

Short answer

A precipitate will not form when 10.0 mL of \(2.0 \times 10^{-3} M\) Cr(NO₃)₃ is added to 10.0 mL of pH=10.0 NaOH solution, as the calculated solubility product (Q) is \(1.25 \times 10^{-15}\), which is less than the solubility product constant (Ksp) of Cr(OH)₃, approximately \(6.3 \times 10^{-32}\).

Step-by-step solution

Calculate the concentration of OH⁻ ions in NaOH solution

Given the pH of the solution, we can find the concentration of OH⁻ ions. pH = 10.0 pOH = 14 - pH = 14 - 10 = 4.0 The concentration of OH⁻ ions is: \[OH^-] = 10^{-pOH} = 10^{-4} = 1.0 \times 10^{-4} M\]

Find the initial concentration of Cr³⁺ and OH⁻ ions

When two solutions with equal volumes (10.0 mL each) are mixed, the total volume becomes 20.0 mL. We can now find the concentration of ions in the mixture: \[Cr^{3+}]_{initial} = \frac {2.0 \times 10^{-3} M \times 10.0 mL}{20.0 mL} = 1.0 \times 10^{-3} M\] \[OH^-]_{initial} = \frac {1.0 \times 10^{-4} M \times 10.0 mL}{20.0 mL} = 0.5 \times 10^{-4} M\]

Write the solubility product expression and find Q

The balanced chemical equation for the possible precipitation reaction is: \[Cr^{3+} + 3OH^- \longleftrightarrow Cr(OH)_3\] Now, let's write the solubility product expression and find the reaction quotient Q: \[Q = [Cr^{3+}][OH^-]^3\] Substitute the initial concentrations: \[Q = (1.0 \times 10^{-3})(0.5 \times 10^{-4})^3 = 1.25 \times 10^{-15}\]

Compare Q with Ksp to determine if precipitation would occur

For a precipitate to form, the value of Q must be greater than or equal to Ksp. The Ksp of Cr(OH)₃ is approximately 6.3 × 10⁻³². Comparing the values: \(Q = 1.25 \times 10^{-15}\) \(Ksp = 6.3 \times 10^{-32}\) Since \(Q > Ksp\), a precipitate will not form. Therefore, adding 10.0 mL of 2.0 × 10⁻³ M Cr(NO₃)₃ to 10.0 mL of pH=10.0 NaOH solution will not result in the formation of a precipitate.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, commonly referred to as Ksp, is a crucial concept in understanding precipitation reactions.
Ksp quantifies the maximum amount of a solid that can dissolve in a solution before it starts forming a precipitate. It is unique to every compound and depends on temperature.
For instance, consider the dissociation of chromium(III) hydroxide, \[ Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq) \]Here, the Ksp expression can be written as:\[ Ksp = [Cr^{3+}][OH^-]^3 \]This expression reflects the concentrations of the ions in a saturated solution at equilibrium. When these concentrations exceed the Ksp value, a precipitate forms because the solution cannot hold any more dissolved ions.
Understanding and utilizing Ksp helps predict whether combining certain solutions will result in precipitation.

Ion Concentration Calculation

Ion concentration calculation is fundamental to understanding how solutions behave when mixed.
It involves determining the molarity of ions present in a solution, which can change when solutions are combined.In our exercise, initially the chromium nitrate solution has a concentration of \[ 2.0 \times 10^{-3} M \] and the NaOH solution, with a pH of 10, has an OH⁻ concentration of \[ 1.0 \times 10^{-4} M \].Upon mixing equal volumes of two solutions, the concentration is adjusted because the total volume has doubled. This is calculated using the formula:\[ [ ext{Ion}] = \frac {[ ext{Ion}]_{initial} \times \text{Volume}_{initial}}{\text{Total Volume}_{final}} \]. This ensures the concentration reflects its diluted form after mixing, which is vital for correct comparisons with the Ksp.

Chemical Equilibria

Chemical equilibria are the state of balance between reactants and products in a chemical reaction.
In precipitation reactions, reaching equilibria involves the saturation point where no more solute can dissolve, and any excess forms a precipitate.This is shown in the reaction:\[ Cr^{3+} + 3OH^- \rightleftharpoons Cr(OH)_3 \]The forward reaction involves the formation of a solid, and the reverse is its dissolution. When a solution reaches equilibrium, the rate of the forward reaction equals the rate of the reverse, and the concentrations of reactants and products remain constant over time.
Assessment of equilibrium in precipitation reactions uses the reaction quotient, Q, which is calculated similarly to Ksp but with the actual ion concentrations in the solution. Comparing Q to Ksp determines the solubility state:

• If \( Q < Ksp \), the solution is unsaturated and more solute can dissolve.
• If \( Q = Ksp \), the solution is at equilibrium; no precipitation or dissolution occurs.
• If \( Q > Ksp \), the solution is supersaturated, leading to the formation of a precipitate.

Understanding equilibria aids in predicting and explaining the results of mixing various chemical solutions.