Question

A solution is prepared by mixing \(35.00 \mathrm{~mL}\) of a \(0.061 \mathrm{M}\) solution of zinc nitrate with \(20.0 \mathrm{~mL}\) of \(\mathrm{KOH}\) with a pH of \(9.00\). Assume that volumes are additive. (a) Will precipitation occur? (b) Calculate \(\left[\mathrm{Zn}^{2+}\right],\left[\mathrm{NO}_{3}^{-}\right],\left[\mathrm{K}^{+}\right]\), and the \(\mathrm{pH}\) after equilibrium is established.

Short answer

If so, what are the final concentrations of the ions and the pH after mixing? Answer: Yes, precipitation will occur when the zinc nitrate and KOH solutions are mixed. The final concentrations of the ions and the pH can be calculated as follows: Final \([\mathrm{Zn}^{2+}] = \frac{Ksp}{[\mathrm{OH}^{-}]^2}\) Final \(\left[\mathrm{OH}^{-}\right] = \sqrt{\frac{Ksp}{\left[\mathrm{Zn}^{2+}\right]}}\) Final \(\left[\mathrm{NO}_{3}^{-}\right]=\frac{0.122\times35.00}{55.0}M\) Final \(\left[\mathrm{K}^{+}\right]=\left[\mathrm{OH}^{-}\right]\) \(pOH = -\log \left[\mathrm{OH}^{-}\right]\) \(pH = 14 - pOH\)

Step-by-step solution

Determine the initial concentration of each ion

Given \(0.061 M\) solution of zinc nitrate \(\left(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\right)\) with a volume of \(35.00 mL\) and a KOH solution with a pH of \(9.00\) in a \(20.0 mL\) volume, we can find the concentrations of each ion in moles per liter. For the zinc nitrate solution: \(\left[\mathrm{Zn}^{2+}\right]=0.061M\) Since there are 2 nitrate ions in one zinc nitrate \(\left(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\right)\), we have: \(\left[\mathrm{NO}_{3}^{-}\right]=2\times0.061M=0.122M\) For the KOH solution: Since pH = \(9.00\), \(\mathrm{pOH} = 14-9.00=5.00\) \(\left[\mathrm{OH}^{-}\right]= 10^{-\mathrm{pOH}}=10^{-5.00}\) Now, we will determine the initial concentrations in the mixture. The total volume = \(35.00 mL + 20.0 mL = 55.0 mL\). Initial \(\left[\mathrm{Zn}^{2+}\right]=\frac{0.061\times35.00}{55.0}M\) Initial \(\left[\mathrm{NO}_{3}^{-}\right]=\frac{0.122\times35.00}{55.0}M\) Initial \(\left[\mathrm{OH}^{-}\right]=\frac{10^{-5.00}\times20.0}{55.0}M\)

Check if precipitation occurs

To check if precipitation occurs, we need to compare the reaction quotient (Q) to the solubility product constant (Ksp). For zinc hydroxide (\(\mathrm{Zn}\left(\mathrm{OH}\right)_{2}\)), \(1 \times 10^{-17}= Ksp \le Q = \left[\mathrm{Zn}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2\) If Q > Ksp, then precipitation will occur. We found the initial concentrations in Step 1: Q = \(\left(\frac{0.061\times35.00}{55.0}\right)\left(\frac{10^{-5.00}\times20.0}{55.0}\right)^2\) Comparing Q and Ksp, we find that Q > Ksp, so precipitation occurs.

Calculate the final concentrations of all ions

Since precipitation occurs, the final equilibrium concentrations are determined by the solubility product constant: \(1 \times 10^{-17}= Ksp = \left[\mathrm{Zn}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2\) Solving for \([\mathrm{Zn}^{2+}]\) and \(\left[\mathrm{OH}^{-}\right]\), we have: \([\mathrm{Zn}^{2+}] = \frac{Ksp}{[\mathrm{OH}^{-}]^2}\) \(\left[\mathrm{OH}^{-}\right] = \sqrt{\frac{Ksp}{\left[\mathrm{Zn}^{2+}\right]}}\) Final \(\left[\mathrm{NO}_{3}^{-}\right]=\frac{0.122\times35.00}{55.0}M\) Final \(\left[\mathrm{K}^{+}\right]=\left[\mathrm{OH}^{-}\right]\)

Calculate the final pH of the mixture

Now that we have the final concentration of the hydroxide ion \(\left(\mathrm{OH}^{-}\right)\), we can calculate the final pH: \(pOH = -\log \left[\mathrm{OH}^{-}\right]\) \(pH = 14 - pOH\) Thus, we have determined if precipitation occurs and calculated the final concentrations of all ions and the pH after equilibrium is established.

Key concepts

Precipitation Reaction

A precipitation reaction is a type of chemical reaction where two soluble reactants combine to form an insoluble product, known as a precipitate. This is observed in the lab as a solid forming in a solution, often upon mixing two clear liquids. Understanding this concept is crucial for students tackling exercises that involve predicting whether a precipitate will form when solutions are mixed.

In the given exercise, a solution of zinc nitrate reacts with potassium hydroxide (KOH) which has a pH of 9.00, indicating that it’s a basic solution. The reaction can potentially produce zinc hydroxide, a compound that is only sparingly soluble in water. Whether or not precipitation occurs is determined by comparing the reaction quotient (Q) to the solubility product constant (Ksp). This process is crucial, as it helps us understand the conditions under which a substance will precipitate and thus, indirectly, the purity and concentration of the substances involved.

If our calculated reaction quotient (Q) exceeds the Ksp for zinc hydroxide, this signals that the solution is supersaturated with respect to zinc hydroxide and a precipitate will form. In our exercise, after performing the calculation, we indeed find that Q > Ksp, meaning that precipitation of zinc hydroxide will occur, making it a useful practical application of theoretical knowledge around precipitation reactions.

Concentration Calculation

The ability to calculate concentrations of ions in a solution is an essential skill in chemistry. This helps to predict the behavior of ions in solutions, such as whether a reaction will go to completion or if a product will precipitate. It involves a clear understanding of molarity, volume, and the dilution principle.

In our exercise, the zinc nitrate solution initially has a concentration of 0.061M. When it’s mixed with KOH, the total volume changes and thus the concentration of each ion must be recalculated according to the new volume. The updated concentrations help us to work out the reaction quotient (Q) and make predictions about the precipitation process.

• To find the initial concentration of zinc ions \(\left[\mathrm{Zn}^{2+}\right]\), we use the formula \(\frac{{\text{{initial molarity}} \times \text{{initial volume}}}}{{\text{{final volume}}}}M\).
• Similarly, for nitrate ions and hydroxide ions, we apply their respective molarity and dilution formulas in the context of the given volumes.

For exercises like this, accurate concentration calculations are pivotal because they directly affect the outcome of precipitation and pH calculations, leading to a concrete understanding of the systems at equilibrium.

pH and pOH Relationship

The pH and pOH relationship is a cornerstone of understanding acidic and basic solutions in chemistry. The pH scale is a measure of the acidity or basicity of an aqueous solution, while pOH is a similar scale that measures the concentration of hydroxide ions.

The pH and pOH of a solution are inversely related and always add up to 14 (at 25°C, which is considered standard temperature). Mathematically, pH + pOH = 14. Using this relationship, one can derive the concentration of hydroxide ions from the pH (or vice versa). In our exercise, given the pH of the KOH solution, we first calculate the pOH and then determine the hydroxide ion concentration.

Here’s how we apply this concept:

• First, subtract the given pH from 14 to get the pOH: \( pOH = 14 - pH \).
• Then, calculate the hydroxide ion concentration using the formula \(\left[\mathrm{OH}^{-}\right] = 10^{-\text{{pOH}}} \).

This calculated hydroxide ion concentration plays a crucial role in the exercise when determining whether a precipitate forms and in calculating the final pH of the solution. A proper understanding of the pH and pOH relationship is fundamental for chemistry students, particularly when dealing with reactions in aqueous solutions and the resulting changes in acidity or basicity.