Question

What are the major species present in a 0.150-M \(\mathrm{NH}_{3}\) solution? Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of this solution.

Short answer

The major species present in a 0.150 M NH\(_{3}\) solution are NH\(_{3}\) and H\(_{2}\)O. The concentration of OH\(^{-}\) ions in the solution is calculated to be \(1.64 \times 10^{-3}\) M. The pH of this solution is found to be 11.216.

Step-by-step solution

Identify the major species in the solution

The major species present in a 0.150 M NH\(_{3}\) solution are: 1. NH\(_{3}\) (ammonia, the solute) 2. H\(_{2}\)O (water, the solvent)

Write the equilibrium expression for the dissociation of NH\(_{3}\) in water

When NH\(_{3}\) reacts with water, it acts as a base, accepting a proton (H\(^{+}\)) from water to form the ammonium ion, NH\(_{4}^{+}\), and the hydroxide ion, OH\(^{-}\). The chemical reaction and the resulting equilibrium expression are: NH\(_{3}\) + H\(_{2}\)O \(\rightleftharpoons\) NH\(_{4}^{+}\) + OH\(^{-}\) \(K_b = \frac{[\text{NH}_{4}^{+}][\text{OH}^{-}]}{[\text{NH}_{3}]}\) Here, \(K_b\) is the base dissociation constant for NH\(_{3}\).

Find the value of \(K_b\) for NH\(_{3}\)

Using a reference table or textbook, find the \(K_b\) value for NH\(_{3}\). The base dissociation constant for NH\(_{3}\) is: \(K_b = 1.8 \times 10^{-5}\)

Calculate the equilibrium concentrations of NH\(_{4}^{+}\) and OH\(^{-}\)

Let \(x\) represent the concentration of NH\(_{4}^{+}\) and OH\(^{-}\) produced at equilibrium. Therefore, the change in NH\(_{3}\) concentration is also \(x\). The equilibrium concentrations can be represented as follows: [NH\(_{4}^{+}\)] = [OH\(^{-}\)] = x [NH\(_{3}\)] = 0.150 - x Now, substitute these concentrations into the \(K_b\) expression and solve for \(x\): \(K_b = 1.8 \times 10^{-5} = \frac{x \cdot x}{0.150 - x}\) Since \(K_b\) is very small, the amount of dissociation is minimal, and we can approximate \(0.150 - x \approx 0.150\). Therefore: \(1.8 \times 10^{-5} = \frac{x^2}{0.150}\)

Calculate the [OH\(^{-}\)] concentration

Solve for \(x\), which represents the equilibrium [OH\(^{-}\)] concentration: \(x = \sqrt{1.8 \times 10^{-5} \times 0.150} = \sqrt{2.7\times 10^{-6}}\) [OH\(^{-}\)] = \(1.64 \times 10^{-3}\) M

Calculate the pH of the solution

Since we have found the concentration of OH\(^{-}\), we can now calculate the pOH and subsequently find the pH of the solution: pOH = -log([OH\(^{-}\)]) = -log(1.64 × 10\(^{-3}\)) pOH = 2.784 Next, use the relationship between pH and pOH: pH + pOH = 14 Now, solve for pH: pH = 14 - pOH = 14 - 2.784 = 11.216 So, the pH of the 0.150 M NH\(_3\) solution is 11.216.

Key concepts

Base Dissociation

In aqueous solutions, bases like aqueous ammonia dissociate by accepting protons. Ammonia (\(\text{NH}_3\)) reacts with water (\(\text{H}_2\text{O}\)) to form ammonium ions (\(\text{NH}_4^+\)) and hydroxide ions (\(\text{OH}^-\)). This process represents the base dissociation reaction:

• \(\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\)

In this context, ammonia acts as a weak base, meaning it does not fully dissociate in water. Instead, it establishes an equilibrium between its various forms, because only a fraction of ammonia molecules turn into\(\text{NH}_4^+\) and \(\text{OH}^-\) ions.
The extent of this dissociation is fundamental in calculating other properties such as pH and ion concentration.
Understanding basic dissociation helps in predicting how much of a solute contributes to the overall ion concentrations in solution.

Equilibrium Constants

The dissociation of ammonia in water is described by the equilibrium constant, specifically the base dissociation constant (\(K_b\)). The expression for \(K_b\) gives insight into the balance between the undissociated and dissociated forms of ammonia:

• \(K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\)

For ammonia, \(K_b\) is \(1.8 \times 10^{-5}\), a small value indicating that ammonia is a weak base.
This small constant suggests that at equilibrium, most of the ammonia remains undissociated, and only a small amount forms ammonium and hydroxide ions.
When calculating the concentrations of these ions, assumptions can often be made, such as approximating \(0.150 - x \approx 0.150\), due to minimal dissociation, which simplifies the computations and allows us to solve for equilibrium concentrations more easily.

pH Calculation

Knowledge of hydroxide ion concentration is essential for pH calculations in basic solutions. The \([\text{OH}^-]\) concentration is used to first find the pOH, which is related to the pH:

• pOH = -log([\text{OH}^-])

For this exercise, \([\text{OH}^-] = 1.64 \times 10^{-3}\)] M, leading to:

• pOH = -log(1.64 \times 10^{-3}) = 2.784

Once the pOH is known, it is used to calculate pH using the relation:

• pH + pOH = 14

Thus, the pH of this ammonia solution is \(14 - 2.784 = 11.216\), characterizing it as basic due to a pH significantly above 7.
This calculation highlights the importance of understanding both pH and pOH in characterizing solutions.

Species in Solution

In a solution of ammonia, several species are present which include dissolved ammonia \(\text{NH}_3\), water \(\text{H}_2\text{O}\), ammonium ions \(\text{NH}_4^+\), and hydroxide ions \(\text{OH}^-\). Each of these has a role in establishing the solution's properties:

• Ammonia (\(\text{NH}_3\)) acts as the base, partially dissociating in water.
• Water (\(\text{H}_2\text{O}\)) acts as the solvent and proton donor in the reaction.
• Ammonium ions (\(\text{NH}_4^+\)) are formed as ammonia accepts protons.
• Hydroxide ions (\(\text{OH}^-\)) contribute to the basicity and help determine the pH.

These species, interacting with one another, mainly depend on their mole fractions to reach equilibrium, which is why knowing their concentrations provides insight into both chemical composition and reactivity.

Hydroxide Ion Concentration

Hydroxide ion concentration is a crucial parameter when characterizing a solution's basic nature. It directly results from the dissociation of the base in solution.
For ammonia, the formula used to find \([\text{OH}^-]\) concentration involves making a simplifying assumption based on \(K_b\)'s small value:

• \([\text{OH}^-] = \sqrt{K_b \times [\text{NH}_3]}\)

Plugging values in, \(\sqrt{1.8 \times 10^{-5} \times 0.150}\) gives \([\text{OH}^-] = 1.64 \times 10^{-3}\)] M.
This concentration indicates the solution's basic character and allows for further calculations of pH, highlighting that despite ammonia's weak dissociation, the presence of \([\text{OH}^-]\) is significant to render the solution basic.