Question

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

Short answer

The major species present in the 0.150 M NH₃ solution are NH₃, NH₄⁺, OH⁻, and H₂O. The concentration of OH⁻ ions is approximately \(1.62 \times 10^{-3} M\), and the pH of the solution is approximately 11.21.

Step-by-step solution

Identify the major species present in the solution

NH₃ is a weak base, which means that it will react with water to form NH₄⁺ and OH⁻ ions. So, the major species present in the solution are NH₃, NH₄⁺, OH⁻, and H₂O.

Write the Kb expression for NH₃

The reaction for NH₃ with water is: \[NH_3(aq) +H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)\] Because NH₃ is a weak base, it has a Kb value associated with it, which represents the equilibrium constant for the reaction with water. The Kb expression for NH₃ is: \[K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}\]

Calculate the initial concentration of NH₃, NH₄⁺, and OH⁻ ions

According to the question, the initial concentration of NH₃ is 0.150 M. Initially, NH₄⁺ and OH⁻ ions are not present in the solution, so their initial concentrations are 0. \[NH_3(aq) \: 0.150M \: \Rightarrow \: NH_4^+(aq) \: 0M \: \: OH^-(aq) \: 0M \]

Determine the change in species concentrations

Since the reaction will proceed forward, some of the NH₃ will react with water and form NH₄⁺ and OH⁻ ions. We will use "x" to represent the amount of NH₃ that reacts and forms NH₄⁺ and OH⁻ ions. \[NH_3(aq) \: 0.150M - x \: \Rightarrow \: NH_4^+(aq) \: x \: \: OH^-(aq) \: x \]

Calculate the equilibrium concentrations and solve for x

At equilibrium, we have the following relation \[K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}\] substituting the values, we get \[K_b=\frac{x^2}{0.150-x}\] For NH₃, its Kb value is 1.8 x 10⁻⁵. We can now solve for x, \[1.8*10^{-5}=\frac{x^2}{0.150-x}\] To simplify calculations, we will assume that x is small compared to 0.150 M. So, we can approximate 0.150 - x ≈ 0.150. \[1.8*10^{-5}=\frac{x^2}{0.150}\] Now, solve for x: \[x=\sqrt{1.8*10^{-5} * 0.150}\approx 1.62 * 10^{-3}\]

Calculate the [OH⁻] concentration and pOH value

Since x represents the concentration of both NH₄⁺ and OH⁻ ions at equilibrium, the concentration of OH⁻ ions is equal to x. Thus: \[[OH^-] \approx 1.62*10^{-3}\: M\] Now, calculate the pOH value: \[pOH=-log_{10} [OH^-]=-log_{10}(1.62*10^{-3})\approx 2.79\]

Calculate the pH value

To find the pH of the solution, we will use the relationship between pH and pOH: \[pH + pOH = 14\] Now, we can determine the pH of the solution: \[pH = 14 - pOH = 14 - 2.79\approx 11.21\] The major species present in the 0.150 M NH₃ solution are NH₃, NH₄⁺, OH⁻, and H₂O. The concentration of OH⁻ ions is approximately 1.62 x 10⁻³ M, and the pH of the solution is approximately 11.21.

Key concepts

NH3 as a Weak Base

Ammonia, or NH3, is known in chemistry as a weak base. It's important to understand that a weak base does not fully dissociate in water, which is unlike strong bases such as sodium hydroxide (NaOH). When ammonia is dissolved in water, it reacts with water molecules to form ammonium ions (NH4+) and hydroxide ions (OH-), but only a small fraction of the ammonia actually undergoes this change.

In a solution, this means that most of the ammonia stays as NH3, coexisting with a small amount of NH4+ and OH-. This behavior significantly impacts the calculation of pH in solutions containing ammonia, as the complete dissociation assumption applied to strong bases is not valid here. Instead, we must consider the equilibrium between the reactants and products to calculate concentration changes.

Calculating pH of Weak Base Solution

The pH of a weak base solution is not as straightforward to calculate as that of a strong base because the base does not fully dissociate. As a result, the amount of OH- produced is significantly less, and this directly affects the solution's pH. To find the pH, we actually calculate the pOH first by taking the negative logarithm of the OH- concentration, and then we use the relation that pH + pOH = 14 to find the pH.

Understanding this two-step process is crucial for accurately determining the pH of weak base solutions. Minor adjustments, such as the approximation that the concentration of the base minus x is roughly equal to the initial concentration, are often valid due to the small degree of dissociation. This simplifies calculations without significantly affecting accuracy, particularly in dilute solutions.

Equilibrium Concentration Calculations

Equilibrium concentration calculations for a weak base involve using an expression where the base's initial concentration and the extent to which it reacts are known variables. Given that the base partially dissociates, we assign a variable, typically x, to represent the unknown concentration of OH- formed.

The initial concentration minus x gives us the equilibrium concentration of the weak base, while x itself represents the equilibrium concentrations of the products, OH- and NH4+. This approach, combined with the equilibrium constant expression, allows us to solve for x using algebraic methods. Although the assumption that x is small helps simplify our calculations, this approximation must be checked for validity, particularly in more concentrated solutions where the ignored x quantity could be significant.

Kb Value and Expression

The equilibrium constant for the dissociation of a base in water is represented by Kb, the base dissociation constant. It's a numerical value that captures how readily a base forms OH- ions when dissolved in water. The higher the Kb, the stronger the base. Weak bases, like NH3, have relatively small Kb values. The general Kb expression for a weak base, A- (where A- stands for the base), reacting with water is of the form:

• Kb = \(\frac{{[BH^+] [OH^-]}}{[B]}\)

Here, [BH+] is the concentration of the conjugate acid, [OH-] is the concentration of hydroxide ions, and [B] is the concentration of the weak base itself. By setting up this expression using initial concentrations and changes in concentration (often denoted as x), we can solve for unknown variables and describe the equilibrium state of the system.