Question

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

Short answer

The major species present in a 0.150 M NH3 solution are NH3, NH4+, and OH-. The concentration of OH- ions is 1.64 x 10^-3 M, and the pH of the solution is approximately 11.22.

Step-by-step solution

Write the chemical equilibrium for the reaction

Write the chemical reaction for ammonia's interaction with water. NH3 (aq) + H2O (l) <=> NH4+ (aq) + OH- (aq)

Write the Kb expression for ammonia

The Kb expression for ammonia (NH3) is given by: Kb = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\) The Kb value for ammonia (NH3) is 1.8 x 10^-5. We will use this value later when we solve for the concentration of OH- ions.

Set up an ICE table

An ICE (Initial, Change, Equilibrium) table will help us keep track of the concentrations of each species throughout the reaction. Since we are given that the initial concentration of NH3 is 0.150 M, our table will look like this: NH3 + H2O <=> NH4+ + OH- Initial: 0.150 - 0 0 Change: -x - x x Equilibrium: 0.150-x - x x

Substitute ICE table values into the Kb expression

Substitute the equilibrium expressions from the ICE table into the Kb expression we wrote in Step 2: \(1.8 \times 10^{-5} = \frac{x^2}{0.150 - x}\) Since Kb is relatively small, we can make a simplifying assumption that x is also small and the concentration of NH3 will not change significantly: \(1.8 \times 10^{-5} = \frac{x^2}{0.150}\)

Solve for x, which represents [OH-]

Now, we will solve for x, which represents the concentration of OH-: \(x^2 = 1.8 \times 10^{-5} \times 0.150\) \(x^2 = 2.7 \times 10^{-6}\) \(x = \sqrt{2.7 \times 10^{-6}}\) \(x = 1.64 \times 10^{-3}\) So, the concentration of OH-, [OH-], is 1.64 x 10^-3 M.

Calculate the pH of the solution

Now that we have the concentration of hydroxide ions, [OH-], we can use it to compute for the pOH, and then find the pH of the solution. We will use the following formula: \(pOH = -\log{[OH^{-}]}\) Then we'll use the relationship between pH and pOH: pH + pOH = 14 First, calculate the pOH: \(pOH = -\log{(1.64 \times 10^{-3})}\) \(pOH = 2.78\) Next, calculate the pH: pH = 14 - pOH = 14 - 2.78 = 11.22 The pH of the 0.150 M NH3 solution is approximately 11.22.

Key concepts

Ammonia

Ammonia, commonly known by its chemical formula, \(\text{NH}_3\), is a weak base. This means it doesn't fully ionize in solution like strong bases do. In aqueous solutions, ammonia accepts a proton \((\text{H}^+)\) from water molecules. This results in the formation of ammonium ions \((\text{NH}_4^+)\) and hydroxide ions \((\text{OH}^-\)). This interaction can be represented by the equilibrium equation:
\[\text{NH}_3 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{NH}_4^+ (aq) + \text{OH}^- (aq)\]

• Ammonia serves as the base, accepting protons.
• Water acts as the acid providing protons (though it's not strong enough to be considered a real acid in this context).

As a weak base, ammonia doesn't dissociate completely, which is crucial for understanding its behavior in solution. The degree of ionization can be quantified using the base dissociation constant, \(K_b\), which for ammonia is \(1.8 \times 10^{-5}\). This small value reflects ammonia's limited ability to ionize in solution.

Hydroxide Ion Concentration

The hydroxide ion concentration \([\text{OH}^-]\) in a solution tells us how basic a solution is. For solutions of weak bases like \(\text{NH}_3\), we can calculate \([\text{OH}^-]\) using the equilibrium constant method.
To find this concentration, we set up what's called an ICE table:
- I stands for Initial concentration.- C represents Change in concentration.- E is for Equilibrium concentration.
The equilibrium concentrations are substituted into the \(K_b\) expression:
\[K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\]Since the \(K_b\) value is small, it allows us a simplification by assuming the change \(x\) (representing \([\text{OH}^-]\)) is negligible compared to the initial concentration of ammonia. This simplifies calculations to
\[K_b = \frac{x^2}{0.150}\]Solving this gives us \([\text{OH}^-]\approx1.64 \times 10^{-3} \space M\), helping us quantify the basic nature of the solution.

pH Calculation

The pH scale measures acidity and basicity of a solution, where 7 is neutral, below 7 is acidic, and above 7 is basic. For solutions like ammonia, we actually first calculate the \(\text{pOH}\) from the hydroxide ion concentration, and then find the \(\text{pH}\).
The formula for \(\text{pOH}\) is:
\[\text{pOH} = -\log{[\text{OH}^-]}\]For our ammonia solution:
\[\text{pOH} = -\log{(1.64 \times 10^{-3})} \approx 2.78\]We know from the formula \(\text{pH} + \text{pOH} = 14\) that:
\[\text{pH} = 14 - \text{pOH} \approx 11.22\]

• This indicates that the ammonia solution is basic.
• Understanding the \(\text{pH}\) helps in assessing the effect ammonia will have in chemical reactions or biological systems.

The process of calculating \(\text{pH}\) is vital in chemistry as it provides insight into the balance of hydrogen and hydroxide ions in solution, determining the solution's chemical nature.