Problem 239 - Introductory Chemistry Atoms First

Question

Determine the ammonia concentration of an aqueous solution that has a pH of \(11.50\). The equation for the dissociation of \(\mathrm{NH}_{3}\) \(\left(\mathrm{K}_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) is \(\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows\) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)\)

Step-by-Step Solution

Answer

The ammonia concentration in the aqueous solution with a pH of 11.50 is \(0.0558 \, M\).

Step 1: Calculate pOH#

To calculate the pOH, use the relationship between pH and pOH: \(pH + pOH = 14\) Now, substitute the given pH value of 11.50 and solve for the pOH: \(pOH = 14 - 11.50 = 2.50\)

Step 2: Calculate the concentration of OH- ions#

Use the pOH value to calculate the concentration of OH- ions: \( [OH^-] = 10^{-pOH} \) Substitute the pOH value (2.50) and solve for [OH^-]: \( [OH^-] = 10^{-2.50} = 3.16 \times 10^{-3} \, M \)

Step 3: Write the Kb expression#

Write the Kb expression for the dissociation of ammonia: \(K_{b} = \frac{[NH_4^+][OH^-]}{[NH_3]}\)

Step 4: Calculate the concentration of NH3#

Notice that the change in concentration for NH4+ and OH- ions is equal. Let x be the concentration of NH3 that ionizes. Then: \(K_{b} = \frac{[(3.16 \times 10^{-3})^2]}{x}\) Substitute the given Kb value (1.8 × 10^(-5)): \(1.8 \times 10^{-5} = \frac{[(3.16 \times 10^{-3})^2]}{x}\) Now, solve for x (the ammonia concentration): \( x = \frac{[(3.16 \times 10^{-3})^2]}{1.8 \times 10^{-5}} \) \( x = 0.0558 \, M \) So, the ammonia concentration in the aqueous solution is 0.0558 M.

Key Concepts

Understanding Ammonia Concentration

In an aqueous solution, ammonia (\(\mathrm{NH}_3\)) can exist in different forms. One important form is the dissolved ammonia that interacts with water molecules to form ammonium ions (\(\mathrm{NH}_4^+\)) and hydroxide ions (\(\mathrm{OH}^-\)). This is known as the ammonia equilibrium in water.

• When providing the concentration of ammonia in a solution, it refers to the amount of ammonia that is dissolved and takes part in these interactions.
• For this specific problem, the concentration of ammonia we are calculating is the initial amount before any dissociation occurs.

By understanding the initial concentration, we can predict the extent of ammonia’s dissociation and its subsequent effects on the solution, like its basicity or pH level.

pH and pOH Relationship

pH and pOH offer insight into the acidic or basic nature of a solution. They are interrelated and together always add up to 14 at 25°C. This relationship means:

• A high pH indicates a basic solution, where the pOH would be lower since\(\text{pH}\ + \text{pOH} = 14\).
• For a solution with a pH of 11.50, as in our exercise, the pOH is calculated as 2.50.

Knowing the pOH helps us determine the concentration of hydroxide ions (\(\mathrm{OH}^-\)), which can be very useful in many chemical reactions and equilibrium calculations in aqueous solutions.

Role of Dissociation Constant

The dissociation constant, specifically the base dissociation constant (\(K_b\)), is crucial in understanding how readily a base like ammonia will dissociate in a solution. In terms of equilibrium:

• The\(K_b\)value quantifies the strength of the base in pulling apart its ions.
• A smaller\(K_b\)indicates a weaker base, meaning that only a small amount of ammonia dissociates into ions.

In the given problem, understanding and using the provided\(K_b\)value was essential to determine the concentration of ammonia. It allowed us to derive the relationship between the concentrations of various species in the solution and calculate the amount of ammonia present initially.

Ion Concentration Calculation

Calculating the concentration of ions like \(\mathrm{OH}^-\) in the solution is a vital step in solving problems involving equilibrium and dissociation.

• As seen in the exercise, once we know the pOH, we can find\([\mathrm{OH}^-]\) using the formula\([\mathrm{OH}^-] = 10^{-\text{pOH}}\), establishing its concentration.
• This concentration helps us further understand the balance of ions in associate equations, like the\(K_b\) expression for ammonia dissociation: \(K_{b} = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]}\).

Using these calculations, we can find how initial concentrations change during the reaction, leading us to discern the initial concentration of the reactants and the equilibrium state of the solution.