Question

Would you expect the pH of a \(0.010 \mathrm{MNH}_{3}\) solution to be higher or lower than \(12.0\) ?

Short answer

The pH of a 0.010 M \(NH_{3}\) solution is expected to be lower than 12.0.

Step-by-step solution

Define Equilibrium Reaction

The first step is to write down the equilibrium reaction of ammonia, \(NH_{3}\), with water, which is expressed as follows: \(NH_{3} + H_{2}O <=> NH_{4^+} + OH^-\)

Formulate the Kb expression

Next, write down the equilibrium expression for Kb (ionization constant for bases at 25°C), which has the general form: Kb = [\(\NH_{4^+}][OH^-]\ / [NH_{3}\]. We are given that the initial molarity of \(NH_{3}\) is 0.010 M and also, we know that Kb for \(NH_{3}\) is \(1.8 x 10^{-5}\) at 25°C. Since \(NH_{3}\) is a weak base, we can assume that it only partially ionizes.

Set up the Equilibrium Concentration and solve for [OH-]

Setting up for the equilibrium concentrations: [\(NH_{3}\)] = 0.010 - x, [\(NH_{4^+}\)] = x and [\(OH^-)\] = x. Assuming x << 0.010 (since \(NH_{3}\) slightly ionizes), [\(NH_{3}\)] can be approximated to 0.010 M and Kb expression becomes: Kb = \(x^2 / 0.010\). From which we then solve for x (which represents concentration of \(OH^-\)). We should find x equal to \(4.24 x 10^{-4}\) M.

Find the pOH

Now, find the pOH by taking the negative logarithm of the [\(OH^-)\] concentration. This will give us pOH \( = - log [OH^-] = -log (4.24 x 10^{-4}\) = 3.37.

Find the pH

Lastly, since we know the pH + pOH = 14 at 25°C, we can solve for pH = 14 - pOH = 14 - 3.37 = 10.63.

Comparison

Finally, we compare the calculated pH with 12.0. The calculated pH of the \(0.010 M NH_{3}\) solution is lower than 12.0.

Key concepts

pH Calculation

Understanding pH calculation is essential when dealing with acid-base chemistry. The pH scale, ranging from 0 to 14, measures the acidity or basicity of a solution. Pure water, which is neutral, has a pH of 7. Solutions with a pH less than 7 are acidic, while those with a pH greater than 7 are basic or alkaline.

To calculate the pH of a solution, one must first determine the concentration of hydrogen ions (H^{+}) or hydroxide ions (OH^{-}) present in the solution. In the case of a basic solution like ammonia (NH_{3}), the key is to find the concentration of hydroxide ions ([OH^{-}]) produced. Through the process known as ionization, NH_{3} reacts with water to produce NH_{4^{+}} and OH^{-} ions. Once the concentration of OH^{-} is known, the pOH is calculated by taking the negative logarithm of the OH^{-} ion concentration. Lastly, to find the pH, you subtract the pOH from 14 (since pH + pOH = 14).

In the given exercise, by solving for the OH^{-} concentration and subsequently calculating the pOH, one can determine the pH of the NH_{3} solution. Applying these steps, a pH of 10.63 is calculated, which is lower than 12.0, indicating that the solution is less basic than a solution with a pH of 12.

Equilibrium Constants

The equilibrium constant for a chemical reaction is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. In the context of acid-base reactions, two equilibrium constants are particularly important: the acid dissociation constant (Ka) and the base dissociation constant (Kb).

For weak bases like ammonia (NH_{3}), Kb provides a measure of the extent to which a base will ionize in water. A high Kb value means the base ionizes significantly, producing more hydroxide ions, and therefore is 'stronger'. Conversely, a low Kb value indicates a 'weaker' base that ionizes less.

In practice, Kb can be used to calculate the equilibrium concentrations of the species involved in the ionization of the base. The equilibrium expression is written based on the balanced equation for the base's reaction with water. Once the ion concentrations are known, other properties like pOH and pH can be determined. Crucially, performing these calculations requires the assumption that the change in concentration of the base due to ionization, designated as x in the solution example, is significantly smaller than the initial concentration, allowing it to be considered negligible in some cases.

Weak Base Ionization

Weak bases, unlike strong bases, do not completely dissociate into ions in an aqueous solution. Instead, they establish an equilibrium between the un-ionized base and the ions produced. Ammonia (NH_{3}) is a classic example of a weak base when dissolved in water, it partially ionizes to form ammonium (NH_{4^{+}}) and hydroxide (OH^{-}) ions.

The degree of ionization for weak bases is represented by their Kb value, which is the equilibrium constant for the base ionization reaction. The small Kb value for NH_{3} implies limited ionization, with a significant proportion of NH_{3} remaining un-ionized at equilibrium.

To calculate the degree of ionization, a typical approach is to start with the initial concentration of the weak base and determine the changes in concentration for all species at equilibrium, often using an ICE table (Initial, Change, Equilibrium) to organize the information. However, simplifications can be made if the ionization is minimal, treating the change in concentration of the weak base as negligible, which allows us to avoid quadratic calculations for x as showcased in the given solved exercise. Understanding this nuance of weak base ionization is crucial for accurately determining properties like the pH of the solution.