Problem 62 - Chemistry and Chemical Reactivity

Question

Calculate the hydronium ion concentration and the \(\mathrm{pH}\) when \(50.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{NH}_{3}\) is mixed with \(50.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{HCl}\).

Step-by-Step Solution

Answer

\( \mathrm{pH} \approx 2.72 \)

Step 1: Identify Reaction Type

When ammonia \( \mathrm{NH}_3 \) (a weak base) is mixed with hydrochloric acid \( \mathrm{HCl} \) (a strong acid), a neutralization reaction occurs. The products are ammonium \( \mathrm{NH}_4^+ \) and chloride \( \mathrm{Cl}^- \).

Step 2: Write Balanced Equation

The balanced chemical equation for the reaction is:\[\mathrm{NH}_3 (aq) + \mathrm{HCl} (aq) \rightarrow \mathrm{NH}_4^+ (aq) + \mathrm{Cl}^- (aq)\].

Step 3: Calculate Moles of Reactants

Find the moles of \( \mathrm{NH}_3 \) and \( \mathrm{HCl} \):\[\text{Moles of } \mathrm{NH}_3 = 0.40 \text{ M} \times 0.050 \text{ L} = 0.020 \text{ moles}\]\[\text{Moles of } \mathrm{HCl} = 0.40 \text{ M} \times 0.050 \text{ L} = 0.020 \text{ moles}\].

Step 4: Determine Limiting Reactant

Both \( \mathrm{NH}_3 \) and \( \mathrm{HCl} \) have 0.020 moles, so neither is the limiting reactant and both are completely consumed to form 0.020 moles of \( \mathrm{NH}_4^+ \).

Step 5: Calculate Total Volume

The total volume of the solution after mixing is:\[50.0 \text{ mL} + 50.0 \text{ mL} = 100.0 \text{ mL} = 0.100 \text{ L} \].

Step 6: Calculate Concentration of \( \mathrm{NH}_4^+ \)

The concentration of \( \mathrm{NH}_4^+ \) is:\[\frac{0.020 \text{ moles}}{0.100 \text{ L}} = 0.20 \text{ M}\].

Step 7: Find \( \mathrm{pH} \) Using \( \mathrm{K}_a \)

\( \mathrm{NH}_4^+ \) acts as a weak acid. Use the ammonia \( \mathrm{pK}_b = 4.75 \) to find \( \mathrm{K}_a \):\[\mathrm{pK}_w = 14.00, \quad \mathrm{K}_a = 10^{-(14.00 - 4.75)} = 1.8 \times 10^{-5}\]Use the formula for a weak acid:\[\mathrm{K}_a = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\]Since initially \([\mathrm{NH}_3] \approx 0\),\[1.8 \times 10^{-5} = \frac{x^2}{0.20} \Rightarrow x = \sqrt{1.8 \times 10^{-5} \times 0.20}\].

Step 8: Solve for \( \mathrm{H}_3\mathrm{O}^+ \) Concentration

Calculate:\[x = \sqrt{3.6 \times 10^{-6}} \approx 1.9 \times 10^{-3} \text{ M}\].

Step 9: Calculate \( \mathrm{pH} \)

The \( \mathrm{pH} \) is:\[\mathrm{pH} = -\log(1.9 \times 10^{-3}) \approx 2.72 \].

Key Concepts

Neutralization Reaction

In chemistry, a neutralization reaction is a type of chemical reaction that occurs when an acid and a base react to form water and a salt. This is a net ionic reaction that is made up when the acid's hydronium ions and the base's hydroxide ions combine. A common form of presenting this reaction is:\[\text{HA (aq) + BOH (aq) } \rightarrow \text{BA (aq) + } H_2O (l)\]In our particular case involving ammonia \(NH_3\) and hydrochloric acid \(HCl\), the reaction forms ammonium \(NH_4^+\) and chloride \(Cl^-\) ions:\[\mathrm{NH}_3 (aq) + \mathrm{HCl} (aq) \rightarrow \mathrm{NH}_4^+ (aq) + \mathrm{Cl}^- (aq)\]A neutralization reaction typically results in a solution that is neither acidic nor basic if mixed in stoichiometric proportions. However, if the acid or base is weak, the product will influence the pH of the final solution because it can act as a weak acid or base itself.

Equilibrium Constant

In the context of acid-base reactions, the equilibrium constant (\(K_{eq}\)) is a value that expresses the ratio of product concentrations to reactant concentrations (each raised to the power of their stoichiometric coefficients) at equilibrium. For a weak acid like ammonium \(NH_4^+\), this constant is known as the acid dissociation constant, \(K_a\).The formula for \(K_a\) is:\[K_a = \frac{[H_3O^+][A^-]}{[HA]}\]Where:

• \([H_3O^+]\) is the hydronium ion concentration.
• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the weak acid.

In our example, ammonium acts as the weak acid, and the equilibrium constant helps us understand how far the reaction moves towards products (how much it dissociates). When using \(K_a\) to find \([H_3O^+]\) in a neutralization scenario, it's important as it enables us to calculate the resulting pH of the solution.

pH Calculation

The pH of a solution provides a measure of its acidity or basicity. It is calculated by the negative logarithm of the hydronium ion concentration:\[pH = -\log[H_3O^+]\]To calculate the pH, you first need to find the \([H_3O^+]\) using the corresponding \(K_a\) and the concentrations of your reactants or products.In the given neutralization between \(NH_3\) and \(HCl\), we calculate \([H_3O^+]\) based on the ammonium ion concentration obtained:\[x = \sqrt{K_a \times [NH_4^+]}\]Then, the pH is calculated as follows:\[pH = -\log(1.9 \times 10^{-3}) \approx 2.72\]This value indicates a slightly acidic solution, as would be expected from a reaction involving a strong acid and a weak base where ammonia ions are not fully neutralizing the acidity.

Weak Acids and Bases

Weak acids and bases do not completely dissociate in water, meaning they reach an equilibrium between the ionized and non-ionized forms. Ammonia \(NH_3\), acting as a weak base, partially accepts a proton in water to form ammonium \(NH_4^+\), which can further act as a weak acid in reactions. In contrast, strong acids like \(HCl\) fully dissociate in water, producing a high concentration of hydronium ions \([H^+\equiv H_3O^+]\).The behavior of weak acids or bases affects the outcome of reactions, influencing properties such as:

• pH: Because they don’t fully dissociate, the pH of solutions containing weak acids or bases is often closer to neutral than those containing strong acids or bases.
• Equilibrium: Their partial dissociation leads to equilibrium between reactants and products, governed by an equilibrium constant \(K_a\) or \(K_b\).

Understanding these properties is key when predicting the behavior and the resulting pH of a solution after an acid-base reaction.