Problem 24 - Chemistry The Molecular Science

Question

You dissolve \(0.425 \mathrm{~g} \mathrm{NaOH}\) in \(2.00 \mathrm{~L}\) of a solution that originally had \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]=\left[\mathrm{HPO}_{4}^{2-}\right]=0.132 \mathrm{M}\). Calcu- late the resulting \(\mathrm{pH}\).

Step-by-Step Solution

Answer

The resulting pH is approximately 7.23.

Step 1: Determine moles of NaOH

First, calculate the number of moles of NaOH. Use its molar mass, which is approximately \(40.00 \, \text{g/mol}\). The formula to find moles is \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Thus, \(\text{moles of NaOH} = \frac{0.425}{40.00} = 0.010625 \, \text{mol}\).

Step 2: Assess the reaction impact

NaOH is a strong base that will completely dissociate in water, contributing to the formation of \(\text{OH}^-\) ions. In the buffer solution, consisting of \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\), the addition of \(\text{OH}^-\) will shift the equilibrium by converting \(\text{H}_2\text{PO}_4^-\) to \(\text{HPO}_4^{2-}\).

Step 3: Calculate resultant concentrations

The initial concentration of \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) is \(0.132 \, \text{M}\) each. All of the \(\text{OH}^-\) will react with \(\text{H}_2\text{PO}_4^-\), reducing its concentration by \(0.010625 \, \text{mol}\) over the \(2 \, \text{L}\). Thus, the new concentration of \(\text{H}_2\text{PO}_4^-\) is \((0.132 - 0.0053125) \, \text{M} = 0.1266875 \, \text{M}\) and \(\text{HPO}_4^{2-}\) becomes \((0.132 + 0.0053125) \, \text{M} = 0.1373125 \, \text{M}\).

Step 4: Use the Henderson-Hasselbalch equation

For the buffer system \(\text{H}_2\text{PO}_4^- / \text{HPO}_4^{2-}\), the \(\text{pK}_a\) of \(\text{H}_2\text{PO}_4^-\) is approximately 7.2. The Henderson-Hasselbalch equation is \(\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\). Substitute the equilibrium concentrations: \[\text{pH} = 7.2 + \log\left(\frac{0.1373125}{0.1266875}\right)\].

Step 5: Calculate the pH

Compute the logarithm and plug it into the equation: \(\log\left(\frac{0.1373125}{0.1266875}\right) \approx 0.0336\). Thus, \(\text{pH} = 7.2 + 0.0336 \approx 7.2336\).

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a fundamental tool in buffer solution calculations. It helps determine the pH of a solution that contains a weak acid and its conjugate base, or vice versa. The equation is derived from the acid dissociation constant expression and is as follows: \[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]where:

• \(\text{pH}\) is the hydrogen ion concentration measure.
• \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant.
• \([\text{A}^-]\) is the concentration of the conjugate base.
• \([\text{HA}]\) is the concentration of the weak acid.

This equation is particularly useful for buffering systems, such as the dihydrogen phosphate \/ hydrogen phosphate system, which helps maintain stable pH levels when small amounts of strong acid or base are introduced. For our calculation, using the equation facilitated finding the pH after the addition of NaOH affected the original buffer equilibrium.

Acid-Base Chemistry

Acid-Base Chemistry explores the nature of acids and bases, and their interactions. Acids release hydrogen ions \(\text{(H}^+\)) in solution, while bases produce hydroxide ions \(\text{(OH}^-)\). In the context of buffer solutions like the one in our problem, it's important to understand how the addition of a strong base like NaOH influences the equilibrium. When NaOH is dissolved, it dissociates completely:\[\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-\]The resulting \(\text{OH}^-\) ions react with \(\text{H}_2\text{PO}_4^-\) to form more \(\text{HPO}_4^{2-}\):\[\text{H}_2\text{PO}_4^- + \text{OH}^- \rightarrow \text{HPO}_4^{2-} + \text{H}_2\text{O}\]These reactions highlight the importance of understanding the principles of acid-base chemistry in predicting and analyzing the changes in solution systems.

Solution Equilibrium

Solution Equilibrium pertains to the balance established in a chemical solution. In a buffer solution, equilibrium is maintained between the weak acid and its conjugate base, preventing significant pH changes. When a strong base or acid is added, equilibrium shifts to accommodate the new ions.For the buffer system \((\text{H}_2\text{PO}_4^- / \text{HPO}_4^{2-})\), equilibrium is impacted by the added NaOH. The \(\text{OH}^-\) ions from NaOH react with the \(\text{H}_2\text{PO}_4^-\). This results in an increase in \([\text{HPO}_4^{2-}]\) and a decrease in \([\text{H}_2\text{PO}_4^-]\). Such changes underline the dynamic nature of solution equilibrium. It illustrates that the system can respond to external changes due to its intrinsic buffering capacity. Maintaining specific ratios in concentrations helps keep the pH stable, a key feature of buffer solutions.

Molarity Calculation

Molarity Calculation is critical when preparing solutions or analyzing their chemical properties. Molarity, denoted by \(\text{M}\), represents the concentration of a solute in a liter of solution.It is calculated using the formula:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]In this exercise, knowing the molarity of both \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) components was essential. Initially, each had a concentration of \(0.132\, \text{M}\). We also calculated the moles of NaOH to determine how it altered these concentrations. This involved using the mass of NaOH and its molar mass to find moles, then adjusting for the total solution volume. Such calculations form the backbone of understanding how adding certain chemicals affects the resulting pH and equilibrium in the solution.