Problem 24 - Chemistry and Chemical Reactivity

Question

You dissolve \(0.425 \mathrm{g}\) of \(\mathrm{NaOH}\) in \(2.00 \mathrm{L}\) of a buffer solution that has \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]=\left[\mathrm{HPO}_{4}^{2-}\right]= 0.132 \mathrm{M} .\) What is the \(\mathrm{pH}\) of the solution before adding NaOH? After adding NaOH?

Step-by-Step Solution

Answer

The pH is initially 7.2 and after adding NaOH, it is 7.236.

Step 1: Understanding the Initial Buffer System

The buffer solution consists of equal concentrations of \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \) and \( \mathrm{HPO}_{4}^{2-} \), each at 0.132 M. These make a conjugate acid-base pair, creating a buffer system. Since their concentrations are equal, the system is at the half-equivalence point in a titration, where \( \text{pH} = \text{pK}_a \) of the acidic component, \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \). For \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \), \( \text{pK}_a \approx 7.2 \). Thus, the initial \( \text{pH} = 7.2 \).

Step 2: Calculating Moles of NaOH

Calculate the moles of \( \mathrm{NaOH} \) added using its mass and molar mass. The molar mass of \( \mathrm{NaOH} \) is approximately \( 40 \, \text{g/mol} \). Therefore, the moles of \( \mathrm{NaOH} \) are \( \frac{0.425}{40} = 0.010625 \ \text{mol} \).

Step 3: Adjusting Buffer Components After Adding NaOH

Adding \( \mathrm{NaOH} \) will cause a reaction with \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \), converting it to \( \mathrm{HPO}_{4}^{2-} \). After reacting with all \( 0.010625 \ \text{mol} \) of \( \mathrm{NaOH} \), the concentration of \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \) decreases by \( 0.00531 \ \text{M} \), and \( \mathrm{HPO}_{4}^{2-} \) increases by the same amount, since this is a 1:1 reaction.

Step 4: Recalculating Concentrations

After the reaction, \[ \left[ \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \right] = 0.132 \, \text{M} - 0.00531 \, \text{M} = 0.12669 \, \text{M} \] and \[ \left[ \mathrm{HPO}_{4}^{2-} \right] = 0.132 \, \text{M} + 0.00531 \, \text{M} = 0.13731 \, \text{M} \].

Step 5: Calculating New pH with Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation for the new pH: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[ \mathrm{HPO}_{4}^{2-} ]}{[ \mathrm{H}_{2}\mathrm{PO}_{4}^{-} ]} \right) \]. Plugging in the values: \[ \text{pH} = 7.2 + \log \left( \frac{0.13731}{0.12669} \right) \].

Step 6: Computing the Result

The ratio \( \frac{0.13731}{0.12669} \approx 1.084 \). Calculate the \( \log (1.084) \approx 0.036 \). Therefore, the new \( \text{pH} = 7.2 + 0.036 = 7.236 \).

Key Concepts

Henderson-Hasselbalch Equation

The concept of the Henderson-Hasselbalch equation is essential for understanding buffer solutions and their role in maintaining pH levels. In essence, this equation provides a way to estimate the pH of a buffer solution based on the concentration of acid and its conjugate base. The formula is expressed as follows:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
What makes the Henderson-Hasselbalch equation so powerful is its ability to account for changes in concentrations when small amounts of acid or base are added to a buffer system. This capacity allows the buffer to maintain a stable pH. Such stability is crucial in many chemical and biological systems. Whether it's maintaining human blood at a pH around 7.4 or a laboratory chemical reaction, buffers play a critical role. Understanding how this equation functions can help predict how a buffer will react in different situations and maintain the desired pH level.

Acid-Base Pair

In a buffer solution, the concept of an acid-base pair is central to its ability to stabilize pH. An acid-base pair consists of two species that differ by a proton. In the context of this exercise, \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \) acts as the acid while \( \mathrm{HPO}_{4}^{2-} \) is the conjugate base. This pair can resist drastic changes in pH upon the addition of small amounts of strong acids or bases.
When a strong base like NaOH is added to the buffer, it will preferentially react with the acid component (\( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \)). This neutralizes the added base and shifts the balance towards the base component (\( \mathrm{HPO}_{4}^{2-} \)). Conversely, if a strong acid were introduced, \( \mathrm{HPO}_{4}^{2-} \) would neutralize it, converting back into \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \). These reactions are simple yet help maintain a constant pH, showcasing the significance of the acid-base pair in buffering solutions.

pH Calculation

Calculating the pH of a solution involves understanding both the principles of acidity and the specific dynamics of the solution's components. Initially, before introducing NaOH, the pH of a buffer is typically set around the \( \text{pK}_a \) value of the acid in the buffer system. Given that equal concentrations of \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \) and \( \mathrm{HPO}_{4}^{2-} \) are present, this means pH initially equals \( 7.2 \).
Once NaOH is added, it reacts, altering the balance of the components. The Henderson-Hasselbalch equation is then employed to derive the new pH:

• Identify the \( \text{pK}_a \) value, here around \( 7.2 \).
• Compute the concentration changes due to the chemical reaction.
• Use the concentrations in the Henderson-Hasselbalch equation to find the updated pH.

The equation accounts for the logarithmic nature of the pH scale, providing a simple yet effective way to capture the buffer's response to addition of acid or base, ensuring that tiny concentration changes don't immediately lead to large shifts in pH.