Problem 24 - Chemistry The Science in Context

Question

What is the pH of a buffer that is \(0.200 M \mathrm{H}_{2} \mathrm{SO}_{3}\) and \(0.250 \mathrm{M} \mathrm{NaHSO}_{3}\) at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: The pH of the buffer solution is approximately 1.91.

Step 1: Write the acid dissociation reaction and equilibrium expression.

The acid dissociation reaction for \(\mathrm{H}_{2}\mathrm{SO}_{3}\) can be written as: \(\mathrm{H}_{2}\mathrm{SO}_{3} \leftrightharpoons \mathrm{H}^{+}+\mathrm{HSO}_{3}^{-}\) The equilibrium expression for this reaction is: \(K_\mathrm{a} = \dfrac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}[\mathrm{H}_{2}\mathrm{SO}_{3}]\)

Step 2: Find the acid dissociation constant for \(\mathrm{H}_{2}\mathrm{SO}_{3}\) at \(25^{\circ}\mathrm{C}\).

Check a reference or database to find the acid dissociation constant at \(25^{\circ}\mathrm{C}\). The \(K_\mathrm{a}\) value for \(\mathrm{H}_{2}\mathrm{SO}_{3}\) is \(1.54 \times 10^{-2}\).

Step 3: Use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is: \(pH = pK_\mathrm{a} + \log \left(\dfrac{[\mathrm{conjugate\,base}]}{[\mathrm{weak\,acid}]}\right)\)

Step 4: Calculate the pH of the buffer.

Substitute the values given in the exercise, along with the \(K_\mathrm{a}\) value found in step 2, into the Henderson-Hasselbalch equation: \(pH = -\log{(1.54 \times 10^{-2})} + \log \left(\dfrac{0.250\,M}{0.200\,M}\right)\) Now, solve for pH: \(pH = 1.81 + \log{(1.25)}\) \(pH \approx 1.81 + 0.097\) \(pH \approx 1.91\) The pH of the buffer is approximately 1.91.

Key Concepts

Acid Dissociation Constant

Understanding the concept of the acid dissociation constant (represented as \(K_a\)) is vital when dealing with buffer solutions and acid-base chemistry. The acid dissociation constant provides insight into the strength of a weak acid by quantifying its degree of ionization in a solution. In essence, the larger the \(K_a\) value, the stronger the acid, since it indicates a greater tendency to donate protons to the solution.

For example, in the reaction:

• \(\mathrm{H}_2\mathrm{SO}_3 \leftrightharpoons \mathrm{H}^+ + \mathrm{HSO}_3^-\)

The equilibrium expression becomes

• \(K_a = \frac{[\mathrm{H}^+][\mathrm{HSO}_3^-]}{[\mathrm{H}_2\mathrm{SO}_3]}\)

This equation tells us how much of the acid (sulfurous acid in this case) dissociates into its ions.

In practical terms, when developing a buffer solution like the one in our exercise, knowing the \(K_a\) value helps chemists predict how the solution will behave when additional acids or bases are introduced. A precise \(K_a\) value allows for accurate calculations of pH, ensuring the buffer can effectively resist changes in pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an essential tool for calculating and understanding the pH of buffer solutions. It correlates pH with \(pK_a\) (or the negative logarithm of the acid dissociation constant) and the ratio of the concentrations of the conjugate base and the weak acid in the solution.

The equation is expressed as:

• \(pH = pK_a + \log \left(\frac{[\mathrm{conjugate\, base}]}{[\mathrm{weak\, acid}]}\right)\)

This relationship between pH, \(pK_a\), and concentration ratios helps chemists control and predict the conditions within a buffer solution.

In practical scenarios, like the exercise mentioned, having both the \(K_a\) and the concentration of substances allows one to determine the pH of a solution efficiently. It helps in planning laboratory experiments where maintaining a specific pH range is critical. Using the Henderson-Hasselbalch equation provides an intuitive understanding of how changes in concentration translate into changes in pH.

pH Calculation

Calculating the pH of a buffer solution involves using information about the buffering agents and applying mathematical formulas like the Henderson-Hasselbalch equation.

pH is a measure of how acidic or basic a solution is. In the context of the exercise, we are tasked with finding the pH of a buffer containing sulfurous acid (\(\mathrm{H}_2\mathrm{SO}_3\)) and its conjugate base sodium bisulfite (\(\mathrm{NaHSO}_3\)).

Using the Henderson-Hasselbalch equation, we plug the known values into the formula:

• \(pH = -\log (1.54 \times 10^{-2}) + \log \left( \frac{0.250\, M}{0.200\, M} \right)\)

First, calculate \(pK_a\) as \(-\log (1.54 \times 10^{-2})\), which equals 1.81. Then for the concentration ratio \(\frac{0.250}{0.200}\), compute the \(\log\) value, which is approximately 0.097.

Adding these values gives us a calculated pH:

• \(pH \approx 1.81 + 0.097 = 1.91\)

The result of this calculation, approximately 1.91, confirms the acidity affected by the presence of both components in the buffer system. Understanding how to perform these calculations is fundamental for maintaining the desired 'resistance to pH change' properties that buffers are known for.