Chapter 16: Problem 24
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 \(25^{\circ} \mathrm{C} ?\)
Short Answer
Expert verified
Answer: The pH of the buffer solution is approximately 1.91.
Step by step solution
01
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:
\(K_\mathrm{a} = \dfrac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}[\mathrm{H}_{2}\mathrm{SO}_{3}]\)
02
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 \(1.54 \times 10^{-2}\).
03
Use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is:
\(pH = pK_\mathrm{a} + \log \left(\dfrac{[\mathrm{conjugate\,base}]}{[\mathrm{weak\,acid}]}\right)\)
04
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
For example, in the reaction:
- \(\mathrm{H}_2\mathrm{SO}_3 \leftrightharpoons \mathrm{H}^+ + \mathrm{HSO}_3^-\)
- \(K_a = \frac{[\mathrm{H}^+][\mathrm{HSO}_3^-]}{[\mathrm{H}_2\mathrm{SO}_3]}\)
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:
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.
The equation is expressed as:
- \(pH = pK_a + \log \left(\frac{[\mathrm{conjugate\, base}]}{[\mathrm{weak\, acid}]}\right)\)
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:
Adding these values gives us a calculated pH:
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:
- \(pH = -\log (1.54 \times 10^{-2}) + \log \left( \frac{0.250\, M}{0.200\, M} \right)\)
Adding these values gives us a calculated pH:
- \(pH \approx 1.81 + 0.097 = 1.91\)