Chapter 10: Problem 24
A \(100.0-\mathrm{mL}\) aliquot of \(0.100 \mathrm{M}\) diprotic acid \(\mathrm{H}_{2} \mathrm{~A}\left(\mathrm{p} K_{1}=4.00\right.\), \(\mathrm{p} K_{2}=8.00\) ) was titrated with \(1.00 \mathrm{M} \mathrm{NaOH}\). Find the \(\mathrm{pH}\) at the following volumes of base added and make a graph of \(\mathrm{pH}\) versus \(V_{\mathrm{b}}: V_{\mathrm{b}}=\) \(0,1,5,9,10,11,15,19,20\), and \(22 \mathrm{~mL}\).
Short Answer
Step by step solution
Calculate Initial pH for Vb = 0 mL
pH for Vb = 1 mL
pH for Vb = 5 mL
pH for Vb = 9 mL
Equivalence Point Vb = 10 mL
Beyond First Equivalence Point Vb = 11 mL
pH for Vb = 15 mL
pH for Vb = 19 mL
Second Equivalence Point Vb = 20 mL
After Second Equivalence Point Vb = 22 mL
Graph Creation of pH versus Vb
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Diprotic Acid
In the example problem, \( \mathrm{H}_2 \mathrm{A} \) has \( \mathrm{p}K_1 = 4.00 \) and \( \mathrm{p}K_2 = 8.00 \), meaning that the first dissociation makes it a relatively strong acid compared to the second, which behaves as a much weaker acid.
The titration curve for a diprotic acid is characterized by two distinct equivalence points because of these two dissociation steps. During titration, you add a strong base like NaOH, and you can visually observe these shifts on a pH vs. volume curve.
Henderson-Hasselbalch Equation
This equation is particularly handy in titration problems, as it predicts the \( \mathrm{pH} \) of a buffer solution formed during the reaction at any given point before reaching the equivalence point. As demonstrated in the titration of a diprotic acid like \( \mathrm{H}_2 \mathrm{A} \), the equation helps in finding the \( \mathrm{pH} \) at different stages of titration:
- First, using \( \mathrm{p}K_1 \) when the acid is partially neutralized to form \( \mathrm{HA}^- \).
- Later, using \( \mathrm{p}K_2 \) when \( \mathrm{HA}^- \) is further neutralized to form \( \mathrm{A}^{2-} \).
Equivalence Point
- The first equivalence point occurs when all of the first acidic hydrogen (\( \mathrm{H}^+ \)) has reacted with the base to form \( \mathrm{HA}^- \).
- The second equivalence point is reached when the remaining hydrogen ion is completely neutralized, converting \( \mathrm{HA}^- \) to \( \mathrm{A}^{2-} \).
At these points, there's a sudden change in the pH, which reveals itself as a steep rise on the titration curve. This transition is because weak acids form at each point, and they don't contribute as many \( \mathrm{H}^+ \) ions to the solution. Recognizing these points is essential for determining the full stoichiometry of the reaction and achieving accurate chemical analysis.
Buffer Solution
- Before the first equivalence point, the solution contains both \( \mathrm{H}_2 \mathrm{A} \) and \( \mathrm{HA}^- \) which together act as a buffer.
- Similarly, after the first equivalence point, a new buffer is formed between \( \mathrm{HA}^- \) and \( \mathrm{A}^{2-} \).
These buffers are characterized by their capacity to maintain the pH within a narrow range, even as more base is added. The Henderson-Hasselbalch equation provides a way to calculate the \( \mathrm{pH} \) in these regions because the logarithmic component quantifies the ratio of buffer components. Understanding buffer capacity and behavior during titration helps in predicting pH changes and controlling the reaction environment.