Problem 100 - General Chemistry: Principles and Modern Applications

Question

A 25.00 -mL sample of \(0.0100 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(\mathrm{K}_{\mathrm{a}}=\right.\) \(\left.6.3 \times 10^{-5}\right)\) is titrated with \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) Calculate the \(\mathrm{pH}\) (a) of the initial acid solution; (b) after the addition of 6.25 mL of \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (c) at the equivalence point; (d) after the addition of a total of \(15.00 \mathrm{mL}\) of \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

Step-by-Step Solution

Answer

The initial pH of the C6H5COOH solution is calculated to be approximately 2.77. After the addition of 6.25mL of 0.0100M Ba(OH)2, the pH is approximately 4.23. At the equivalence point, the pH is approximately 8.85. After the addition of a total of 15.00 mL of 0.0100 M Ba(OH)2, the pH is approximately 12.74.

Step 1 - Calculation of initial pH of the acid solution

For a weak acid \(C6H5COOH \rightarrow C6H5COO- + H+ \), the initial \(H+\) ion concentration can be given by \(\sqrt{{ka * [acid]_{initial}}}\), where \(ka = 6.3*10^-5 \) and \([acid]_{initial} = 0.01 M\). Then, using the definition of pH = -log[H+], we get the initial pH value.

Step 2 - Calculation of pH after the addition of 6.25 mL of Ba(OH)2

After adding 6.25 mL of 0.01 M Ba(OH)2, the moles of \(OH-\) ions will neutralize an equivalent quantity of \(H+\) ions and forming \(C6H5COO-\) ion. The remaining \(C6H5COOH\) and \(C6H5COO-\) ion can form a buffer solution. The pH can be calculated by the acid and its conjugate base pair's ratio via the Henderson–Hasselbalch equation:\(pH = pKa + log ( [C6H5COO-]/[C6H5COOH])\).

Step 3 - Calculation of pH at the equivalence point

At the equivalence point, all the weak acid has reacted with the strong base, and only \(C6H5COO-\) ion (conjugate base) remains, which will undergo hydrolysis with water forming \(OH-\) ions and thus forming a basic solution. The \(pOH\) can be calculated by \(\sqrt{{Kw/Ka}} \), where \(Kw = 1x10^-14 \) at 25 degree Celsius. Then, using pH + pOH = 14, we get the pH value.

Step 4 - Calculation of pH after the addition of a total of 15.00ml of 0.0100 M Ba(OH)2

After adding 15ml of 0.01M Ba(OH)2, the strong base is in excess. The \(OH-\) ion concentration can be calculated by the excess moles/volume. Then, pOH = -log[OH-], and using pH + pOH = 14, we get the pH value.

Key Concepts

Weak Acid

Weak acids are acids that do not fully dissociate in water. Instead, they only partially dissociate into their ions. This means that a solution of a weak acid contains both the undissociated acid and its ions. For example, in the exercise, \( ext{C}_6 ext{H}_5 ext{COOH}\) (benzoic acid) is a weak acid. When it is in the solution, it partially breaks down into \( ext{C}_6 ext{H}_5 ext{COO}^-\) and \(H^+\) ions.

Understanding weak acids is crucial because they do not change the pH of a solution as much as strong acids. The extent of its dissociation is given by the acid dissociation constant, \(K_a\). For weak acids, \(K_a\) is typically a small number, indicating that most of the acid molecules remain in the undissociated form. In our example, the \(K_a\) of benzoic acid is \(6.3 \times 10^{-5}\), showing it is indeed a weak acid.

Weak acids are important in titration because they form buffer solutions when mixed with their conjugate base.

Equivalence Point

The equivalence point in a titration is when the amount of titrant added completely reacts with the substance being titrated. In the titration of a weak acid like \( ext{C}_6 ext{H}_5 ext{COOH} \) with a strong base like \( ext{Ba(OH)}_2 \), the equivalence point is achieved when all the acid molecules are neutralized by the base. This means that the moles of acid are equal to the moles of base added.

At this point, the solution contains only the conjugate base of the acid, which is \( ext{C}_6 ext{H}_5 ext{COO}^- \), and no excess \( H^+ \) ions from the weak acid since they have been neutralized. Consequently, the pH of the solution is not neutral (pH 7) as one might expect, but instead is basic. This happens because the conjugate base undergoes hydrolysis, forming \(OH^-\) ions in solution, leading to a higher pH.

Understanding the equivalence point helps us comprehend the behavior of solutions during titrations and is essential for accurate calculations of concentrations and outcomes in laboratory settings.

Henderson–Hasselbalch Equation

The Henderson–Hasselbalch equation is a handy tool for calculating the pH of a buffer solution. A buffer solution resists changes in pH when small amounts of acid or base are added. This is particularly important in titration when the weak acid and its conjugate base coexist.

The equation is expressed as: \( ext{pH} = ext{pK}_a + ext{log}rac{[ ext{A}^-]}{[ ext{HA}]}\), where \

• \( ext{pH} \) is the measure of the acidity of the solution,
• \( ext{pK}_a \) is the negative logarithm of the acid dissociation constant, \( K_a \),
• \( [ ext{A}^-] \) is the concentration of the conjugate base,
• \( [ ext{HA}] \) is the concentration of the weak acid.

Using this equation allows a chemist to calculate the pH of the solution when both the weak acid (\( ext{C}_6 ext{H}_5 ext{COOH} \)) and its conjugate base (\( ext{C}_6 ext{H}_5 ext{COO}^- \)) are present. The ratio between them determines the efficiency of the buffering action, and thus the resulting pH. Understanding this concept helps make sense of buffer actions during titrations, particularly before and after reaching the equivalence point.