Problem 30 - Chemistry and Chemical Reactivity

Question

Construct a rough plot of \(\mathrm{pH}\) versus volume of base for the titration of \(25.0 \mathrm{mL}\) of \(0.050 \mathrm{M} \mathrm{HCN}\) with \(0.075 \mathrm{M}\) \(\mathrm{NaOH}.\) (a) What is the pH before any \(\mathrm{NaOH}\) is added? (b) What is the pH at the halfway point of the titration? (c) What is the pH when \(95 \%\) of the required \(\mathrm{NaOH}\) has been added? (d) What volume of base, in milliliters, is required to reach the equivalence point? (e) What is the \(\mathrm{pH}\) at the equivalence point? (f) What indicator would be most suitable for this titration? (See Figure 18.10 .) (g) What is the pH when \(105 \%\) of the required base has been added?

Step-by-Step Solution

Answer

(a) pH = 4.8, (b) pH = 9.31, (c) pH = 10.99, (d) 16.67 mL, (e) pH = 10.89, (f) phenolphthalein, (g) pH = 11.18.

Step 1: Initial pH Calculation

Calculate the initial pH before any NaOH is added. The initial concentration of HCN is 0.050 M. Use the expression for pH of weak acid, where \([H^+]=\sqrt{K_a \cdot [HCN]}\) with \(K_a = 4.9 \times 10^{-10}\). \[[H^+] = \sqrt{4.9 \times 10^{-10} \times 0.050} \approx 1.57 \times 10^{-5}\]So, \(\text{pH} = -\log(1.57 \times 10^{-5}) \approx 4.8\).

Step 2: Halfway Point pH

At the halfway point, half of the HCN has been converted to its conjugate base. Here, \(\text{pH} = \text{pK}_a\) because \([HCN] = [CN^-]\). Hence, \(\text{pH} = -\log(4.9 \times 10^{-10}) \approx 9.31\).

Step 3: 95% Titration Point pH

Calculate the moles of HCN initially: \(0.025 \times 0.050 = 0.00125\ \text{mol}\). At 95%, moles of NaOH = 0.95 \times 0.00125 = 0.0011875 mol. Moles of CN- (conjugate base) = moles of NaOH added. Find \([CN^-]= 0.0011875 / (25.0 + 0.0011875/0.075) \approx 0.0468\ M\).Using the formula for a weak base, \([OH^-]=\sqrt{K_b [CN^-]}\) with \(K_b = \frac{1.0 \times 10^{-14}}{4.9 \times 10^{-10}}\):\[[OH^-] \approx \sqrt{2.04 \times 10^{-5} \times 0.0468} \approx 9.8 \times 10^{-4}\]\(\text{pOH} = -\log(9.8 \times 10^{-4}) \approx 3.01\).\(\text{pH} = 14 - 3.01 = 10.99\).

Step 4: Calculate Volume at Equivalence Point

Use the relation \(n_{HCN} = n_{NaOH}\).Initial moles of HCN = 0.00125 mol. \(V_{eq} = \frac{0.00125}{0.075} = 0.01667 \ \text{L} \times 1000 \ \text{mL/L} = 16.67\ \text{mL}\).

Step 5: pH at Equivalence Point

At the equivalence point, assume complete conversion to CN-. Use \([CN^-]=\frac{0.00125}{0.04167} \approx 0.03\ M\).With \(K_b\), \([OH^-] = \sqrt{2.04 \times 10^{-5} \times 0.03} \approx 7.8 \times 10^{-4}\)\(\text{pOH} = -\log(7.8 \times 10^{-4}) \approx 3.11\).\(\text{pH} = 14 - 3.11 = 10.89\).

Step 6: Choose Suitable Indicator

The best indicator will change color in the range of the pH at the equivalence point. With \(\text{pH}\) near 10.89, phenolphthalein is suitable, as it changes color in the range of 8.2 to 10.0.

Step 7: pH with 105% NaOH Added

Moles of excess NaOH added = 0.00125 \times 0.05 = 0.0000625 mol. Find concentration: \([OH^-] = \frac{0.0000625}{0.04167} \approx 0.0015\ M\).\(\text{pOH} = -\log(0.0015) \approx 2.82\).\(\text{pH} = 14 - 2.82 = 11.18\).

Key Concepts

pH Calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions, which is the basis for determining how acidic or basic a solution is.
For weak acids like HCN, the pH is not straightforward because they do not completely dissociate in water. Instead, we use the formula \([H^+] = \sqrt{K_a \cdot [HCN]}\) to find the concentration of hydrogen ions, where \\(K_a\) is the acid dissociation constant.
The pH is then calculated using the formula: \( \text{pH} = -\log([H^+]) \). It provides a sense of how acidic the solution is before any base is added. In our example, the initial pH is 4.8, showing that the solution is weakly acidic.

Weak Acid and Base

Weak acids and bases are fundamental in many titration problems. They don’t fully ionize in solution, which means their reactions are slightly more complex than strong acids and bases.
For HCN, a weak acid, and NaOH, a strong base, understanding their interaction helps us determine the pH at different titration stages.

• Weak acids partially release hydrogen ions, which makes their pH calculation different from strong acids.
• Similarly, weak bases partially accept protons, so their pH impact is calculated by the formula for equilibrium, considering the base dissociation constant \(K_b\).

Understanding the characteristics of these substances helps predict the changes in pH throughout the titration process.

Equivalence Point

The equivalence point in a titration marks the moment when the amount of acid equals the amount of base added.
This is crucial because it tells us about the composition of the solution at that precise time.
For weak acids being titrated by strong bases, the equivalence point will generally have a pH greater than 7.
In this exercise, using the equation \( n_{\text{acid}} = n_{\text{base}} \), we find that 16.67 mL of NaOH is required to reach the equivalence point.
This understanding aids in determining which indicator to use and interpreting the behavior of the reaction during titration.

Indicators in Titration

Indicators are crucial for visually identifying the completion point of a titration, particularly when direct measurement is challenging.
They are compounds that change color at certain pH levels and help signal when a reaction reaches the equivalence point.

• In the exercise, the best indicator for a pH around 10.89 is phenolphthalein, which changes color between a pH of 8.2 and 10.0.
• This selection is important to ensure that the conclusion of titration aligns closely with the actual equivalence point.

Choosing the right indicator ensures accurate results and helps in vividly observing the pH progression.