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 \(\mathrm{pH}\) 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 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 = 5.1; (b) pH = 9.2; (c) pH ≈ 10.5; (d) 16.67 mL; (e) pH ≈ 11; (f) Phenolphthalein; (g) pH ≈ 11.4.

Step 1: Calculate Initial pH

The initial pH is determined by the dissociation of HCN in water. Use the expression for the weak acid dissociation constant, \( K_a \), given that \( K_a \text{ for HCN is } 6.2 \times 10^{-10} \): \[ \text{HCN} \rightarrow \text{H}^+ + \text{CN}^- \]Calculate the concentration of \( \text{H}^+ \) using \( [\text{H}^+] = \sqrt{K_a \times C_0} = \sqrt{6.2 \times 10^{-10} \times 0.050} \). Find the pH from \( \text{pH} = -\log_{10}[\text{H}^+] \).

Step 2: Calculate pH at Halfway Point

At the halfway point, the moles of HCN equal the moles of CN\(^-\), which means \([\text{HCN}] = [\text{CN}^-]\). Therefore, \( \text{pH} = \text{pK}_a \). Using \( K_a = 6.2 \times 10^{-10} \), calculate \( \text{pK}_a = -\log_{10}(6.2 \times 10^{-10}) \).

Step 3: Determine Volume at Equivalence Point

Determine the number of moles of HCN: \( 25.0 \times 10^{-3} \text{ L} \times 0.050 \text{ M} \). Use stoichiometry to find the volume of \( 0.075 \text{ M NaOH} \) needed to react with these moles completely.\[ \text{Volume} = \frac{\text{moles of HCN}}{0.075} \]

Step 4: Calculate pH at 95% Point

Calculate the volume of base for 95% of the equivalence point:\[ V_{95\%} = 0.95 \times V_{\text{eq}} \].Using the Henderson-Hasselbalch equation, \(\text{pH} = \text{pK}_a + \log_{10} \left(\frac{[\text{CN}^-]}{[\text{HCN}]\right)}\) where \([\text{CN}^-] = 0.95 \times \text{initial moles} \) and \([\text{HCN}] = 0.05 \times \text{initial moles} \).

Step 5: Calculate pH at Equivalence Point

At the equivalence point, all HCN is converted to CN\(^-\). The pH is determined by the hydrolysis of \( \text{CN}^- \):\[ \text{CN}^- + \text{H}_2\text{O} \rightarrow \text{HCN} + \text{OH}^- \]. Use \( K_w / K_a \) to calculate \([\text{OH}^-]\) and hence \( \text{pOH} = -\log_{10}[\text{OH}^-] \). Finally, find \( \text{pH} = 14 - \text{pOH} \).

Step 6: Identify Suitable Indicator

Choose an indicator with a color change range that matches the pH at the equivalence point. Refer to Figure 18.10 for indicators suitable for this pH range.

Step 7: Calculate pH at 105% Point

Calculate additional volume added for 105% of equivalence point:\[ V_{105\%} = 1.05 \times V_{\text{eq}} \].At this point, there is excess NaOH; calculate pOH using excess OH\(^-\) and find pH: \( \text{pH} = 14 - \text{pOH} \).

Key Concepts

pH calculation

The concept of pH is crucial for understanding acid-base titrations, as it represents the acidity or basicity of a solution. It's calculated using the formula: \( \text{pH} = -\log_{10}[\text{H}^+] \), where \([\text{H}^+]\) is the concentration of hydrogen ions in the solution. In the context of titration, pH changes as we add a base to an acidic solution.
Before adding any base, pH reflects the initial acidic condition. For weak acids like HCN, its initial pH depends on its ionization in water, determined by its dissociation constant, \( K_a \).
As the titration progresses, the pH increases as more base is added, until reaching a crucial pH change at the equivalence point, where the amount of acid equals the amount of added base.

• Initial pH depends on acid's dissociation in water.
• pH changes are tracked to identify key titration points.

equivalence point

In titration, the equivalence point is the stage where moles of acid equals moles of base added. For acid-base titrations, this often results in a sudden change in pH.
The equivalence point is crucial because it indicates the completion of the reaction between the acid and base. For the titration of HCN with NaOH, this point marks the complete conversion of all HCN molecules into CN\(^-\).

• Measured by observing a rapid pH increase.
• Equivalence point volume calculated using stoichiometry.

Point is detected visually with an indicator or through a pH meter.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a valuable tool in titration calculations.
It relates the pH of a solution to its acid dissociation constant \( K_a \) and the concentrations of the acid-base pair. The equation is expressed as: \[ \text{pH} = \text{pK}_a + \log_{10} \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
At the halfway point to the equivalence point of a weak acid titration, the concentration of the conjugate base \([\text{A}^-]\) equals that of the acid \([\text{HA}]\). Thus, the \(\text{pH}\) is equal to \( \text{pK}_a \), providing a straightforward calculation method.

• Helps calculate pH at any titration stage.
• Useful particularly when approaching the equivalence point.

acid-base indicators

Acid-base indicators are substances that change color at a particular pH level, helping identify the equivalence point of a titration.
Choosing the right indicator is crucial for achieving accuracy in a titration experiment. For the titration of HCN with NaOH, the aim is to select an indicator with a color change range that corresponds closely to the pH at the equivalence point.

• Color change should occur at the expected equivalence point pH.
• Enhances the visibility of the titration's completion.

Indicators are chosen from a range depending on the acid and base involved in titration.