Question

Consider the following four titrations. i. \(100.0 \mathrm{~mL}\) of \(0.10 M \mathrm{HCl}\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) ii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NaOH}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iii. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) titrated by \(0.10 \mathrm{M} \mathrm{HCl}\) iv. \(100.0 \mathrm{~mL}\) of \(0.10 M\) HF titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) Rank the titrations in order of: a. increasing volume of titrant added to reach the equivalence point. b. increasing pH initially before any titrant has been added. c. increasing \(\mathrm{pH}\) at the halfway point in equivalence. d. increasing \(\mathrm{pH}\) at the equivalence point. How would the rankings change if \(\mathrm{C}_{5} \mathrm{H}_{3} \mathrm{~N}\) replaced \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and if \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) replaced \(\mathrm{HF}\) ?

Short answer

The original rankings of the titrations are: a. Increasing volume of titrant added to reach the equivalence point: i = ii = iii = iv b. Increasing initial pH: i < iv < iii < ii c. Increasing pH at the halfway point of equivalence: i < iv < iii < ii d. Increasing pH at the equivalence point: iii < i = ii < iv With the replacement of C₅H₃N for CH₃NH₂ and HO-C₆H₅ for HF, the modified rankings are: a. i = ii = iii = iv (unchanged) b. i < HO-C₆H₅ < iv < C₅H₃N < ii c. i < HO-C₆H₅ < iv < C₅H₃N < ii d. C₅H₃N < i = ii < iii < HO-C₆H₅ < iv

Step-by-step solution

Volume of Titrant to Reach Equivalence Point

In all four titrations, the given volume of titrant and the molarity are the same: \(100.0\mathrm{~mL }\) and \(0.10 \mathrm{M}\). Therefore, the volume of titrant needed to reach the equivalence point will be the same in all four cases. Ranking: i = ii = iii = iv

Initial pH

To calculate the initial pH, we will use the pH values before any titrant is added. i. HCl is a strong acid, so its initial pH will be low. ii. NaOH is a strong base, so its initial pH will be high. iii. CH₃NH₂ is a weak base, so its initial pH will be greater than 7. iv. HF is a weak acid, so its initial pH will be less than 7. Ranking: i < iv < iii < ii

pH at Halfway Point of Equivalence

To calculate the pH at the halfway point, we need to consider the concentrations and properties of the original compound and the titrant: i. As HCl is a strong acid, its pH at the halfway point won't change significantly. ii. As NaOH is a strong base, its pH at the halfway point won't change significantly. iii. CH₃NH₂ is a weak base, so at its halfway point, the pH will be close to its pKa value. iv. HF is a weak acid, and at its halfway point, the pH will be equal to its pKa value. Ranking: i < iv < iii < ii

pH at Equivalence Point

To find the pH at the equivalence point, we need to consider the species formed during the titration: i. HCl + NaOH -> NaCl + H2O: NaCl is a neutral salt, so pH = 7. ii. NaOH + HCl -> NaCl + H2O: NaCl is a neutral salt, so pH = 7. iii. CH₃NH₂ + HCl -> CH₃NH₃Cl: CH₃NH₃Cl is an acidic salt, so pH < 7. iv. HF + NaOH -> NaF + H2O: NaF is a basic salt, so pH > 7. Ranking: iii < i = ii < iv If C₅H₃N replaced CH₃NH₂ and HO-C₆H₅ replaced HF: - C₅H₃N is a more acidic weak base, so its initial pH, halfway point pH, and equivalence point pH would be lower than CH₃NH₂. - HO-C₆H₅ is a stronger acid than HF, so its initial pH would be lower, and its halfway point pH might be slightly lower too. Its equivalence point pH will still be greater than 7 because it still forms a basic salt with NaOH. Modified rankings: a. i = ii = iii = iv (unchanged) b. i < HO-C₆H₅ < iv < C₅H₃N < ii c. i < HO-C₆H₅ < iv < C₅H₃N < ii d. C₅H₃N < i = ii < iii < HO-C₆H₅ < iv

Key concepts

Acid-Base Titration

Acid-base titration is a laboratory method used to determine the concentration of an acid or a base by reacting it with a standard solution of the opposite nature. The process involves the gradual addition of a known solution (the titrant) to the unknown solution (the analyte) until the chemical reaction between them is complete. This point of completion is called the equivalence point, which can be identified by a noticeable change in the physical properties of the solution, often indicated by a pH-sensitive dye called an indicator.
In our exercise, we investigate four different titrations involving strong and weak acids and bases. In the first two scenarios, we have the titration of hydrochloric acid (HCl) with sodium hydroxide (NaOH) and vice versa, representing a titration of strong acid with strong base. In both cases, the equivalence point occurs when the number of moles of HCl equals the number of moles of NaOH, resulting in a neutral pH of 7. The remaining scenarios involve a weak base, methylamine (CH₃NH₂), and a weak acid, hydrofluoric acid (HF), titrated by strong opposites. Remember, the nature of the substances involved impacts the initial pH, the pH at the halfway point, and the pH at the equivalence point, which are crucial in understanding the titration curve for each reaction.
During the titration, several stages are noticeable. The initial pH reflects the nature of the solution being titrated before the addition of any titrant. As the titrant is added, the solution gradually approaches the equivalence point, which is interestingly not when half the titrant has been added but rather at a point where stoichiometrically the amount of acid equals the amount of base. The pH at the halfway point to the equivalence point is particularly important for weak acids or bases, as it approximates to their pKa or pKb value respectively.

Equivalence Point

The equivalence point in a titration is the moment when the amount of titrant added is stoichiometrically equivalent to the quantity of the substance being titrated. It represents the completion of the reaction, where the number of moles of hydrogen ions (H⁺) equals the number of moles of hydroxide ions (OH⁻) for acid-base reactions.
Detecting the equivalence point is crucial as it signals that the precise amount of titrant has been added to react completely with the substance in the solution. In the case of strong acid-strong base titrations, like our HCl and NaOH example, the equivalence point occurs at a pH of 7, because the resulting salt (NaCl) and water do not affect the pH. However, when titrating a weak acid with a strong base (or vice versa), the equivalence point occurs at a pH above or below 7 due to the formation of a salt derived from a weak component. For instance, the HF titrated with NaOH ends with NaF, a salt with basic properties, resulting in a pH higher than 7.
It's important to understand that the equivalence point is not necessarily the same as the endpoint, which is an approximation observed in a practical sense, often indicated by a color change with the help of an indicator. The true equivalence point can be determined precisely using a pH meter or by constructing a titration curve that shows the continuous pH change as the titrant is added.

pH Calculation

Calculating pH is a fundamental aspect of chemistry, especially in titration studies. pH is a logarithmic scale used to quantify the acidity or basicity of an aqueous solution. It is defined as the negative base 10 logarithm of the hydrogen ion concentration (\( pH = -\text{log}_{10}[H⁺] \)).
In the context of the titration exercise, initial pH levels are determined by the nature of the analyte—strong acids/bases will have low/high initial pH values, respectively, while weak acids/bases will have pH values that depend on their dissociation in water. To calculate the pH of a solution at any point during a titration, we must know the concentration of the H⁺ or OH⁻ ions at that stage.
For a strong acid or base, this is straightforward as they fully dissociate in solution. For example, a 0.10 M solution of HCl will have a pH of about 1, given that \( pH = -\text{log}_{10}(0.10) \). However, weak acids and bases only partially dissociate, requiring the use of the acid dissociation constant (Ka) or base dissociation constant (Kb) to find the pH. In a buffer solution, or at the halfway point to the equivalence point in a titration of a weak acid/base, the pH approximately equals the pKa or pKb of the substance. The Henderson-Hasselbalch equation is often used in these situations for pH calculation.
Keep in mind, the pH at the equivalence point is more complex to calculate for weak acids/bases because it depends on the resulting salt's hydrolysis in water. Ultimately, profound understanding of these pH calculations is essential for analyzing titration curves and understanding the buffer capacity of solutions, both of which are imperative for real-world applications in chemistry.