Question

In the titration of \(50.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right)\), with \(0.50 \mathrm{M} \mathrm{HCl}\), calculate the \(\mathrm{pH}\) under the following conditions. a. after \(50.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) has been added b. at the stoichiometric point

Short answer

In the titration of methylamine with HCl, the pH after the addition of 50.0 mL of 0.50 M HCl is 3.36, and the pH at the stoichiometric point is 5.47.

Step-by-step solution

Calculate moles of methylamine and HCl

First, we'll calculate the initial number of moles of methylamine and HCl at each point in the titration. Initial moles of methylamine: \( (50.0 \mathrm{~mL}) \times (1.0 \mathrm{M}) = 50.0 \mathrm{~mmol} \) a. Moles of HCl after 50.0 mL has been added: \( (50.0 \mathrm{~mL}) \times (0.50 \mathrm{M}) = 25.0 \mathrm{~mmol} \) b. At the stoichiometric point, all the methylamine has reacted with the HCl. The stoichiometric ratio between methylamine and HCl is 1:1, so the moles of HCl needed to reach the stoichiometric point is equal to the initial moles of methylamine. There are 50.0 mmol of HCl at the stoichiometric point.

Write the balanced chemical equation

Methylamine is a weak base, and it reacts with HCl, a strong acid, to form a salt and water: \[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]

Set up an ICE table

We'll set up an ICE table to find the concentration of the species involved in the reaction: ### CONDITION A ### | | CH3NH2 | HCl | CH3NH3+ | Cl- | |-------|--------|-----|---------|-----| | I | 50.0 |25.0 | 0 | 0 | | C | -25.0 | -25.0| +25.0 |+25.0| | E | 25.0 | 0 | 25.0 | 25.0| ### CONDITION B ### | | CH3NH2 | HCl | CH3NH3+ | Cl- | |-------|--------|-----|---------|-----| | I | 50.0 | 50.0| 0 | 0 | | C | -50.0 | -50.0| +50.0 |+50.0| | E | 0 | 0 | 50.0 | 50.0|

Calculate the pH at each point

a. After 50.0 mL of 0.50 M HCl has been added: There are still some unreacted methylamine and an equal moles of its conjugate acid, CH3NH3+. We have to consider the buffer solution made by CH3NH2 and CH3NH3+ and use the Henderson-Hasselbalch equation to find the pH: \( pH = pK_b + \log\frac{[CH_3NH_2]}{[CH_3NH_3^+]} \) \( pK_b = -\log(4.4 \times 10^{-4}) = 3.36 \) \( pH = 3.36 + \log \frac{25.0}{25.0} = 3.36 \) b. At the stoichiometric point: There is no CH3NH2 remaining at this point. The solution contains the CH3NH3+ ion and its subsequent hydrolysis needs to be considered: \[ CH_3NH_3^+ + H_2O \rightleftharpoons CH_3NH_2 + H_3O^+ \] To find the pH, we can use the equation: \( K_a = \frac{[H_3O^+][CH_3NH_2]}{[CH_3NH_3^+]} \) \( K_a \) can be found using the relationship: \(K_w = K_a \times K_b \) \( K_a = \frac{1 \times 10^{-14}}{4.4 \times 10^{-4}} = 2.27 \times 10^{-11}\) Let the concentration of H3O+ be x. Then, we can rewrite the K_a equation as: \( 2.27 \times 10^{-11} = \frac{(x)(x)}{50.0 - x} \) The ionization is very small compared to the initial concentration of CH3NH3+, so x is very small. Hence, we can assume that 50.0 - x ≈ 50.0. Solving for x, we get: \( x = [H_3O^+] = 3.38 \times 10^{-6} M \) Finally, the pH can be calculated as: \( pH = - \log{[H_3O^+]} = - \log{3.38 \times 10^{-6}} = 5.47 \) So, the pH under condition a is 3.36 and under condition b is 5.47.

Key concepts

Henderson-Hasselbalch equation

In the analysis of acid-base titrations, utilizing the Henderson-Hasselbalch equation is a key tool, especially when dealing with buffer solutions. This equation provides us an easy way to estimate the pH of a buffer solution using the concentrations of the acid and its conjugate base. When a weak base, like methylamine, and its conjugate acid, methylammonium ion, form a buffer solution, the Henderson-Hasselbalch equation comes into play: \[ pH = pK_a + \log \left(\frac{[A^-]}{[HA]}\right) \]where: - \([A^-]\) is the concentration of the conjugate base,- \([HA]\) is the concentration of the acid, and- \(pK_a\) is the negative logarithm of the acid dissociation constant of the conjugate acid. For a solution containing a weak base like methylamine (\(CH_3NH_2\)), the equation is slightly adapted to suit a base, using \(pK_b\). The equation becomes: \[ pH = pK_b + \log \left(\frac{[Base]}{[Acid]}\right) \]Applying the Henderson-Hasselbalch equation makes it apparent that the logarithmic term becomes zero when the concentrations of the base and acid are equal, hence the pH equals the \(pK_b\) of the weak base. This is exactly what happens after certain amounts of acid are added to the solution, as seen in titration processes.

Stoichiometric point

In a titration process, the stoichiometric point is a significant milestone. It represents the exact stage when the moles of acid equates to the moles of base, achieving a precise balance between the reacting species. At this point, neither component is in excess, which leads to the complete neutralization of the base by the acid or vice versa.For example, in the titration of a weak base such as methylamine with hydrochloric acid, reaching the stoichiometric point means that the weak base has been entirely converted into its conjugate acid. The chemical equation involved would look like this:\[ CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^- \]This marks the complete reaction of methylamine with hydrogen chloride, meaning all the original weak base is now in the form of its conjugate acid. The stoichiometric point is crucial because it affects the resultant pH of the solution, which generally relies on the leftover species in the solution, such as CH_3NH_3^+. Understanding and determining the stoichiometric point allows accurate calculations of the solution’s pH at this juncture.

Buffer solution

A buffer solution is a special blend of substances that allows the solution to resist drastic pH changes upon adding small amounts of acids or bases. This is crucial during titrations involving weak acids or weak bases since it helps maintain the pH within a narrow range.In the given problem, methylamine (\(CH_3NH_2\)) and its conjugate acid formed by the addition of HCl (methylammonium ion, \(CH_3NH_3^+\)) create a buffer system. This buffer system can be idealized using the Henderson-Hasselbalch equation to determine the resulting pH as a function of the concentrations of the weak base and conjugate acid.For example, when 50.0 mL of 0.50 M HCl is added to the methylamine solution, the solution behaves as a buffer. At this point, even though we have added an acid, the solution doesn't undergo a significant pH change. Instead, the buffer manages to maintain the pH close to that of the weak base's \(pK_b\), illustrating the associative behavior as predicted by the equation. This is a unique property of buffer systems, making them incredibly valuable in chemical processes that demand pH stability.

Weak base

A weak base is a chemical compound that does not fully ionize in an aqueous solution, meaning it produces relatively few hydroxide ions. This partial ionization results in a less pronounced increase in the pH compared to strong bases, which fully dissociate.Methylamine (\(CH_3NH_2\)) exemplifies a weak base, as it only partially reacts with water under typical conditions, producing a small concentration of hydroxide ions:\[ CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^- \]This incomplete ionization is quantified by the base dissociation constant, \(K_b\), an indicator of the base's strength. For methylamine, \(K_b = 4.4 \times 10^{-4}\), suggesting it does not dissociate extensively. Understanding the properties of a weak base is crucial during titrations. Their specific behavior during reactions helps us determine how much of a strong acid like HCl is needed to reach the stoichiometric point. Taking into account the weak base's properties allows for precise calculations of reaction conditions and potential changes in pH, crucial for anticipating results during scientific experiments and practical applications.