Question

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

Short answer

For a methylamine titration with hydrochloric acid, the pH values under the two conditions are: a. After adding 50.0 mL of 0.50 M HCl, the pH is 10.64. b. At the stoichiometric point, the pH is 10.64.

Step-by-step solution

Write the balanced equation for the reaction

First, we need to write the balanced equation for the reaction between methylamine and hydrochloric acid: \[CH_{3}NH_{2}(aq) + HCl(aq) \rightarrow CH_{3}NH_{3}^{+}(aq) + Cl^{-}(aq)\]

Calculate the moles of both methylamine and HCl

We will calculate the initial moles of both methylamine and HCl: - Moles of methylamine: 1.0 M × 0.050 L = 0.050 mol - Moles of HCl (condition a): 0.050 L × 0.50 M = 0.025 mol

Calculate the pH for condition a (after adding 50.0 mL of 0.50 M HCl)

In this condition, there is still an excess of methylamine since only half the moles of HCl are added. First, we need to calculate the moles of methylamine after the reaction with HCl: - Moles of methylamine after the reaction: 0.050 mol - 0.025 mol = 0.025 mol Now, we can set up an equilibrium expression for the remaining methylamine: \[CH_{3}NH_{2}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}NH_{3}^{+}(aq) + OH^{-}(aq)\] Let x represent the concentration of OH⁻ ions at equilibrium. Then the equilibrium equation becomes: \[\frac{x(0.025+x)}{0.025-x} = 4.4 \times 10^{-4}\] Since x is significantly smaller than 0.025, we can approximate the equation: \[\frac{x(0.025)}{0.025} = 4.4 \times 10^{-4}\] Solving for x: \[x = 4.4 \times 10^{-4}\] Now, find the pOH and pH: \[pOH = -\log{(4.4\times 10^{-4})} = 3.36\] \[pH = 14 - pOH = 14 - 3.36 = 10.64\] Thus, the pH after adding 50.0 mL of 0.50 M HCl is 10.64.

Calculate the pH for condition b (at the stoichiometric point)

At the stoichiometric point, all of the methylamine has reacted with an equal amount of HCl. In this case, the pH is determined by the equilibrium of the methyl ammonium ion (CH₃NH₃⁺). Knowing that the total volume has doubled, we can find the concentration of methyl ammonium ion: \[C_{CH_{3}NH_{3}^{+}} = \frac{0.050 \, \text{mol}}{2 \cdot 0.050 \, \text{L}} = 0.50 \, \text{M}\] Now, we need to set up an equilibrium expression for the dissociation of the methyl ammonium ion: \[CH_{3}NH_{3}^{+}(aq) \rightleftharpoons CH_{3}NH_{2}(aq) + H^{+}(aq)\] Let x represent the concentration of H⁺ ions at equilibrium. Then the equilibrium equation becomes: \[\frac{x(0.50+x)}{0.50-x} = \frac{K_{w}}{K_{b}} = \frac{1.0 \times 10^{-14}}{4.4\times10^{-4}}\] Since x is significantly smaller than 0.50, we can approximate the equation: \[\frac{x(0.50)}{0.50} = \frac{1.0 \times 10^{-14}}{4.4\times10^{-4}}\] Solving for x: \[x = \frac{1.0 \times 10^{-14}}{4.4\times10^{-4}} = 2.27 \times 10^{-11}\] Now, find the pH: \[pH = -\log{(2.27\times 10^{-11})} = 10.64\] Thus, the pH at the stoichiometric point is 10.64.

Key concepts

Acid-Base Reactions

Acid-base reactions are fundamental chemical processes where an acid donates a proton to a base. These reactions are characterized by the transfer of electrons and usually result in the formation of water and a salt. Understanding these reactions is crucial in fields like chemistry, biology, and environmental science.
In the case of our exercise, methylamine (\( \text{CH}_3\text{NH}_2 \)) acts as a base and hydrochloric acid (\( \text{HCl} \)) is the acid. When these two substances react, \( \text{CH}_3\text{NH}_2 \) accepts a proton from \( \text{HCl} \), forming methylammonium ion (\( \text{CH}_3\text{NH}_3^+ \)) and chloride ion (\( \text{Cl}^- \)):
\[\text{CH}_3\text{NH}_2(aq) + \text{HCl}(aq) \rightarrow \text{CH}_3\text{NH}_3^+(aq) + \text{Cl}^-(aq)\]

• This exchange is typical of acid-base reactions.
• Methylamine gains a hydrogen ion and becomes positively charged.
• The reaction can be represented by a balanced ionic equation.

Such reactions are critical for understanding titrations and other chemical processes.

pH Calculation

pH is a measure of the acidity or basicity of a solution. It is determined by the concentration of hydrogen ions (\( \text{H}^+ \)) present. The formula to find pH is:
\[pH = -\log{[\text{H}^+]}\]
In the exercise, the reaction initially involves methylamine converting to its conjugate acid, causing changes in pH.
To calculate pH, we often calculate the pOH (related to hydroxide ions, \( \text{OH}^- \)) and convert it using:
\[pH = 14 - pOH\]
Here's a simplified approach for condition a (after adding certain amount of HCl):

• Based on the moles of methylamine and HCl, calculate the remaining base.
• Set up and solve the equilibrium expression for the base.
• Determine the concentration of \( \text{OH}^- \) and find pOH.
• Subtract pOH from 14 to get the pH.

Understanding pH is crucial in titration, affecting how reactions proceed.

Methylamine and Hydrochloric Acid

Methylamine (\( \text{CH}_3\text{NH}_2 \)) is an organic base, often involved in chemical synthesis and reactions. Hydrochloric acid (\( \text{HCl} \)) is a strong acid widely used in laboratory and industrial processes.
When these two compounds react, they embark on a classic acid-base reaction:

• Methylamine, being a base, accepts a proton from hydrochloric acid.
• The result is methylammonium chloride, a salt.
• The products can influence the pH of the solution significantly.

This reaction is part of a larger class of reactions where weak bases react with strong acids:
\[\text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ + \text{Cl}^-\]
Studying these interactions helps in understanding their roles in buffer solutions and various chemical applications.

Stoichiometric Point

The stoichiometric point, often referred to as the equivalence point, is reached during a titration when the added reactant completely reacts with the analyte present. Essentially, at this point, the amount of acid equals the amount of base in a reaction, resulting in the neutralization of each component.
In our exercise:

• The stoichiometric point is when the moles of \( \text{CH}_3\text{NH}_2 \) equal the moles of \( \text{HCl} \) added.
• No \( \text{CH}_3\text{NH}_2 \) remains unreacted.
• The pH at this point is determined by the methylammonium ion (\( \text{CH}_3\text{NH}_3^+ \)).

Reaching the stoichiometric point is critical in titration experiments, facilitating the accurate determination of concentration and reaction completeness.

Equilibrium Expressions

Equilibrium expressions are mathematical equations used to describe the ratio of the concentrations of products to reactants in a reversible reaction at equilibrium. For the reaction in our exercise:

• Involves the methylammonium equilibrium:\[\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-\]
• Equilibrium is characterized by the constant \( K_b \) for methylamine.

The concept of equilibrium is crucial when:

• Exploring how the forward and reverse reactions balance each other.
• Understanding how concentrations of species change over time.
• Using approximations to simplify calculations in complex reactions.

In titrations, equilibrium expressions help in predicting the pH and understanding reaction dynamics.