Problem 77 - Chemistry: An Atoms-Focused Approach

Question

Methylamine is a weak base. a. Use information in Appendix 5 to sketch the titration curve for the titration of \(125 \mathrm{mL}\) of a \(0.015 M\) solution of methylamine with \(0.100 M\) HCl. b. Label the curve with the \(\mathrm{pH}\) of the analyte solution, the \(\mathrm{p} K_{\mathrm{a}}\) of the analyte, and with the \(\mathrm{pH}\) and titrant volumes halfway to the equivalence point and at the equivalence point. c. Draw the structures of the species present in the solution at the equivalence point.

Step-by-Step Solution

Answer

Question: Sketch the titration curve and label specific points for the titration of a methylamine solution with HCl. Include the initial pH, halfway point, and equivalence point. Also, draw the structures of the species in the solution at the equivalence point. Answer: Based on the step-by-step solution provided, the initial pH of the solution can be calculated using the base dissociation constant (Kb) and the relationship between pH and pOH. At the halfway point, the pH is equal to the pKa of the analyte (methylamine), which can be found using the relationship between Ka, Kb, and Kw. At the equivalence point, calculate the pH using the concentration of H+ ions and the stoichiometric relationship between base and acid. The dominant species in the solution at the equivalence point are water, CH3NH2, and CH3NH3+. Draw the structures of these species using appropriate chemical notations.

Step 1: Gather Information from Appendix 5

From Appendix 5, we can find the following relevant information for methylamine: - \(K_\mathrm{b}\) (base dissociation constant) = \(3.7 \times 10^{-4}\)

Step 2: Calculate the Initial pH of the Solution

First, we need to calculate the initial pH of the methylamine solution. Using the \(K_\mathrm{b}\) value, we can find the concentration of OH⁻ ions. Methylamine dissociation: \(\mathrm{CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-}\) \(K_\mathrm{b} = \frac{[\mathrm{CH_3NH_3^+}][\mathrm{OH^-}]}{[\mathrm{CH_3NH_2}]}\) Assuming \(x\) is the concentration of dissociated ions, we get: \(K_\mathrm{b} = \frac{(\mathrm{0.015 - x})(x)}{x}\) Solving for \(x\) (try using approximation method if necessary), we will get the concentration of OH⁻ ions. Calculate the \(pOH\) using the following formula: \(pOH = -\log_{10} [\mathrm{OH^-}]\) Then, find the pH using the relationship: \(pH = 14 - pOH\)

Step 3: Determine the pH at Halfway Point

At the halfway point of titration, half the amount of the weak base has reacted with the acid. The pH is equal to the \(pK_a (=-\log_{10} K_a)\) of the analyte (methylamine) at this point. We can find \(K_a\) using the relationship between \(K_a\), \(K_b\), and \(K_w\): \(K_w = K_a \times K_b\) Solve for \(K_a\) and calculate \(pK_a = -\log_{10} K_a\).

Step 4: Determine the pH at the Equivalence Point

At the equivalence point, we have fully titrated the weak base. To find the volume of titrant needed, we can use the stoichiometric relationship between the concentration and volume of base and acid: \([\mathrm{CH_3NH_2}]V_\mathrm{b} = [\mathrm{HCl}]V_\mathrm{a}\) Where \(V_\mathrm{b}\) is the volume of the base (methylamine) and \(V_\mathrm{a}\) is the volume of the acid (HCl). Solve for \(V_\mathrm{a}\). In the equivalence point, the solution is dominated by the conjugate acid \(\mathrm{CH_3NH_3^+}\), which will react with water to release H⁺ ions: \(\mathrm{CH_3NH_3^+ + H_2O \rightleftharpoons CH_3NH_2 + H_3O^+}\) Use the \(K_a\) value for this equation to determine the concentration of H⁺ ions. Then, calculate the pH using the formula: \(pH = -\log_{10} [\mathrm{H^+}]\)

Step 5: Draw Species Structures at the Equivalence Point

At the equivalence point, the dominant species in the solution are water, \(\mathrm{CH_3NH_2}\) (uncharged), and \(\mathrm{CH_3NH_3^+}\) (conjugate acid). Draw their structures using appropriate chemical notations.

Key Concepts

Weak Bases

Methylamine is an example of a weak base, which partially dissociates in solution. Unlike strong bases, which dissociate completely, weak bases like methylamine only release a small amount of hydroxide ions (\(\text{OH}^-\)) in water. The dissociation can be expressed by the equilibrium reaction:
\[\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-\]To quantify this dissociation, we use the base dissociation constant, \(K_b\). For methylamine, \(K_b = 3.7 \times 10^{-4}\). This value indicates that the compound is a weak base because \(K_b\) is relatively small compared to strong bases. In calculations, weak bases often require an approximation technique due to their incomplete dissociation.

Titration Curve

A titration curve graphically represents the change in pH during the titration process. For a weak base like methylamine titrated with a strong acid such as HCl, the titration curve takes on a characteristic shape.

• The curve starts at a basic pH, as methylamine is a weak base.
• As the titration progresses, the pH gradually decreases as more \(\text{HCl}\) is added. This shows the neutralization of the base.
• The halfway point in the curve, where half of the base is neutralized, aligns with the \(pK_a\) of the conjugate acid formed.
• The equivalence point is reached when the amount of \(\text{HCl}\) added equals the amount of \(\text{CH}_3\text{NH}_2\) in the solution, resulting in a sharp drop in the curve.

Overall, the titration curve for methylamine demonstrates how the titration process transitions from basic to a more neutral pH.

pH Calculation

Calculating the pH at different stages of titration involves knowing the dissociation constants and using logarithms. Initially, calculate the pH by finding the concentration of hydroxide ions (\( [\text{OH}^-]\)) using the equilibrium expression for \(K_b\):
\[K_b = \frac{[\text{CH}_3\text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3\text{NH}_2]}\]Once \([\text{OH}^-]\) is known, the \(pOH\) can be calculated as:
\[pOH = -\log_{10} [\text{OH}^-]\]The initial pH is found using:
\[pH = 14 - pOH\]At the halfway point, the \(pH\) equals \(pK_a\), which is derived from the relationship:
\[pK_a = -\log_{10} \left(\frac{K_w}{K_b}\right)\]Finally, at the equivalence point, calculate pH by setting up the equilibrium for the conjugate acid \(\text{CH}_3\text{NH}_3^+\) in water, releasing \(\text{H}^+\) ions.

Equivalence Point

The equivalence point in the titration of a weak base with a strong acid is crucial. It signifies the point at which the amount of acid added exactly neutralizes the base in the solution. The equivalence point is identified by:

• Using the stoichiometric formula: \([\text{CH}_3\text{NH}_2]V_b = [\text{HCl}]V_a\) to determine the volume of titrant needed.
• At this stage, the solution is composed mostly of the conjugate acid \(\text{CH}_3\text{NH}_3^+\), which slightly dissociates, causing the pH to drop below 7.

Nevertheless, due to the presence of the conjugate acid, the solution becomes weakly acidic. At the equivalence point, draw structures for all species in the solution, such as \(\text{CH}_3\text{NH}_3^+\), \(\text{H}_2\text{O}\), and the remnants of \(\text{CH}_3\text{NH}_2\). It signifies a full transformation from the original weak base to its conjugate acid.