Question

Calculate the \(\mathrm{pH}\) at the equivalence point for the following titration: \(0.20 M \mathrm{HCl}\) versus \(0.20 \mathrm{M}\) methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\).

Short answer

The pH at the equivalence point is approximately 5.17.

Step-by-step solution

Identify the Type of Titration

This is a titration between a strong acid (HCl) and a weak base (methylamine, \(\mathrm{CH}_{3}\mathrm{NH}_{2}\)). At the equivalence point, the solution contains only the salt, \(\mathrm{CH}_{3}\mathrm{NH}_{3}\mathrm{Cl}\).

Understand the Equivalence Point

At the equivalence point, all \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) has reacted with \(\mathrm{HCl}\) to form \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) and \(\mathrm{Cl}^{-}\). The solution is composed of \(\mathrm{CH}_{3}\mathrm{NH}_{3}\mathrm{Cl}\), which dissociates into \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) and \(\mathrm{Cl}^{-}\). The \(\mathrm{Cl}^{-}\) does not affect the pH significantly, but \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) will affect it.

Calculate the Concentration of \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\)

The volume of acid and base are equal since their concentrations were equal at the start of the titration (0.20 M). Therefore, the concentration of \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) at the equivalence point is the same as the initial molarity of the reactants, which is 0.20 M.

Write the Hydrolysis Reaction

\(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) hydrolyzes in water as follows:\[\mathrm{CH}_{3}\mathrm{NH}_{3}^{+} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{NH}_2 + \mathrm{H^{+}}\]\

Calculate \(pK_b\) for \(\mathrm{CH}_{3}\mathrm{NH}_{2}\)

For methylamine:\[K_b = 4.4 \times 10^{-4}\]. Using \(pK_b = -\log(K_b)\):\[pK_b = -\log(4.4 \times 10^{-4}) \approx 3.36\]

Use the Relationship between \(K_b\) and \(K_a\)

The relationship is:\[K_w = K_a \times K_b\] where \(K_w = 1.0 \times 10^{-14}\). Thus,\[K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}} = 2.27 \times 10^{-11}\]

Calculate \([H^+]\) using the Hydrolysis Equation

Using the expression for \(K_a\):\[K_a = \frac{[\mathrm{CH}_{3}\mathrm{NH}_2][\mathrm{H^+}]}{[\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}]} = 2.27 \times 10^{-11}\] Assume \([\mathrm{CH}_{3}\mathrm{NH}_2] = [\mathrm{H^+}] = x\) and \([\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}] = 0.20 M - x\), which simplifies to \([\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}] \approx 0.20\) for small \(x\). Thus,\[2.27 \times 10^{-11} = \frac{x^2}{0.20}\]\[x^2 = 4.54 \times 10^{-12}\]\[x = \sqrt{4.54 \times 10^{-12}} \approx 6.74 \times 10^{-6} \]

Calculate the pH

Since \(x\) represents \([H^+]\), compute the pH using \(\mathrm{pH} = -\log([H^+])\): \[\mathrm{pH} = -\log(6.74 \times 10^{-6}) \approx 5.17\].

Key concepts

Equivalence Point

In acid-base titrations, the equivalence point is a critical stage where the amount of titrant added exactly neutralizes the reactant. At this point, the moles of acid are equal to the moles of base. For the titration of hydrochloric acid (HCl) with methylamine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)), we reach the equivalence point when all the methylamine has reacted with the strong acid present.
At the equivalence point in this reaction, we are left with a solution of \(\mathrm{CH}_{3}\mathrm{NH}_{3}\mathrm{Cl}\), a salt formed from the reaction products. The chloride ion (\(\mathrm{Cl}^{-}\)) has negligible impact on pH, while the methylammonium ion (\(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\)) influences the pH of the solution. This is because it can dissolve in water to produce a weakly acidic solution through hydrolysis.

Strong Acid

A strong acid is characterized by its complete ionization or dissociation in water. This implies that the acid releases a high concentration of hydrogen ions (\(\mathrm{H}^+\)) into the solution. Hydrochloric acid (HCl), used in this titration exercise, is a classic example of a strong acid.
Because HCl fully dissociates in the solution, it contributes significantly to the ion concentration balance throughout the titration process. During the reaction with methylamine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)), it ensures the formation of methylammonium ion (\(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\)) until the equilibrium point. Even before reaching this point, the strong acid dominates the reaction environment, changing the pH significantly as the titration continues. By fully understanding the role of HCl, you can better comprehend the changes leading to the equivalence point.

Weak Base

Unlike strong bases, weak bases only partially ionize in solution. Methylamine (\(\mathrm{CH}_{3}\mathrm{NH}_{2}\)) is an example of a weak base. It does not fully dissociate in water, which affects the overall outcome of the titration.
During the titration, methylamine reacts with the strong acid HCl, transforming into its conjugate acid, methylammonium ion (\(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\)). This ion is still capable of accepting protons from water, resulting in a slight release of hydroxide ions (\(\mathrm{OH}^-\)). The fact that methylamine is a weak base and doesn't completely ionize makes the calculation of the solution's pH more intricate, especially around the equivalence point. This is where understanding the weak base's behavior and impact is essential for accurate pH determination.

pH Calculation

The pH calculation at the equivalence point in a titration involving a strong acid and a weak base involves various steps and considerations. Once you have established the point where the acid and base are neutralized, you need to compute how the remaining components affect the solution's pH.

• Start by identifying the concentration of the conjugate acid form, which in this case is \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\). We deduce this from the initial concentrations of the strong acid and weak base, leading to a concentration of 0.20 M at the equivalence point.
• The \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) hydrolyzes to release \(\mathrm{H^+}\) ions in the water, contributing to the acidity of the solution.
• Using the base dissociation constant (\(K_b\)), we calculate \(pK_b\) and derive \(K_a\) through the water ion product \(K_w = 1.0 \times 10^{-14}\)
• With the calculated \(K_a\), we solve for the concentration of \(\mathrm{H^+}\), allowing us to find the pH.

Ultimately, pH is determined using the formula \(pH = -\log([H^+])\). For the reaction between HCl and methylamine, this leads us to a calculated pH around 5.17, emphasizing the solution's acidic nature at the equivalence point.