Problem 130 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (ionization constant \(\mathrm{K}_{\mathrm{b}}=5 \times 10^{-4}\) ) is mixed with \(0.08 \mathrm{~mol} \mathrm{HCl}\) and the volume is made up of 1 litre. Find the \(\left[\mathrm{H}^{+}\right]\)of resulting solution. (a) \(8 \times 10^{-2}\) (b) \(2 \times 10^{-11}\) (c) \(1.23 \times 10^{-4}\) (d) \(8 \times 10^{-11}\)

Step-by-Step Solution

Answer

The \([\mathrm{H}^+]\) of the resulting solution is \(8 \times 10^{-11}\).

Step 1: Determine the Nature of the Solution

When methylamine (\(\mathrm{CH}_3 \mathrm{NH}_2\)) is mixed with hydrochloric acid (\(\mathrm{HCl}\)), it forms a salt solution. Methylamine is a weak base and hydrochloric acid is a strong acid.

Step 2: Calculate Moles of Salt Formation

Methylamine reacts with \(\mathrm{HCl}\) to form methylammonium chloride. The reaction is:\[\mathrm{CH}_3 \mathrm{NH}_2 + \mathrm{HCl} \rightarrow \mathrm{CH}_3 \mathrm{NH}_3^+ + \mathrm{Cl}^-\]Since \(0.1\) mole of \(\mathrm{CH}_3 \mathrm{NH}_2\) reacts with \(0.08\) mole of \(\mathrm{HCl}\), the limiting reagent is \(\mathrm{HCl}\). Thus, \(0.08\) moles of the salt \(\mathrm{CH}_3 \mathrm{NH}_3^+\) are formed. The leftover methylamine will be \(0.1 - 0.08 = 0.02\) moles.

Step 3: Use the Henderson-Hasselbalch Equation

Since we have a buffer solution containing the weak base (leftover methylamine) and its conjugate acid (\(\mathrm{CH}_3 \mathrm{NH}_3^+\)), we can use the Henderson-Hasselbalch equation:\[p\mathrm{H} = p\mathrm{K}_\mathrm{a} + \log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\]First, calculate \(\mathrm{pK}_\mathrm{a}\) from \(\mathrm{K}_\mathrm{b}\) using the relation \(\mathrm{K}_\mathrm{w} = \mathrm{K}_\mathrm{a} \times \mathrm{K}_\mathrm{b}\)\ \(\mathrm{K}_\mathrm{w} = 1 \times 10^{-14}\), hence\[\mathrm{K}_\mathrm{a} = \frac{10^{-14}}{5 \times 10^{-4}} = 2 \times 10^{-11}\]and\[\mathrm{pK}_\mathrm{a} = -\log(2 \times 10^{-11}) = 10.7\]Substitute the values into the Henderson-Hasselbalch equation:\[p\mathrm{H} = 10.7 + \log\left(\frac{0.02}{0.08}\right)\]\[p\mathrm{H} = 10.7 + \log(0.25)\]\[p\mathrm{H} = 10.7 - 0.6 = 10.1\]

Step 4: Calculate H⁺ Concentration

Convert \(\mathrm{pH}\) back to \([\mathrm{H}^+]\):\[\mathrm{pH} = -\log[\mathrm{H}^+]\]So,\[[\mathrm{H}^+] = 10^{-10.1}\approx 8 \times 10^{-11}\]

Key Concepts

Buffer Solution

A buffer solution is a special type of solution that resists changes in pH when small amounts of an acid or a base are added. This remarkable property makes buffer solutions incredibly important in both industrial and biological systems. The key components of a buffer solution are a weak acid and its conjugate base, or a weak base and its conjugate acid. In our example, a buffer solution is formed because there is a mixture of methylamine (a weak base) and its conjugate acid, methylammonium ion, in the solution.

After the reaction between methylamine and hydrochloric acid, there are 0.08 moles of methylammonium ions formed, and 0.02 moles of unreacted methylamine remain. This combination acts as a buffer, maintaining the pH of the solution in a specific range. The buffering action comes from the ability of the weak base and its conjugate acid to neutralize any added acids or bases, keeping the pH stable.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental formula used to estimate the pH of buffer solutions. It's particularly useful because it provides a straightforward way to calculate the pH based on concentrations of the acid and base involved. The equation is represented as follows:
\[p\mathrm{H} = p\mathrm{K}_\mathrm{a} + \log\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\]
This equation helps to determine the pH when you have a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

In our example, once we know the pKa, which was calculated from the given Kb for methylamine, we can substitute it into the equation along with the concentrations of the methylamine and methylammonium. This substitution gives us a quick and easy way to compute the pH, allowing us to find the pH of this buffer system as 10.1.

Ionization Constant

The ionization constant, denoted as Kb for bases and Ka for acids, is a measure of the strength of an acid or a base in aqueous solution. It indicates how well an acid or base ionizes in water.

For bases, Kb is the equilibrium constant for the ionization reaction, which shows how a base reacts with water to form its conjugate acid and hydroxide ions. In the given exercise, the ionization constant of methylamine is provided as Kb = \(5 \times 10^{-4}\). Using the relation between Ka and Kb, coupled by the ionization constant of water \(K_w\), you can switch between the strength of bases and their conjugate acids. This is critical in determining pKa from Kb, using:
\[K_\mathrm{w} = K_\mathrm{a} \times K_\mathrm{b}\]

The pKa can then be calculated using:
\[\mathrm{pK}_\mathrm{a} = -\log(K_\mathrm{a})\]
Knowing the ionization constant is essential in buffer solution calculations as it directly influences the pH value. It helps us figure out how much of a compound will ionize, setting the stage for the practical application of the Henderson-Hasselbalch equation.