Problem 46 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

An amount of \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=5 \times 10^{-4}\right)\) is mixed with \(0.08\) mole of \(\mathrm{HCl}\) and diluted to one litre. What will be the \(\mathrm{H}^{+}\) concentration in the solution? (a) \(1.25 \times 10^{-4} \mathrm{M}\) (b) \(8 \times 10^{-11} \mathrm{M}\) (c) \(1.6 \times 10^{-11} \mathrm{M}\) (d) \(2 \times 10^{-3} \mathrm{M}\)

Step-by-Step Solution

Answer

\(8 \times 10^{-2} \, \mathrm{M}\) (none of the provided options)

Step 1: Identify the Reaction

Understand that when \(\mathrm{CH}_3\mathrm{NH}_2\) (a weak base) is mixed with \(\mathrm{HCl}\) (a strong acid), they will react to form \(\mathrm{CH}_3\mathrm{NH}_3^+\) and \(\mathrm{Cl}^-\).

Step 2: Calculate Reaction Extent

Since \(\mathrm{HCl}\) is a strong acid, it will completely dissociate and react with the weak base. The limiting reactant will be \(\mathrm{CH}_3\mathrm{NH}_2\), as there is less of it. Calculate the moles of \(\mathrm{CH}_3\mathrm{NH}_3^+\) produced, which will be equivalent to the moles of \(\mathrm{CH}_3\mathrm{NH}_2\) reacted, that is 0.1 mole. Then subtract the mole of \(\mathrm{CH}_3\mathrm{NH}_2\) reacted from the total \(\mathrm{CH}_3\mathrm{NH}_2\) and \(\mathrm{HCl}\).

Step 3: Find the Moles of Unreacted \(\mathrm{HCl}\)

The moles of unreacted \(\mathrm{HCl}\) can be found by subtracting the moles of reacted \(\mathrm{CH}_3\mathrm{NH}_2\) from the initial moles of \(\mathrm{HCl}\), thus \(0.08 - 0.1 = -0.02\). Since the moles cannot be negative, we realize that all of \(\mathrm{HCl}\) reacts and leaves an excess of \(\mathrm{CH}_3\mathrm{NH}_2\), which won't affect the \(\mathrm{H}^+\) concentration.

Step 4: Calculate the Concentration of \(\mathrm{H}^+\) Ions

Since all the \(\mathrm{HCl}\) has reacted, it is completely converted into \(\mathrm{H}^+\) ions. Therefore, the concentration of \(\mathrm{H}^+\) is equal to the number of moles of initially present \(\mathrm{HCl}\) in one litre of solution, which is \(0.08 \, \mathrm{M}\).

Key Concepts

Acid-Base Reaction

Understanding the basics of an acid-base reaction is crucial for solving chemistry problems, particularly those involving neutralization reactions. An acid-base reaction involves the transfer of a proton (hydrogen ion, \(H^+\)) from an acid to a base. Strong acids like hydrochloric acid (\(HCl\)) ionize completely in water, releasing \(H^+\) ions with ease. Weak bases, such as methylamine (\(CH_3NH_2\)), accept protons less readily and do not completely ionize in a solution. When a weak base and a strong acid react, the acid's protons are transferred to the base, forming a weak conjugate acid (for example, \(CH_3NH_3^+\)) and a conjugate base (such as \(Cl^-\)).

A clear understanding of the nature of the reactants—knowing whether they are strong or weak acids or bases—allows students to efficiently predict the outcome of the reaction and, just as importantly, calculate the concentration of ions in the resulting solution. The reaction between \(CH_3NH_2\) and \(HCl\) forms \(CH_3NH_3^+\) and \(Cl^-\), with the \(CH_3NH_2\) acting as a base by accepting \(H^+\) from the \(HCl\), which behaves as an acid.

Equilibrium Concentration

The concept of equilibrium concentration is important in understanding reactions that do not proceed to completion. However, in our case with strong acid-weak base reactions, the situation is slightly different. \(HCl\), being a strong acid, dissociates completely and reacts fully with \(CH_3NH_2\). The \(H^+\) concentration is tied directly to the amount of strong acid initially present, as it releases its protons completely into the solution. Since all the \(HCl\) reacts, there's no 'equilibrium' in the traditional sense where both reactants and products exist simultaneously; the \(H^+\) concentration after the reaction is simply the initial concentration of \(HCl\).

In cases where an equilibrium is established (which would happen if both reactants were weak), the Equilibrium Constant Expression (for acids \(K_a\) or bases \(K_b\)) would be used to calculate the equilibrium concentrations of all species in solution. This is crucial for understanding the behavior of weak acids or bases, which do not fully dissociate and for which the equilibrium concentrations are significantly less straightforward to calculate.

Weak Base with Strong Acid

When a weak base is mixed with a strong acid, the acidity of the solution is determined by the strong acid due to its complete dissociation. In our example, \(CH_3NH_2\) is the weak base and \(HCl\) is the strong acid. The strong acid \(HCl\) relinquishes its \(H^+\) ions to the base upon contact, resulting in a new species, \(CH_3NH_3^+\), effectively removing \(H^+\) ions from the solution. Given the strong acid dictates the pH, and knowing the mole-to-mole reaction between \(HCl\) and \(CH_3NH_2\), calculating the remaining concentration of \(H^+\) ions becomes a simple matter of accounting for the initial number of moles of \(HCl\) present.

It is important to identify the limiting reactant in these reactions to accurately calculate the expected concentrations. If \(CH_3NH_2\) is in excess, it will not influence the \(H^+\) concentration post-reaction, since all of the \(HCl\) would have been consumed. The remaining \(CH_3NH_2\) only impacts the solution if additional protons are introduced, which isn’t the case here. The phenomenon is a key aspect of titration problems and is essential for accurate calculations in laboratory settings.