Question

When \(0.1\) mole of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (ionization constant \(\left.\mathrm{K}_{\mathrm{b}}=5 \times 10^{-4}\right)\) 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) \(8 \times 10^{-11}\)

Short answer

The hydrogen ion concentration is \(8 \times 10^{-11}\) M.

Step-by-step solution

Analyze the Reaction

Methylamine (\( \text{CH}_3 \text{NH}_2 \)) is a weak base that reacts with hydrochloric acid (HCl) in a 1:1 stoichiometry to form methylammonium chloride (\( \text{CH}_3 \text{NH}_3^+\text{Cl}^- \)). The initial moles of methylamine are 0.1 moles, and HCl is 0.08 moles.

Determine Limiting Reactant

Since you have 0.08 moles of HCl and 0.1 moles of methylamine, HCl is the limiting reactant. It will react completely.

Calculate Remaining Moles of Methylamine

After the reaction, calculate the moles of unreacted methylamine: \[ 0.1 - 0.08 = 0.02 \text{ moles} \] of methylamine remain unreacted.

Calculate Moles of Conjugate Acid

The reaction produced an equal amount of conjugate acid, \( \text{CH}_3 \text{NH}_3^+ \), which is 0.08 moles.

Set Up the Equilibrium Expression

Use the equilibrium constant expression for the ionization of methylamine. The reaction is: \( \text{CH}_3 \text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3 \text{NH}_3^+ + \text{OH}^- \)The expression for \( \text{K}_b \) is: \[\text{K}_b = \frac{[\text{CH}_3 \text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3 \text{NH}_2]} \]Values:- \([\text{CH}_3 \text{NH}_2] = 0.02 \text{ M}\)- \([\text{CH}_3 \text{NH}_3^+] = 0.08 \text{ M}\).

Solve for Hydroxide Concentration

Assume \([\text{OH}^-] = x \). Substitute the known values into the \( \text{K}_b \) expression:\[5 \times 10^{-4} = \frac{0.08 \times x}{0.02}\]Solve for \( x \):\[x = \frac{5 \times 10^{-4} \times 0.02}{0.08} = 1.25 \times 10^{-4} \]thus \([\text{OH}^-] = 1.25 \times 10^{-4} \text{ M}\).

Calculate Hydrogen Ion Concentration

Use the relation between \([\text{H}^+]\) and \([\text{OH}^-]\) in water:\[[\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\] Substitute \([\text{OH}^-] = 1.25 \times 10^{-4}\): \[[\text{H}^+] = \frac{1.0 \times 10^{-14}}{1.25 \times 10^{-4}} = 8 \times 10^{-11}\] Thus, the concentration of \([\text{H}^+] = 8 \times 10^{-11} \text{ M}\).

Key concepts

Ionization Constant

The ionization constant, often referred to as the equilibrium constant for ionization processes, characterizes the extent to which a substance dissociates into ions in a solution. When dealing with weak bases like methylamine, the constant is denoted as \( K_b \), specifically indicating its basic nature. Methylamine reacts in a manner represented by the equation:

• \( \text{CH}_3 \text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3 \text{NH}_3^+ + \text{OH}^- \)

Here, \( K_b \) is the ratio of the concentration of the products over reactants at equilibrium. In our example, given \( K_b = 5 \times 10^{-4} \), this value indicates how much methylamine becomes ionized in solution under equilibrium conditions. It informs us of the base strength: the larger the \( K_b \), the stronger the base. This principle is fundamental in calculating the equilibrium concentrations, which further helps in determining related ion concentrations like \([\text{OH}^-]\) and eventually \([\text{H}^+]\). Understanding these concepts helps in practical applications like calculating the pH of mixtures.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, limiting the amount of product formed. Identification of the limiting reactant is crucial because it dictates the maximum extent of product formation. Consider the reaction between methylamine and hydrochloric acid. Initially, we mix 0.1 moles of methylamine with 0.08 moles of HCl. Because the reaction proceeds with a 1:1 mole ratio, the hydrochloric acid, being lesser in quantity, becomes the limiting reactant. It will react completely with methylamine, defining how much methylammonium chloride can be formed. The limiting reactant concept holds significant implications in calculations. It allows chemists to predict leftover materials, like the 0.02 moles of remaining methylamine after HCl is all used up. Failing to identify it can lead to incorrect estimates of reaction yields and concentrations of substances in the product mixture.

Conjugate Acid

A conjugate acid is formed when a base gains a hydrogen ion. In our chemical reaction scenario, methylamine (\( \text{CH}_3 \text{NH}_2 \)) acts as a weak base, and when it accepts a hydrogen ion from HCl (acting as an acid), it forms its conjugate acid, methylammonium ion \( \text{CH}_3 \text{NH}_3^+ \).The conjugate acid-base concept helps in understanding the behavior of substances in water, showcasing how substances can shift their roles between acids and bases during a reaction. In our example, the conjugate acid, \( \text{CH}_3 \text{NH}_3^+ \), equilibrates within the solution influencing the overall pH.Understanding the formation of conjugate acids not only contributes to our knowledge of the dynamic equilibrium in a given reaction but also aids in calculating important values like the hydrogen ion concentration \([\text{H}^+]\). Conjugate acids play an integral role as they directly relate to the \( K_b \) calculations and subsequent understanding of the solution's pH and acidity.