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Chapter 8: Problem 130

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) \(8 \times 10^{-11}\)

Short Answer

Expert verified
The [H^+] is approximately (d) \(8 \times 10^{-11}\).

Step by step solution

01

Analyze the Problem

First, notice that methylamine (2H_3NH_2) reacts with hydrochloric acid (HCl) to form the methylammonium ion (4-CH_3NH_3^+) and chloride ion (Cl^-). The reaction is: 4-CH_3NH_2 + HCl b 4-CH_3NH_3^+ + Cl^-.
02

Identify Limiting Reactant

Calculate the moles of each reactant. Methylamine: 0.1 moles, HCl: 0.08 moles. Since HCl is less in amount, it is the limiting reactant, meaning all 0.08 moles of HCl will react with the same amount of 4-CH_3NH_2.
03

Calculate Moles After Reaction

After the reaction, 0.08 moles of HCl are used, leaving 0.02 moles of 4-CH_3NH_2 unreacted and producing 0.08 moles of 4-CH_3NH_3^+.
04

Set Up the Equilibrium of Methylammonium Ion

Use the ionization constant, N_{b} = 5 x 10^{-4} , for methylammonium ion: 4-CH_3NH_3^+ + H_2O b 4-CH_3NH_2 + 4-H_3O^+. Use Kb: [4-OH^-][4-CH_3NH_2]/[4-CH_3NH_3^+].
05

Derive the [OH^-] Concentration

Because there is a small change in concentration, assume it is approximately [OH^-] = [ 4-CH_3NH_2 4+. So, [OH^-]^2 4 / 0.08 - x = 5 4x 10^{-4}. Solve for x to find [OH^-].
06

Calculate the [H^+] Using [OH^-]

Using the equation ,K_w = [H^+][OH^-] where 4-K_w = 1 4x 10^{-14}4, find [H^+]. Substituting the value of [OH^-] obtained from the previous step gives you the [H^+].
07

Compare the Calculated pH with the Answer Options

After calculating [H^+], check which option among (a) (b) (c) (d) matches the found value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Methylamine
Methylamine is a weak base, often encountered in chemistry problems where it reacts with strong acids like hydrochloric acid (HCl). It is an organic compound and a derivative of ammonia, characterized by the presence of a methyl group. When methylamine, denoted as \( \text {CH}_3 \text{NH}_2 \), is mixed with HCl, they undergo a neutralization reaction. This reaction leads to the formation of methylammonium ions \( \text{CH}_3 \text{NH}_3^+ \) and chloride ions \( \text{Cl}^- \).

Let's explore some properties:
  • As a weak base, methylamine does not fully dissociate in water, which is why its reactions with acids are important in predicting solution behavior.
  • Understanding the interaction of methylamine with acids is crucial because it helps determine the concentrations of various species in the solution, impacting pH values.
  • Methylamine's reactivity with HCl is a common approach to examining acid-base equilibria, essential for solving equilibrium problems in chemistry.
Ionization Constant (Kb)
The ionization constant, or \( K_b \), is a critical concept when discussing bases like methylamine. It quantifies the extent to which a base dissociates in water.

For methylamine, the ionization constant \( K_b \) is given as \( 5 \times 10^{-4} \). Here's how it works and why it's relevant:
  • To set up the equilibrium, we use the equation \( \text{CH}_3 \text{NH}_2 \) + \( \text{H}_2 \text{O} \rightarrow \text{CH}_3 \text{NH}_3^+ + \text{OH}^- \).
  • \( K_b \) reflects the equilibrium concentrations, represented as \( K_b = \frac{[\text{CH}_3 \text{NH}_3^+][\text{OH}^-]}{[\text{CH}_3 \text{NH}_2]} \).
  • This constant is crucial for predicting the concentration of hydroxide ions \([\text{OH}^-]\) and thus the degree of basicity. The value of \( K_b \) helps calculate important properties like pH and \([\text{H}^+]\) in a solution.
By utilizing \( K_b \), you can effectively analyze and predict the behavior of weak base systems in various chemical contexts.
Limiting Reactant Analysis
When performing chemical reactions, identifying the limiting reactant is key, especially in reactions involving acids and bases, like our methylamine and HCl reaction.

Here's how limiting reactant analysis is conducted and why it matters:
  • The limiting reactant is the substance that is completely consumed first in a reaction, halting the process since there's no more substance to react with.
  • In this problem, we start with 0.1 moles of methylamine and 0.08 moles of HCl. Since HCl is in lesser quantity, it becomes the limiting reactant.
  • After all the 0.08 moles of HCl react, you are left with 0.02 moles of methylamine and 0.08 moles of methylammonium, creating a new equilibrium.
  • Understanding the limiting reactant permits calculation of residual or remaining species in the solution, essential for determining pH and other equilibrium properties.
Conducting limiting reactant analysis not only clarifies the stoichiometry of the reaction but also aids in accurately predicting the changes occurring in the solution post-reaction.

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Most popular questions from this chapter

The Ka value of formic acid and acetic acid are respectively \(1.77 \times 10^{-4}\) and \(1.75 \times 10^{-5}\). the ratio of the acid strength of \(0.1 \mathrm{~N}\) acids is (a) \(0.1\) (b) \(0.3\) (c) \(3.178\) (d) 100

The dissociation constant of acetic acid is \(1.6 \times 10^{-5}\). The degree of dissociation \((\alpha)\) of \(0.01 \mathrm{M}\) acetic acid in the presence of \(0.1 \mathrm{M} \mathrm{HCl}\) is equal to (a) \(0.4\) (b) \(0.026\) (c) \(1.6\) (d) \(0.016\)

Ionization constant of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.7 \times 10^{-5}\) and concentration of \(\mathrm{H}^{+}\)ion is \(3.4 \times 10^{-4}\). Then initial concentration of \(\mathrm{CH}_{3} \mathrm{COOH}\) is (a) \(3.4 \times 10^{-4}\) (b) \(3.4 \times 10^{-3}\) (c) \(6.8 \times 10^{-4}\) (d) \(6.8 \times 10^{-3}\)

\(100 \mathrm{ml}\) of \(0.3 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\) is mixed with \(100 \mathrm{ml}\) of \(0.2\) M \(\mathrm{NaOH} . \mathrm{K}_{b}\) of \(\mathrm{NH}_{4} \mathrm{OH}\) is \(1.8 \times 10^{-5} .\) The degree of dissociation of \(\mathrm{NH}_{4} \mathrm{OH}\) is (a) \(1.02 \times 10^{-2}\) (b) \(1.8 \times 10^{-5}\) (c) \(1.8 \times 10^{-4}\) (d) \(1.02 \times 10^{-4}\)

If molar concentrations of two weak acids are the same, their relative strengths can be compared by (a) \(\frac{\alpha_{1}}{\alpha_{2}}\) (b) \(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\) (c) \(\sqrt{\mathrm{K}_{1} / \mathrm{K}_{2}}\) (d) \(\frac{\left[\mathrm{H}^{+}\right]_{1}}{\left[\mathrm{H}^{+}\right]_{2}}\)
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