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Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of \(0.40 M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table 13-3). a. aniline b. methylamine

Short Answer

Expert verified
For aniline, \(\left[\mathrm{OH}^{-}\right] = x 鈮 1.3 \times 10^{-5} M\), \(\left[\mathrm{H}^{+}\right] 鈮 7.7 \times 10^{-10} M\), and \(\mathrm{pH} 鈮 9.11\). For methylamine, \(\left[\mathrm{OH}^{-}\right] = x 鈮 6.63 \times 10^{-3} M\), \(\left[\mathrm{H}^{+}\right] 鈮 1.5 \times 10^{-12} M\), and \(\mathrm{pH} 鈮 11.82\).

Step by step solution

01

Write the equilibrium expressions for both amines

We will write the equilibrium expressions for both amines as they react with water to form hydroxide ions and their respective conjugate acids. The general reaction is: Amine + H2O \(\rightleftharpoons\) Conjugate Acid + OH- For aniline (C6H5NH2): C6H5NH2 + H2O \(\rightleftharpoons\) C6H5NH3+ + OH- For methylamine (CH3NH2): CH3NH2 + H2O \(\rightleftharpoons\) CH3NH3+ + OH-
02

Set up the ICE table for the equilibrium expressions

We will set up an ICE (Initial, Change, Equilibrium) table for each amine to determine how the concentrations change during the reaction. For aniline: \[ \begin{array}{cccc} &{} & C6 H5 N H2 & & + & & H2 O & \rightleftharpoons & C6 H5 N H3^{+} & + & O H^{-} \\ I & & 0.40 & M & - & - & - & & 0 & M & & 0 & M \\ C & & - x & & - & - & - & & + x & & + x \\ E & & 0.40 - x & & - & - & - & & x & & x \end{array} \] For methylamine: \[ \begin{array}{cccc} &{} & C H3 N H2 & & + & & H2 O & \rightleftharpoons & C H3 N H3^{+} & + & O H^{-} \\ I & & 0.40 & M & - & - & - & & 0 & M & & 0 & M \\ C & & - x & & - & - & - & & + x & & + x \\ E & & 0.40 - x & & - & - & - & & x & & x \end{array} \]
03

Use the Kb values to solve for x, and then find [OH-]

For aniline, given \(K_b = 4.2 \times 10^{-10}\): \[K_b = \frac{[C6H5NH3^+][OH鈦籡}{[C6H5NH2]} = \frac{x^2}{0.40-x}\] For methylamine, given \(K_b = 4.4 \times 10^{-4}\): \[K_b = \frac{[CH3NH3^+][OH鈦籡}{[CH3NH2]} = \frac{x^2}{0.40-x}\] For each amine, we will solve for x, which represents the equilibrium concentration of OH鈦. Since \(K_b\) values are small, we can assume that \(x << 0.40\) and simplify the equations. For aniline: \[4.2 \times 10^{-10} \approx \frac{x^2}{0.40}\] Solve for x to find \(\left[OH^{-}\right]\). For methylamine: \[4.4 \times 10^{-4} \approx \frac{x^2}{0.40}\] Solve for x to find \(\left[OH^{-}\right]\).
04

Calculate [H+] and pH

Using the obtained OH鈦 concentration, we can find the H鈦 concentration using the relationship: \(K_w = \left[H^+\right]\left[OH^{-}\right]\), where \(K_w = 1.0 \times 10^{-14}\). Solve for the [H+] for both amines and use the formula to find the pH: \(pH = -\log\left([H^+]\right)\) Calculate the pH of both the aniline and methylamine solutions.

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Key Concepts

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

ICE table
The ICE table is a handy tool in chemistry for tracking the concentrations of reactants and products in a chemical reaction as it reaches equilibrium. Let's break it down:

An ICE table consists of three main parts:
  • Initial (I): The starting concentrations of the reactants and products.
  • Change (C): The change in concentrations as the reaction moves towards equilibrium.
  • Equilibrium (E): The concentrations of the reactants and products when the system is at equilibrium.
For aniline and methylamine, we begin with 0.40 M concentrations of each amine and 0 M of their respective conjugate acids or hydroxide ions. As the reaction proceeds, an amount 'x' of the reactive species is changed, resulting in the formation of hydroxide ions and the conjugate acid. The equilibrium expressions add clarity to how concentrations adjust as equilibrium is achieved. Understanding each part of the ICE framework is vital for solving equilibrium problems efficiently.
Equilibrium expressions
Equilibrium expressions are equations derived from the balanced chemical reactions. They help us understand how the concentrations of reactants and products are related at equilibrium. For a base dissociation reaction, the general expression can be written from the equilibrium law:

For amines, the equilibrium reactions are:
  • Aniline: \(C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^-\)
  • Methylamine: \(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\)
The equilibrium constant expression for the dissociation of these weak bases (denoted by the base dissociation constant, \(K_b\)) is crucial. It compares the concentration of the products to the concentration of the reactants at equilibrium. For example, for aniline, the expression is:\[ K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} \]This equation shows us the relationship between the concentration of aniline, its conjugate acid, and hydroxide ions once equilibrium is reached.
Base dissociation constant (Kb)
The base dissociation constant, \(K_b\), quantifies the strength of a base in solution. It indicates the extent to which a base dissociates to form hydroxide ions in water. A larger \(K_b\) value means the base is stronger and will dissociate more to produce OH鈦 ions.

Both aniline and methylamine have their own \(K_b\) values, reflecting their base strength:
  • Aniline: \(4.2 \times 10^{-10}\)
  • Methylamine: \(4.4 \times 10^{-4}\)
Given these \(K_b\) values, we calculate the concentration of hydroxide ions, \([OH^-]\), in the solution using the formula:\[ K_b = \frac{x^2}{[Initial \, Concentration] - x} \]For simplicity, since \(K_b\) is quite small relative to the initial concentration (0.40 M), we often approximate by assuming \(x\) is much smaller than 0.40 M. This approximation enables us to simplify the calculation and solve for \(x\), which represents the \([OH^-]\) concentration.
pH and pOH relationship
Understanding the relationship between pH and pOH is crucial in calculating the acidity or basicity of a solution. It stems from the ion product constant of water, \(K_w\), which is always \(1.0 \times 10^{-14}\) at 25掳C. This constant maintains a balance:\[ [H^+][OH^-] = K_w \]Knowing \([OH^-]\), we can find \([H^+]\) using the formula:\[ [H^+] = \frac{K_w}{[OH^-]} \]Once \([H^+]\) is calculated, the pH is determined using:\[ pH = -\log([H^+]) \]These steps showcase how hydroxide ion concentrations lead to pH calculations and extend into real-world applications, like understanding soil acidity, water treatment, and more. This method of calculation allows us to bridge the gap between the theoretical expressions and practical assessments of acidity or basicity in a given solution.

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Most popular questions from this chapter

A solution is prepared by adding \(50.0 \mathrm{mL}\) of \(0.050 \mathrm{M}\) HBr to 150.0 mL of 0.10 \(M\) HI. Calculate \(\left[\mathrm{H}^{+}\right]\) and the pH of this solution. HBr and HI are both considered strong acids.

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{M} \mathrm{HCl}\) b. 5.0 M HCl c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\)

A sample containing 0.0500 mole of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{Fe}^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is 6.73 atm at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+} .\) (To do this calculation, you must assume that none of the ions go through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)

A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4}\) ) has a pH of \(2.70 .\) Calculate the initial concentration of formic acid in this solution.

Consider the species \(\mathrm{PO}_{4}^{3-}, \mathrm{HPO}_{4}^{2-},\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} .\) Each ion can act as a base in water. Determine the \(K_{\mathrm{b}}\) value for each of these species. Which species is the strongest base?

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