Question

Calculate the pH of each of the following solutions. a. \(0.100 \mathrm{M} \mathrm{HONH}_{2}\left(K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\) b. \(0.100 \mathrm{M} \mathrm{HONH}_{3} \mathrm{Cl}\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HONH}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{HONH}_{3} \mathrm{Cl}\)

Short answer

a. Using the given \(K_b\) value and the dissociation reaction for HONH2, we find that the solution's pH is approximately \(10.40\). b. For the hydrolysis of HONH3Cl, the solution's pH is approximately \(3.48\). c. The pH of pure water is \(7.0\). d. Using the Henderson-Hasselbalch equation for the mixture containing HONH2 and HONH3Cl, the pH is equal to the pKa value, which is approximately \(5.2\).

Step-by-step solution

Write down the dissociation reaction for the weak base

The weak base dissociation reaction for (HONH2) is: HONH2 + H2O ⇌ HONH3+ + OH-

Write down the Kb expression for the reaction

For the given reaction, we can write down the Kb expression as: Kb = [HONH3+][OH-] / [HONH2] We are given the Kb value as \(1.1 \times 10^{-8}\).

Set up the equilibrium concentrations and solve for x

Let [OH-] = x, [HONH3+] = x, and [HONH2] = 0.100 - x. Plugging these values into the Kb expression: \(1.1 \times 10^{-8}\) = x * x / (0.100 - x) Note that x<<0.100. So, we can simplify the equation as: \(1.1 \times 10^{-8}\) = x * x / 0.100 x = sqrt(\(1.1 \times 10^{-8}\) * 0.100)

Calculate the pH of the solution

Now to find the pH, we first need to find pOH: pOH = -log10[OH-] = -log10(x) Then, pH = 14 - pOH b) Salt of weak base and strong acid

Write down the hydrolysis reaction for the salt

The hydrolysis reaction for HONH3Cl is: HONH3+ + H2O ⇌ HONH2 + H3O+

Use Kb to find Ka for the base and write down the Ka expression for the salt

Knowing that \(K_{w} = K_{a} \cdot K_{b}\), where \(K_{w} = 1.0 \times 10^{-14}\), and given \(K_{b}\): Ka = \(K_{w}/K_{b} = (1.0 \times 10^{-14})/ (1.1 \times 10^{-8})\) Now we write the expression for Ka, Ka = [HONH2][H3O+] / [HONH3+]

Set up equilibrium concentrations and solve for x

Let [H3O+] = x, [HONH2] = x, and [HONH3+] = 0.100 - x. Plugging these values into the Ka expression: Ka = x * x / (0.100 - x) Assuming x << 0.100: Ka = x * x / 0.100 x = sqrt(Ka * 0.100)

Calculate the pH of the solution

Now to find the pH: pH = -log10[H3O+] = -log10(x) c) Pure water

Use Kw expression to find the pH of pure water

For pure water, [H3O+] = [OH-], and the \(K_{w}\) expression is given by: \(K_{w}\) = [H3O+][OH-] From this expression, we can calculate the pH directly: pH = 7.0 d) Mixture containing the base and the salt For the mixture, we need to use the Henderson-Hasselbalch (HH) equation.

Analyze the base and the salt in the mixture

The buffer in this problem is made from a weak base (HONH2) and its conjugate acid (HONH3+). The concentrations of the base and its conjugate acid are 0.100 M.

Use the HH equation to calculate the pH

The HH equation for a weak base and its conjugate acid is: pH = pKa + log10([base] / [acid]) We know the Ka value from the part b). pH then becomes: pH = -log10(Ka) + log10([0.100] / [0.100]) Since the base and its conjugate acid are present in equal amounts, their ratio is 1, and log10(1) = 0. The pH of the mixture is equal to the pKa value: pH = -log10(Ka)

Key concepts

Weak Base Dissociation

When we talk about the dissociation of a weak base, we're referring to the partial separation of the base into its ions in an aqueous solution. Unlike strong bases, which fully dissociate, weak bases only partially dissociate, and an equilibrium is established between the undissociated base and the products of dissociation. This process is vital for understanding how bases affect the acidity or basicity of a solution.

In the original exercise, the dissociation of the weak base HONH2 was considered. The equilibrium reaction for its dissociation in water can be written as follows:

HONH2 + H2O ⇌ HONH3+ + OH-

The equilibrium concentrations of the ions produced are key to determining the pH of the solution. The expression for the base dissociation constant, Kb, can help calculate these concentrations:

Hydrolysis Reaction

Hydrolysis is the reaction in which water molecules react with ions or molecules to effect a chemical change. In the context of acids and bases, when a salt is dissolved in water, the ions can react with water in a hydrolysis reaction. The extent of hydrolysis and the resulting pH of the solution depend on the strengths of the conjugate acid or base from which the salt is derived.

For example, in part b of the exercise, the salt HONH3Cl, derived from the weak base HONH2 and the strong acid HCl, undergoes hydrolysis to produce the weak base and H3O+ ions:

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch (HH) equation provides a straightforward way to calculate the pH of buffer solutions, which are mixtures of a weak acid and its conjugate base or a weak base and its conjugate acid. The equation is built upon the relationship between the pH, the pKa (or pKb), and the ratio of concentrations of the conjugate base and acid:

pH = pKa + log10([base]/[acid])

In part d), the HH equation was used to find the pH of a buffer system composed of HONH2 and its conjugate acid, HONH3+. Since their concentrations are equal, the log term cancels out, making the pH equal to the pKa of the conjugate acid:

Equilibrium Concentrations

Understanding equilibrium concentrations is critical in pH calculations involving weak acids, bases, or salts. In chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and concentrations of reactants and products remain constant, but not necessarily equal. For weak bases, the equilibrium concentration of OH- ions, for example, can be found using the base dissociation constant and initial concentrations.

In the provided solutions, equilibrium concentrations were used to calculate the values of [OH-] and [H3O+], which were then used to determine the pH. It's noteworthy that in weak acid or base equilibria, it is often valid to assume that the dissociation is small relative to the initial concentration, simplifying the calculations as shown in the exercise solutions.