Question

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

Short answer

The pH of the 0.050 M diethylamine \((C_2H_5)_2NH\) solution is 10.51.

Step-by-step solution

Write the acid-base reaction and the Kb expression

The reaction for the dissociation of diethylamine (C2H5)2NH is given by: \((C_2H_5)_2NH+H_2O \Leftrightarrow (C_2H_5)_2NH_2^+ + OH^-\) Now, we can write the Kb expression based on the given value: \(K_b = \frac{[(C_2H_5)_2NH_2^+][OH^-]}{[(C_2H_5)_2NH]}\) We know that the initial concentration of diethylamine is 0.050 M, and initially, the concentrations of (C2H5)2NH2+ and OH- are 0.

Create an ICE table

To determine the concentrations of the species at equilibrium, set up an ICE table: \[\begin{array}{cccc} & -x & -x & +x \\ (C_2H_5)_2NH & 0.050 & 0 & 0 \\ H_2O & - & - & - \\ (C_2H_5)_2NH_2^+ & 0 & +x & \\ OH^- & 0 & +x & \\ \end{array}\]

Substitute the equilibrium concentrations into the Kb expression

Now, we have the following algebraic equation for x: \(K_b = \frac{x^2}{0.050 - x}\)

Solve for x

Since Kb is small, we can approximate x as being much smaller than 0.050, so: \(K_b = \frac{x^2}{0.050}\) Solving for x, we get: \(x = [\ce{OH^{-}}] = \sqrt{0.050 * K_b} = \sqrt{0.050 * 1.3 * 10^{-3}} = 3.22 * 10^{-4} M\)

Calculate pOH

Now that we have found the concentration of OH-, we can calculate the pOH using the formula: \(pOH = -\log_{10} [\ce{OH^{-}}]\) \(pOH = -\log_{10} (3.22 * 10^{-4}) = 3.49\)

Calculate pH

Finally, we can find the pH using the relationship between pH, pOH, and the ion product of water (Kw): \(pH + pOH = 14\) Thus, the pH is: \(pH = 14 - pOH = 14 - 3.49 = 10.51\) So, the pH of the given diethylamine solution is 10.51.

Key concepts

Acid-Base Reaction

When dealing with acid-base reactions, it's important to understand there are two fundamental parts: acids, which donate protons (H+), and bases, which accept protons. In our example exercise, we are examining the behavior of a base, diethylamine ext{(C$_2$H$_5$)$_2$NH}. When diethylamine is dissolved in water, it undergoes an interaction often called dissociation. In simple terms, it accepts a proton from water, resulting in the formation of its conjugate acid, ext{(C$_2$H$_5$)$_2$NH$_2^+$}, and hydroxide ions, ext{OH$^-$}. This process is as follows: ext{(C$_2$H$_5$)$_2$NH + H$_2$O ⇌ (C$_2$H$_5$)$_2$NH$_2^+$ + OH$^-$}. Understanding this reaction is key because it allows us to measure the concentration of hydroxide ions which directly impacts our ability to calculate the pOH of the solution.

• Diethylamine acts as a base by accepting a proton from water.
• The process forms hydroxide ions, crucial for pOH calculations.
• The formed conjugate acid, ext{(C$_2$H$_5$)$_2$NH$_2^+$}, is a product of this reaction.

Equilibrium Concentrations

In an acid-base reaction such as the one described, at equilibrium, the concentration of reactants and products are constant. Equilibrium is a state where the rate of the forward reaction equals the rate of the backward reaction. Here, it is important because it tells us how much of each species is present at equilibrium. For the dissociation of diethylamine, the equilibrium concentrations are necessary to work with the base dissociation constant ( K$_b$) expression ext{K$_b$ = rac{[(C$_2$H$_5$)$_2$NH$_2^+$][OH$^-$]}{[(C$_2$H$_5$)$_2$NH]}} to find the concentration of ext{OH$^-$}.

• We begin with an initial concentration of 0.050 M diethylamine.
• The concentration of each species changes as the reaction reaches equilibrium.
• Understanding equilibrium concentrations is critical in using the K$_b$ expression effectively.

ICE Table

One of the most organized ways to track what happens in a chemical reaction is using an ICE table. ICE stands for Initial, Change, and Equilibrium. Essentially, it's a tool that helps us visualize and calculate changes in concentration during the reaction.
Let's walk through each column:
- **Initial**: Here, we list the starting concentrations for all species. For our diethylamine, we begin with a concentration of 0.050 M. The other species, hydroxide ions ext{OH$^-$} and the conjugate acid ext{(C$_2$H$_5$)$_2$NH$_2^+$}, start at a concentration of zero.
- **Change**: As the reaction progresses, the concentration of reactants decreases by an unknown value, x, while the products' concentrations increase by x. This results in expressions like x and -x in the changes row.
- **Equilibrium**: These values represent the concentrations at equilibrium, which are then substituted into the K$_b$ equation for calculations.

• ICE tables simplify complex equilibrium problems.
• They break down the steps into Initial, Change, and Equilibrium stages.
• This helps determine the equilibrium concentrations essential for K$_b$ calculations.

pOH Calculation

Calculating the ext{pOH} is straightforward once you have the hydroxide ion concentration. The ext{pOH} is a measure of hydroxide ion concentration in a solution and is calculated using the formula: pOH = - ext{log}_{10}[ ext{OH$^-$}]. In our example, after determining the concentration of hydroxide ions to be about 3.22 x 10$^{-4}$ M, inserting this value into the formula yields a ext{pOH} of 3.49.
Knowing the ext{pOH} allows us to easily find the ext{pH} of the solution because of the relationship: ext{pH} + ext{pOH} = 14. Therefore, by subtracting the calculated ext{pOH} from 14, we obtain the ext{pH}, reflecting the solution's acidity or basicity.

• pOH calculation is vital for determining the alkalinity of a solution.
• The relationship between ext{pH} and ext{pOH} (pH + pOH = 14) simplifies this process.
• Knowing ext{pOH} helps calculate the ext{pH} efficiently.