Question

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

Short answer

The pH of the \(0.050\)-M \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution is approximately 12.46.

Step-by-step solution

Writing the ionization equation for the base

Ethylamine, \((\mathrm{C}_{2}\mathrm{H}_{5})_{2} \mathrm{NH}\), ionizes in water as follows: \[ (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH} + \mathrm{H}_{2}\mathrm{O} \rightleftarrows (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+} + \mathrm{OH}^{-} \]

Writing the expression for Kb

The base ionization constant, Kb, is given by: \[ K_{b} = \frac{[ (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+}] [\mathrm{OH}^{-}]}{[(\mathrm{C}_{2}\mathrm{H}_{5})_{2} \mathrm{NH}]} \]

Setting up the ICE table

An ICE table helps us write the initial concentration, the change in concentration, and the equilibrium concentration for all species involved. Initial (M): \(0.050 \qquad \qquad 0 \qquad \qquad \quad 0\) Change (M): \(-x \qquad \qquad \quad +x \qquad \qquad +x\) Equilibrium (M): \(0.050 - x \qquad x \qquad \qquad x\)

Substituting values into the Kb expression

Substituting the values from the ICE table into the Kb expression: \[ K_{b} = \frac{x \times x}{0.050 - x} = 1.3 \times 10^{-3} \]

Solving for hydroxide ion concentration

Because Kb is small, we can assume that x is much smaller than 0.050, simplifying the equation: \[ x^2 \approx 1.3 \times 10^{-3} \times 0.050 \] Solving for x (concentration of \(\mathrm{OH}^-\)): \[ x \approx \sqrt{1.3 \times 10^{-3} \times 0.050} \approx 2.88 \times 10^{-2} \ \mathrm{M} \]

Calculating pOH

Now, we can find the pOH by taking the negative log of the hydroxide ion concentration: \[ \mathrm{pOH} = -\log [ \mathrm{OH}^- ]= -\log ( 2.88 \times 10^{-2}) \approx 1.54 \]

Calculating pH

Finally, we can calculate the pH using the relationship between pOH and pH: \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.54 \approx 12.46 \] The pH of the \(0.050\)-M \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution is approximately 12.46.

Key concepts

Base Ionization

When bases dissolve in water, they undergo a process known as ionization, whereby they accept protons and produce hydroxide ions (OH⁻).
In the case of the given problem, the base Ethylamine - represented by the formula \((\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}\) - ionizes in water. This leads to the formation of its conjugate acid, \((\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+}\), and hydroxide ions.
The ionization equation is set up as follows:\[(\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH} + \mathrm{H}_{2}\mathrm{O} \rightleftarrows (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+} + \mathrm{OH}^{-}\]This reversible reaction is key because it allows us to describe and calculate the extent of ionization through Kb, the base ionization constant. Understanding this ionization also provides insight into how bases increase the pH of solutions, which is crucial for many chemical calculations.

Kb Expression

The base ionization constant, represented as \(K_b\), is pivotal in quantifying the strength of a base in solution. It expresses the extent of ionization, indicating how much a base dissociates in water to form ions. In our exercise, Ethylamine has a given \(K_b\) value of \(1.3 \times 10^{-3}\).
The expression for \(K_b\) based on Ethylamine's ionization is:\[K_{b} = \frac{[(\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+}] [\mathrm{OH}^{-}]}{[(\mathrm{C}_{2}\mathrm{H}_{5})_{2} \mathrm{NH}]}\]Remember:- The numerator represents the concentrations of products: the conjugate acid and hydroxide ions.- The denominator represents the base's concentration left in its un-ionized form.This equilibrium expression helps determine the equilibrium concentrations based on initial amounts and changes using the ICE table.

ICE Table

The ICE table is a handy tool in chemistry, especially useful in calculating equilibrium concentrations through initial, change, and equilibrium stages. Let's break this useful tool down further:- **Initial Concentrations**: Begin with the concentration before any reaction occurs. For Ethylamine: \(0.050\) M for the base, and zero for the products.- **Change in Concentration**: As the reaction proceeds, the concentration of the base decreases by \(x\), while those of the products increase by \(x\).- **Equilibrium Concentrations**: Finally, the concentrations at equilibrium are \\(0.050 - x\) for Ethylamine and \(x\) for both the conjugate acid and hydroxide ions.
This systematic approach allows us to substitute values into equilibrium expressions like \(K_b\), providing a straightforward path to find unknown concentrations.

Hydroxide Ion Concentration

Calculating the hydroxide ion concentration \([\mathrm{OH}^-]\) is a critical step in determining the pH of basic solutions.
From the simplified derived equation \(x^2 \approx 1.3 \times 10^{-3} \times 0.050\), we can solve for \(x\), which represents \([\mathrm{OH}^-]\).
Upon solving, \(x \approx 2.88 \times 10^{-2} \, \mathrm{M}\).This concentration is central to calculating pOH using \(-\log [ \mathrm{OH}^- ]\). The calculated pOH can then convert to pH using the relationship \(\mathrm{pH} = 14 - \mathrm{pOH}\). Understanding this relationship and calculation is important for evaluating the alkalinity of solutions, crucial in fields like medicine and environmental science.