Question

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.5.6 \times 10^{-4}\right)\)

Short answer

The pH of the 0.20 M C2H5NH2 solution is \(12.03\).

Step-by-step solution

Write the ionization reaction of the weak base in water

When the weak base ethylamine (C2H5NH2) ionizes in water, it accepts a proton (H+) from water and forms an ethylammonium ion (C2H5NH3+) and hydroxide ion (OH-). We can represent it as follows: \[C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-\]

Write the expression for the base ionization constant (Kb)

The Kb expression for the above reaction is given by: \[K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}\] Where, Kb - base ionization constant, values provided as 5.6 x 10^(-4). Initially, ignore the contribution of water to the ionization.

Calculate the change in the concentration of the ions

Let x be the concentration of C2H5NH3+ and OH- at equilibrium. Since the initial concentration of C2H5NH2 (weak base) is given as 0.20 M, the change in concentration during ionization can be represented as follows: C2H5NH2: 0.20 - x C2H5NH3+: x OH-: x

Substitute the concentrations back into the Kb expression

Now substitute the concentration values into the Kb expression and solve for x: \[5.6 \times 10^{-4} = \frac{x \times x}{(0.20-x)}\]

Make an approximation and solve for x

Since ethylamine is a weak base, we assume x is very small and therefore can be neglected in the denominator. So the expression now becomes: \[5.6 \times 10^{-4} = \frac{x^2}{0.20}\] Solve for x (which represents both [C2H5NH3+] and [OH-] concentrations): \[ x = \sqrt{5.6 \times 10^{-4} \times 0.20} = \sqrt{1.12 \times 10^{-4}} = 1.06 \times 10^{-2} \, M\]

Calculate pOH and pH

Now that we have found the concentration of OH-, we can calculate the pOH and then the pH of the solution. The pOH = -log[OH-] \[pOH = -\log{(1.06 \times 10^{-2})} = 1.97\] The relationship between pH and pOH is given by: pH + pOH = 14 So, the pH of the solution is: \[pH = 14 - pOH = 14 - 1.97 = 12.03\] Thus, the final pH of the 0.20 M C2H5NH2 solution is 12.03.

Key concepts

Weak Base Ionization

The process of weak base ionization refers to the partial dissociation of a weak base in water solutions. Unlike strong bases, which completely ionize, weak bases partially ionize, leading to an equilibrium between the undissociated base and the ions produced. For example, in the case of ethylamine (C2H5NH2), the ionization in water is represented by the equation:
\[C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-\]
This equation shows that ethylamine accepts a hydrogen ion (H+) from water, thus forming an ethylammonium ion (C2H5NH3+) and a hydroxide ion (OH-). The backwards arrow signifies that the reaction doesn't go to completion and establishes an equilibrium between reactants and products.

It's essential for students to understand that the degree of ionization is crucial in determining the pH of the solution. A weak base will only slightly increase the concentration of hydroxide ions, hence affecting the pH to a lesser degree compared to a strong base.

Base Ionization Constant (Kb)

The base ionization constant (Kb) is a key concept in understanding how weak bases behave in solution. It measures the strength of a weak base, or its tendency to ionize in solution. The expression for Kb is related to the equilibrium concentrations of the products and reactants involved in the ionization of the base, as shown in the provided equation:
\[K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}\]
For ethylamine, Kb is provided as \(5.6 \times 10^{-4}\). To calculate the concentration of the ions at equilibrium, it is often useful to establish an equivalence between the change in concentration of the ionized form and the hydroxide ion. Through a series of steps involving algebraic manipulation and approximation, one can solve for the hydroxide ion concentration. This process hinges on the understanding that the initial concentration of the weak base sets the stage for how much ionization can occur. Remember, in solutions of weak bases, the concentration of hydroxide ions is typically low, reflecting the incomplete ionization and thus a moderate Kb value compared to those of strong bases.

pOH to pH Conversion

After determining the concentration of hydroxide ions, we need to convert this into something more universally understood: the pH. The pOH is a measure of the hydroxide ion concentration and is calculated using the logarithm of the inverse of OH- concentration. For the given exercise, this is achieved as follows:
\[pOH = -\log{[OH^-]}\]
After calculating the pOH, the conversion to pH is straightforward, thanks to the relationship between pH and pOH, which states that the sum of pH and pOH is always equal to 14 (in pure water at 25°C). So, to find the pH from the pOH, you subtract the pOH from 14:
\[pH = 14 - pOH\]
Understanding this relationship is critical for students as it connects the concentration of hydroxide ions directly to the acidity or basicity of a solution. It's also a reminder that pH and pOH are interconnected and are simply different ways to express the concentrations of the hydrogen and hydroxide ions in a solution. This conversion is particularly important in chemistry and biology, where pH levels can dramatically affect reactions and living systems, respectively.