Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a 0.075 \(\mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .\) Calculate the pH of this solution.

Short answer

The molar concentration of \(\mathrm{OH}^{-}\) in the 0.075 M solution of ethylamine is approximately \(6.87 \times 10^{-3}\mathrm{M}\), and the pH of the solution is approximately 11.84.

Step-by-step solution

Write down the base ionization reaction for ethylamine

Ethylamine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), is a weak base. It will react with water to form the respective cation and hydroxide ion. The base ionization reaction is as follows: \[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} (\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} (\mathrm{aq}) + \mathrm{OH}^{-} (\mathrm{aq})\]

Set up the equilibrium table

Next, we will set up an equilibrium table (ICE table) to track the change in concentrations of each species involved in the ionization reaction. Since the initial concentration of ethylamine is given as 0.075 M: | | Initial | Change | Equilibrium | |-------------------|----------|---------|-------------| | \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2\) | 0.075 M | -x | 0.075-x | | \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+\) | 0 | +x | x | | \(\mathrm{OH}^-\) | 0 | +x | x |

Use the \(K_b\) expression

Now, we will write the \(K_b\) expression for the base ionization reaction and substitute the equilibrium concentrations of each species: \[K_b = \frac{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^{-}]}{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2]}\] Substituting the equilibrium concentrations: \(6.4 \times 10^{-4} = \frac{x \cdot x}{0.075 - x}\)

Solve for x

To simplify the problem, the assumption can be made that \(x \ll 0.075\), so the equation becomes: \(6.4 \times 10^{-4} = \frac{x^2}{0.075}\) Solving for x: \(x = [\mathrm{OH}^{-}] = \sqrt{0.075 \times 6.4 \times 10^{-4}} \approx 6.87 \times 10^{-3}\mathrm{M}\)

Calculate the pH

Finally, we will convert the concentration of \(\mathrm{OH}^{-}\) to pH. First, we find the \(pOH\): \(pOH = -\log{[\mathrm{OH}^{-}]} = -\log{(6.87 \times 10^{-3})} \approx 2.16\) Using the relationship between pH, \(pOH\), and the temperature-dependent value \(K_w\), we can find the pH: \(pH + pOH = 14\) \(pH = 14 - pOH = 14 - 2.16 = 11.84\) The pH of the solution is approximately 11.84.

Key concepts

Understanding Molar Concentration

When studying chemistry, the concept of molar concentration is a critical aspect that students encounter. Molar concentration, often expressed as molarity and denoted by the symbol M, measures the number of moles of a solute present in one liter of solution. Essentially, it tells us how concentrated a solution is.

For instance, if we have a 0.075 M solution of ethylamine, this means that there are 0.075 moles of ethylamine per liter of the solution. Understanding how to calculate molar concentration is fundamental when delving into reactions in solution, as the concentrations of reactants can directly affect reaction rates and the position of equilibrium.

To further elucidate, imagine you have a pitcher of lemonade. The amount of lemon concentrate in the pitcher represents the 'solute,' and the entire pitcher's volume represents the 'solution.' The 'concentration' is a measure of how strong or weak the lemonade is, which parallels how molar concentration gives us a measure of how 'strong' or 'weak' a chemical solution is.

Navigating Base Ionization Reactions

The process of a base ionization reaction is essentially a base reacting with water and producing its conjugate acid and hydroxide ions (OH^-). This reaction is important because it is the fundamental concept for understanding how basic solutions can be.

Let's look at our example, ethylamine (C₂H₅NH₂). When in water, it reacts to form ethylammonium ions (C₂H₅NH₃⁺) and hydroxide ions. The hydroxide ions are responsible for making the solution basic. The base's ability to produce hydroxide ions is quantified by its base ionization constant (K_b).

It is helpful to visualize this: when you drop a piece of effervescent tablet into water, it fizzes and dissolves, creating bubbles—this resembles how ethylamine releases hydroxide ions into the water,

Equilibrium Constant Expression and Its Role

In chemical reactions, equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. This dynamic state can be described quantitatively with an equilibrium constant expression.

For our base ionization reaction, the equilibrium constant is represented by K_b, the base ionization constant. This constant is a ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. In our example, the expression looks like this:
\[K_b = \frac{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^{-}]}{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2]}\]
Understanding this expression is key as it allows us to solve for unknown concentrations, like the concentration of hydroxide ions in the solution. The equilibrium constant not only gives insight into the extent of the reaction but also helps us to achieve the quantitative aspects of chemical reactions in a solution, such as pH calculations.

It's like a tug of war where both teams are equally strong. The rope doesn't move because the forces are balanced, paralleling how at equilibrium, the rate of product formation is matched by the rate of the reverse reaction.