Question

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(C_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) ii.. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}+\) b. Calculate the \(\mathrm{pH}\) of this solution.

Short answer

a. Major species in the solution are: i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), iii. \(\mathrm{OH}^{-}\), iv. \(\mathrm{H}_{2} \mathrm{O}\), v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^+\) b. The pH of the solution is approximately 12.29.

Step-by-step solution

Write the Reaction of Ethylamine in Water

Ethylamine (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)) acts as a weak base in water, as it can accept a proton (\(\mathrm{H}^{+}\)) from water. The balanced reaction is given by: \[\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2(aq) + \mathrm{H}_2\mathrm{O}(l) \leftrightarrows \mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^{+}(aq) + \mathrm{OH}^{-}(aq)\]

Identify the Major Species

Major species are those which have significant concentrations in the solution. Based on the reaction, we can identify following major species: i. \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2\): As 0.67 M solution, it is the major species. ii. \(\mathrm{H}^{+}\): Not a major species, as it comes from the autoionization of water which is very small. iii. \(\mathrm{OH}^{-}\): A major species as ethylamine is a weak base and reacts with water to create hydroxide ions. iv. \(\mathrm{H}_2\mathrm{O}\): A major species, as it is a solvent. v. \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^+\): A major species, as it is formed when ethylamine reacts with water.

Calculate the Concentration of Hydroxide Ions

To calculate the concentration of hydroxide ions, we can set up an equilibrium expression using \(K_b\) and an ICE (Initial, Change, Equilibrium) table: \[K_b = \frac{[\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2]}\] Initial concentrations: \([C_{2}H_5NH_2] = 0.67 M\), \([C_{2}H_5NH_3^+] = 0 M\), \([\mathrm{OH}^{-}] = 0 M\) Change in concentrations: \(+x\) moles of \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2\) and \(-x\) moles of \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^+\) and \(\mathrm{OH}^{-}\) will be in equilibrium. Equilibrium concentrations: \([C_{2}H_5NH_2] = 0.67 - x M\), \([C_{2}H_5NH_3^+] = x M\), \([\mathrm{OH}^{-}] = x M\) Plugging in concentrations into the \(K_b\) expression: \[5.6\times10^{-4} = \frac{x^2}{0.67-x}\] Since \(K_b\) is small, \(x\) will be small compared to 0.67. Thus, we can approximate: \[5.6\times10^{-4} \approx \frac{x^2}{0.67}\] Solving for \(x\): \[x = [\mathrm{OH}^{-}] = \sqrt{5.6\times10^{-4}\times0.67} \approx 0.0193\ M\]

Calculate the pH of the Solution

Now that we have the concentration of hydroxide ions, we can calculate the concentration of hydrogen ions using the relationship: \[\mathrm{pK_w} = \mathrm{pH} + \mathrm{pOH} = 14\] First, calculate pOH: \[\mathrm{pOH} = -\log(0.0193) \approx 1.71\] Next, calculate pH: \[\mathrm{pH} = \mathrm{pK_w} - \mathrm{pOH} = 14 - 1.71 \approx 12.29\] So, the pH of the ethylamine solution is approximately 12.29.

Key concepts

Ethylamine in Water

Ethylamine is a type of weak base that can interact with water. When ethylamine dissolves in water, it tends to accept protons from water molecules. This process is why it's considered a base. The chemical equation for this interaction is:

• \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2(aq) + \mathrm{H}_2\mathrm{O}(l) \leftrightarrows \mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^{+}(aq) + \mathrm{OH}^{-}(aq)\)

This reaction shows that when ethylamine is in water, it forms two new species: ethylammonium ions (\(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^+\)) and hydroxide ions (\(\mathrm{OH}^-\)).
In this context, water acts as an acid because it donates a proton to ethylamine, while ethylamine acts as a base by accepting that proton.
It's essential to understand these interactions to grasp how weak bases alter the pH of a solution.

Equilibrium in Chemistry

In chemistry, equilibrium refers to the state where the rates of the forward and reverse reactions are equal.
Meaning the concentrations of reactants and products remain constant over time.
In the case of ethylamine in water, equilibrium is established when ethylamine and water interact to form ethylammonium and hydroxide ions.
Once equilibrium is reached, the reaction doesn't stop; both forward and reverse reactions continue to occur, but at equal rates.
This dynamic balance is sensitive to changes in concentration, pressure, or temperature, which is explained by Le Chatelier's Principle.
If you add more ethylamine, for example, the system will shift to produce more products until a new equilibrium is established.

pH Calculation

Calculating pH is crucial to understanding the chemical environment of a solution.
The pH scale ranges from 0 to 14, with 7 being neutral. Solutions with a pH below 7 are acidic, while those above 7 are basic.
For weak bases like ethylamine, the hydroxide ion concentration is key.First, find the pOH using the concentration of \(\mathrm{OH}^-\):

• pOH = -\(\log(0.0193)\) which is approximately 1.71
• pH and pOH are related through the equation: \(\mathrm{pH} + \mathrm{pOH} = 14\)

So, knowing the pOH allows us to find the pH:

• pH = 14 - pOH, resulting in a pH around 12.29

This calculation shows that ethylamine solutions are basic due to the formation of \(\mathrm{OH}^-\) ions.

Chemical Species Identification

Identifying the chemical species in a solution helps predict its behavior and properties.
In ethylamine solutions, several species can be present, and it's essential to determine which are significant. Major species typically include compounds with the highest concentrations and those contributing to the solution's chemical behavior.For a 0.67 M solution of ethylamine:

• \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2\) is the dominant species, being the starting base in the solution.
• \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^+\) forms as ethylamine accepts a proton; therefore, it's significant.
• \(\mathrm{OH}^-\) ions are a product of the reaction with water, enhancing the solution's basic nature.
• Water (\(\mathrm{H}_2\mathrm{O}\)) acts as a solvent and is a major component.

Minor species like \(\mathrm{H}^+\) are typically negligible due to the very low concentration resulting from water's autoionization compared to the reaction's rate.