Problem 22 - Chemistry: Principles and Reactions

Question

Calculate the pH of a solution prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.20 \mathrm{M}\) ethanolamine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{2}\), with \(50.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{HCl} . \mathrm{K}_{\mathrm{a}}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{3}+\) is \(3.6 \times 10^{-10}\)

Step-by-Step Solution

Answer

Question: Calculate the pH of a solution formed by mixing 100.0 mL of 1.20 M ethanolamine \( (\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{2}) \) with 50.0 mL of 1.0 M HCl, given that the \( K_{a} \) of the conjugate acid \( (\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{3}^+) \) is \(3.6 \times 10^{-10}\). Answer: The pH of the solution is approximately 9.59.

Step 1: Calculate moles of reactants

To determine the initial moles of ethanolamine and HCl, we can use the formula: moles = volume × concentration Moles of ethanolamine: moles = \((100.0 \mathrm{~mL}) \times (1.20 \mathrm{M})\) = \(120.0 \mathrm{~mmol}\) Moles of HCl: moles = \((50.0 \mathrm{~mL}) \times (1.0 \mathrm{M})\) = \(50.0 \mathrm{~mmol}\)

Step 2: Determine limiting reactant

Since we know the moles of ethanolamine and HCl, we can determine which reactant will be completely consumed in the reaction: Ethanolamine + HCl → C2H5ONH3⁺ + Cl⁻ HCl is the limiting reactant because there are fewer moles of HCl (50.0 mmol) than ethanolamine (120.0 mmol).

Step 3: Calculate the moles of species at equilibrium

To calculate the moles of species at equilibrium, we will consider the stoichiometry of the reaction. Since HCl is the limiting reactant, 50.0 mmol of it will react completely with the ethanolamine, resulting in 50.0 mmol of C2H5ONH3⁺ and Cl⁻. Moles of ethanolamine remaining at equilibrium = 120.0 mmol - 50.0 mmol = 70.0 mmol Moles of C2H5ONH3⁺ at equilibrium = 0.0 mmol + 50.0 mmol = 50.0 mmol

Step 4: Calculate the equilibrium concentration

Now that we have the moles of each species at equilibrium, we can divide by the total volume of the mixture to obtain the equilibrium concentration. Total volume = volume of ethanolamine + volume of HCl = 100.0 mL + 50.0 mL = 150.0 mL = 0.150 L Concentration of ethanolamine = \(\frac{70.0 \mathrm{~mmol}}{0.150 \mathrm{~L}}\) = 0.467 M Concentration of C2H5ONH3⁺ = \(\frac{50.0 \mathrm{~mmol}}{0.150 \mathrm{~L}}\) = 0.333 M

Step 5: Calculate the pH of the solution

Since we have the equilibrium concentrations for ethanolamine and C2H5ONH3⁺, we can now use the Ka equation: \(\mathrm{K}_{\mathrm{a}} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{2}]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{3}^+]}\) Rearrange the equation to solve for [H⁺]: \([\mathrm{H}^{+}] = \frac{\mathrm{K}_{\mathrm{a}} \times [\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{3}^+]}{[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONH}_{2}]} = \frac{(3.6 \times 10^{-10}) \times (0.333 \mathrm{M})}{(0.467 \mathrm{M})} = 2.57 \times 10^{-10} \mathrm{M}\) Finally, calculate the pH by taking the negative logarithm of the H⁺ concentration: \(\mathrm{pH} = -\log([\mathrm{H}^{+}]) = -\log(2.57 \times 10^{-10}) \approx 9.59\) The pH of the solution is approximately 9.59.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is pivotal when analyzing acid-base reactions and calculating pH levels. It refers to the state of a reaction where 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 doesn't mean the amounts of reactants and products are equal but that their concentrations remain constant. When a solution reacts, reaching this balance is essential before we can accurately determine the pH.

Certain factors can affect equilibrium, including concentration changes, temperature, and pressure, but in the context of our exercise, the focus is on the shift that occurs when reactants are mixed and react to form products. Equilibrium is vital to comprehend because it dictates the final amounts of acids, bases, and ions in a solution, directly influencing the solution's acidity or alkalinity.

Acid-Base Reactions

Acid-base reactions are a type of chemical reaction that involves the transfer of protons (H⁺ ions) from an acid to a base. These reactions are broadly characterized by the formation of water and a salt and are pivotal in the context of understanding how to calculate pH.

When we talk about acids, like HCl in our exercise, we refer to them donating an H⁺ ion to a base—in this case, ethanolamine. This is central to calculating the change in pH because the interaction between acids and bases affects the concentration of hydrogen ions [H⁺] in a solution. A higher concentration of [H⁺] makes a solution more acidic (lower pH), while a lower concentration results in a more basic (higher pH) solution. The calculation of pH hinges on the equilibrium concentrations of the acid and base after the reaction has occurred.

Stoichiometry

Stoichiometry is the study of the quantitative relationships, or ratios, in chemical reactions. It can be used to predict the amount of product formed or the amount of reactants needed for a reaction. For instance, in our exercise, we used stoichiometry to determine the moles of ethanolamine and HCl before figuring out the limiting reactant, the one that will be completely consumed in the reaction.

The stoichiometric relationship between the acid and the base in the balanced chemical equation tells us how much of one reactant will react with a given amount of another. These proportional relationships are crucial to calculate the moles of species at equilibrium, and effectively, the equilibrium concentrations needed to ascertain the pH of the solution.

Equilibrium Concentration

Equilibrium concentration, an essential aspect of chemical equilibrium, is the concentration of a reactant or product in a reaction mixture when the reaction has reached equilibrium. It's expressed in molarity (moles per liter) and is used along with the equilibrium constant to calculate the pH of a solution.

In our exercise, after establishing the limiting reactant and the stoichiometry of the reaction, we calculated the equilibrium concentrations of remaining ethanolamine and the produced C2H5ONH3⁺. These concentrations were then used in conjunction with the acid dissociation constant (Ka) to determine the hydrogen ion concentration [H⁺], which is directly related to pH. Remember, the equilibrium concentration of reactants and products is inherently linked to the final attributes of the solution, including its pH.