Problem 21 - Chemistry: Principles and Reactions

Question

Calculate the pH of a solution prepared by mixing \(2.00 \mathrm{~g}\) of butyric acid \(\left(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\right)\) with \(0.50 \mathrm{~g}\) of \(\mathrm{NaOH}\) in water \(\left(K_{\mathrm{a}}\right.\) butyric acid \(\left.=1.5 \times 10^{-5}\right)\)

Step-by-Step Solution

Answer

Answer: The pH of the solution is 5.91.

Step 1: Write the dissociation equation of butyric acid and the reaction equation with sodium hydroxide

The butyric acid dissociation equation is as follows: \[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\leftrightharpoons\mathrm{H}^+ + \mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-\] And the reaction between butyric acid and sodium hydroxide is: \[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2 + \mathrm{NaOH} \rightarrow \mathrm{NaC}_4\mathrm{H}_7\mathrm{O}_2 + \mathrm{H}_2\mathrm{O}\]

Step 2: Calculate the moles of butyric acid and sodium hydroxide

First, we need to find the molar mass of butyric acid and sodium hydroxide. Molar mass of butyric acid (\(\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\)) = \(12.01 × 4 + 1.01 × 8 + 16 × 2 = 88.11\mathrm{~g/mol}\) Now, calculate the moles of butyric acid: \[\frac{2.00\mathrm{~g}}{88.11\mathrm{~g/mol}} = 0.02271\mathrm{~mol}\] Molar mass of sodium hydroxide (\(\mathrm{NaOH}\)) = \(22.99 + 16 + 1.01 = 40.00\mathrm{~g/mol}\) Now, calculate the moles of sodium hydroxide: \[\frac{0.50\mathrm{~g}}{40.00\mathrm{~g/mol}} = 0.01250\mathrm{~mol}\]

Step 3: Determine the moles of each species after the reaction between butyric acid and sodium hydroxide

Since sodium hydroxide is a strong base, it will fully react with butyric acid. The limiting reagent is sodium hydroxide (0.01250 mol), so the reaction equation becomes: \(0.02271\mathrm{~mol}~\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2 - 0.01250\mathrm{~mol}~\mathrm{NaOH} = 0.01021\mathrm{~mol}~\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\) (remaining) Now, we have 0.01021 mol of butyric acid and 0.01250 mol of butyrate ion (\(\mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-\)) in the solution.

Step 4: Calculate the pH of the solution using the equilibrium constant and an equilibrium expression

Using the equilibrium expression, we can write: \[\frac{[\mathrm{H}^+][\mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-]}{[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2]} = K_a = 1.5 \times 10^{-5}\] We can use the initial concentrations of the species and set up an equilibrium table to find the final concentrations: \[\frac{[\mathrm{H}^+][0.01250\mathrm{~mol}+x]}{[0.01021\mathrm{~mol}-x]} = 1.5 \times 10^{-5}\] Since \(K_a\) is very small, we can assume that \(x\) will be negligible compared to the initial concentrations: \[\frac{[\mathrm{H}^+][0.01250\mathrm{~mol}]}{[0.01021\mathrm{~mol}]} = 1.5 \times 10^{-5}\] Now, solve for \([\mathrm{H}^+]\): \[[\mathrm{H}^+] = \frac{1.5 \times 10^{-5} \times 0.01021\mathrm{~mol}}{0.01250\mathrm{~mol}} = 1.229 \times 10^{-6}\] Finally, calculate the pH – pH = -\(\log {[\mathrm{H}^+]}\): \[\mathrm{pH} = -\log{(1.229 \times 10^{-6})} = 5.91\] So, the pH of the solution is 5.91.

Key Concepts

Acid-Base Reaction

An acid-base reaction occurs when an acid and a base interact in an aqueous solution. This is one of the basic types of chemical reactions often encountered in chemistry.
In the example provided, butyric acid (\(\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\)) and sodium hydroxide (\(\mathrm{NaOH}\)) are mixed together in water.

• The acid here, butyric acid, can donate a proton (\(\mathrm{H}^+\)) to a base.
• Sodium hydroxide, as a strong base, can accept a proton from the acid to produce water.

This produces sodium butyrate (\(\mathrm{NaC}_4\mathrm{H}_7\mathrm{O}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). The hydroxide ion from sodium hydroxide neutralizes the hydrogen ion from butyric acid.
In this particular reaction, sodium hydroxide is a limiting reagent, meaning that it will be entirely used up, while some butyric acid remains.

Equilibrium Expression

The equilibrium expression is a mathematical way to express the concentrations of reactants and products in a chemical reaction at equilibrium.
For an acid, like butyric acid, it is described using the Ka, known as the acid dissociation constant. The formula is:\[\frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} = K_a\]where \(\mathrm{HA}\) is the undissociated acid, \(\mathrm{H}^+\) is the hydrogen ion, and \(\mathrm{A}^-\) is the dissociated ion.
In the exercise, initial amounts of the acid and base dictate the equilibrium state. Since the Ka is quite small (\(1.5 \times 10^{-5}\)), it indicates that only a small fraction of the butyric acid dissociates into ions.
By solving the expression, you find the concentration of hydrogen ions, which is essential for calculating pH.

Dissociation Equation

The dissociation equation represents the process by which an acid breaks down into ions in solution.
For butyric acid, the dissociation equation is:\[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\leftrightharpoons\mathrm{H}^+ + \mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-\]This shows how butyric acid dissociates in water to yield butyrate ion (\(\mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-\)) and a hydrogen ion (\(\mathrm{H}^+\)). In equilibrium, the forward and reverse reactions occur at the same rate.
In practical terms, dissociation is crucial when calculating the pH of a solution because it determines how many hydrogen ions are available.
The dissociation process is influenced by both the strength of the acid and the presence of a strong base, as seen in the example where \(\mathrm{NaOH}\) leads to further reactions.

Molar Mass Calculation

Molar mass calculation is a fundamental step in reactions involving moles. The molar mass of a compound tells you how much one mole of the substance weighs.
This is crucial when converting the mass of a substance to moles in chemical calculations.

• For butyric acid (\(\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\)), you calculate its molar mass (88.11 g/mol) by adding up the atomic masses: 4 carbons, 8 hydrogens, and 2 oxygens.
• For sodium hydroxide (\(\mathrm{NaOH}\)), the molar mass is calculated by adding the masses of 1 sodium, 1 oxygen, and 1 hydrogen, resulting in 40.00 g/mol.

By knowing the molar mass, you can determine how many moles of each substance are involved in the reaction. This allows you to track how much of each reagent is used and how much remains, which is critical for finding the solution's pH.