Question

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) and \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to enough water to form \(2.00 \mathrm{~L}\) of solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

Short answer

The pH of the buffer is 4.64. The complete ionic equation for the reaction with HCl is \(CH_3COO^- (aq) + H^+ (aq) \rightarrow CH_3COOH (aq)\). The complete ionic equation for the reaction with NaOH is \(CH_3COOH (aq) + OH^- (aq) \rightarrow CH_3COO^- (aq) + H_2O (l)\).

Step-by-step solution

Calculate the molar mass of acetic acid and sodium acetate.

First, we need to determine the molar mass of acetic acid (CH3COOH) and sodium acetate (CH3COONa). Molar mass of CH3COOH: \(12.01 × 2 + 1.01 × 4 + 16.00 × 2 = 60.05 \, g/mol\) Molar mass of CH3COONa: \(22.99 + 12.01 × 2 + 1.01 × 3 + 16.00 × 2 = 82.03 \, g/mol\)

Calculate the moles and concentrations of CH3COOH and CH3COO- in the buffer.

Given that we have 20.0 g of both CH3COOH and CH3COONa, we can now calculate the moles of each component: Moles of CH3COOH: \(20.0 \, g \div 60.05 \, g/mol = 0.333 \, mol\) Moles of CH3COO- (CH3COONa dissociates completely in water): \(20.0 \, g \div 82.03 \, g/mol = 0.244 \, mol\) Now, we can find the concentrations of CH3COOH and CH3COO- in the buffer: Concentration of CH3COOH: \(0.333 \, mol / 2.0 \, L = 0.167 \, M\) Concentration of CH3COO-: \(0.244 \, mol / 2.0 \, L = 0.122 \, M\)

Determine the pH of the buffer.

We can now use the Henderson-Hasselbalch equation to find the pH of the buffer: \(pH = pK_a + \log\frac{[A-]}{[HA]}\), where [A-] is the concentration of the conjugate base (CH3COO-) and [HA] is the concentration of the weak acid (CH3COOH). The pKa of acetic acid is 4.76. Substituting the values, we get: \(pH = 4.76 + \log\frac{0.122}{0.167} = 4.64\). The pH of the buffer is 4.64.

Write the complete ionic equation for the reaction with HCl.

When hydrochloric acid (HCl) is added to the buffer, the following reaction occurs: \(CH_3COO^- + H^+ \rightarrow CH_3COOH\) The complete ionic equation for this reaction is: \(CH_3COO^- (aq) + H^+ (aq) \rightarrow CH_3COOH (aq)\)

Write the complete ionic equation for the reaction with NaOH.

When sodium hydroxide (NaOH) is added to the buffer, the following reaction occurs: \(CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O\) The complete ionic equation for this reaction is: \(CH_3COOH (aq) + OH^- (aq) \rightarrow CH_3COO^- (aq) + H_2O (l)\) In summary, the pH of the buffer is 4.64, the complete ionic equation for the reaction with HCl is \(CH_3COO^- (aq) + H^+ (aq) \rightarrow CH_3COOH (aq)\), and the complete ionic equation for the reaction with NaOH is \(CH_3COOH (aq) + OH^- (aq) \rightarrow CH_3COO^- (aq) + H_2O (l)\).

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a mathematical expression that relates the pH of a buffer solution to the pKa (acid dissociation constant) of the acid and the ratio of the concentrations of its conjugate base and the acid. Recognizing this equation is vital for students, as it provides a straightforward method for estimating the pH of buffer solutions. To apply this equation, you need to identify the acid component (HA) and its conjugate base (A-).

The formula is expressed as:
\[ pH = pKa + \log\frac{[A^-]}{[HA]} \]
It is important for students to remember that the pKa value is unique to each weak acid. For exercises that involve the Henderson-Hasselbalch equation, it's essential to always ensure the correct pKa value for the acid in question is used. Practical exercises typically include calculating the pH of a buffer after adding specific amounts of acid or base, promoting the comprehension of the buffer's capacity to maintain pH.

pH Calculation

Understanding pH Levels

Understanding the concept of pH is integral to chemistry and biology, as it measures the acidity or basicity of an aqueous solution. The pH scale ranges from 0 to 14, where pH less than 7 indicates acidity, 7 is neutral, and greater than 7 indicates alkalinity.

To calculate the pH, one must understand the logarithmic scale used to define it. The pH of a solution is given by the negative logarithm of the hydrogen ion concentration:
\[ pH = -\log[H^+] \]
This relationship means that as the hydrogen ion concentration increases, the pH decreases, indicating a more acidic solution. Through practice problems involving pH calculation, students learn to connect the theoretical knowledge of pH with practical examples, such as determining the pH of a given buffer solution.

Ionic Equations

Navigating Ionic Reactions

Ionic equations are a simplified representation of chemical reactions that only show the ions involved in the reaction. This is particularly helpful for understanding reactions in solution, where compounds dissociate into their constituent ions. When writing complete ionic equations, it is important to include all of the species present in the reaction in their ionic form, indicating the physical states (such as (aq) for aqueous or (s) for solid).

In buffer reactions, these equations demonstrate how added acids or bases react with the buffer components. For instance, when a strong acid like hydrochloric acid (HCl) is added, it dissociates completely into hydrogen ions and chloride ions. These hydrogen ions react with the acetic ion (CH3COO-) part of the buffer, maintaining the system's pH. Explaining this through ionic equations enables students to visualize how buffers work on a molecular level and reinforces the concept of ion exchange in chemical reactions.