Question

Calculate the \(\mathrm{pH}\) of a \(0.36 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\) solution. \(\left(K_{\mathrm{a}}\right.\) for acetic acid \(\left.=1.8 \times 10^{-5} .\right)\)

Short answer

The pH of the solution is approximately 9.15.

Step-by-step solution

Identify the Nature of the Solution

First, note that the solution contains sodium acetate ( \(\text{CH}_3\text{COONa}\)), which is a salt. It dissociates in water into \(\text{CH}_3\text{COO}^-\) and \(\text{Na}^+\). Sodium ions (\(\text{Na}^+\) are neutral, so the significant species left to affect pH is the acetate ion \(\text{CH}_3\text{COO}^-\), which can react with water to form acetic acid and hydroxide ions, creating a basic solution.

Set Up the Equilibrium Equation

Realize that \(\text{CH}_3\text{COO}^-\) (acetate ion) acts as a base. The equilibrium reaction in water is: \[\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-\] We must find the \(\text{K}_b\) to determine the pH.

Calculate the Base Dissociation Constant (Kb)

Use the relationship between \(\text{K}_a\) and \(\text{K}_b\) through the ion-product constant for water, \(\text{K}_w = 1.0 \times 10^{-14}\):\[\text{K}_b = \frac{\text{K}_w}{\text{K}_a}\]Substitute \(\text{K}_a = 1.8 \times 10^{-5}\):\[\text{K}_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\]

Describe the Initial Conditions

The initial concentration of \(\text{CH}_3\text{COO}^-\) is the same as the solution concentration, which is 0.36 M.With no initial amounts of \(\text{CH}_3\text{COOH}\) and \(\text{OH}^-\).

Apply the ICE Table Method

Set up an ICE table for the equilibrium concentrations:- **I**nitial: \[ \text{CH}_3\text{COO}^- = 0.36, \; \text{CH}_3\text{COOH} = 0, \; \text{OH}^- = 0 \]- **C**hange: \[ -x, \; x, \; x \]- **E**quilibrium: \[ 0.36-x, \; x, \; x \]

Formulate the Equilibrium Expression and Solve for x

Set up the expression for \(\text{K}_b\):\[ \text{K}_b = \frac{[\text{CH}_3\text{COOH}][\text{OH}^-]}{[\text{CH}_3\text{COO}^-]} \]Substitute equilibrium values:\[ 5.56 \times 10^{-10} = \frac{x^2}{0.36-x} \approx \frac{x^2}{0.36} \]Assume \(x\) is small compared to 0.36, therefore:\[x^2 = 5.56 \times 10^{-10} \times 0.36 \]\[x = \sqrt{2.0 \times 10^{-10}} \approx 1.4 \times 10^{-5}\]

Calculate the pOH and Then the pH

Using the value of \(x\) which represents the concentration of \(\text{OH}^-\), calculate the \(\text{pOH}\):\[ \text{pOH} = -\log[\text{OH}^-] = -\log(1.4 \times 10^{-5}) \approx 4.85 \]Since \(\text{pH} + \text{pOH} = 14\), calculate \(\text{pH}\):\[ \text{pH} = 14 - 4.85 = 9.15 \]

Key concepts

Sodium Acetate

Sodium acetate is a common chemical compound often used in laboratory settings and has the formula \( ext{CH}_3 ext{COONa}\). In water, sodium acetate dissociates entirely into sodium ions (\( ext{Na}^+\)) and acetate ions (\( ext{CH}_3 ext{COO}^-\)). While sodium ions are neutral and do not affect the pH of the solution significantly, the acetate ion can undergo a reaction with water that influences the solution's pH.
This process involving acetate ions is crucial because it contributes to the basic nature of the solution. When dissolved, acetate ions act as a base and interact with water to produce acetic acid (\( ext{CH}_3 ext{COOH}\)) and hydroxide ions (\( ext{OH}^-\)), making the solution basic. This basicity arises from the creation of these hydroxide ions, which is essential to understanding how sodium acetate solutions can increase the pH of pure water.

• Recognize that sodium acetate's effect on pH is due to acetate ions.
• Remember that sodium ions are spectators and do not affect the pH.
• Consider the acetate ion's ability to form hydroxide ions in water, increasing alkalinity.

Equilibrium Constant

The equilibrium constant is a fundamental concept in chemistry that represents the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. For acid-base reactions involving weak acids and bases, we often encounter two types of equilibrium constants: \(K_a\) and \(K_b\).
The equilibrium constant \(K_b\) for a base provides insight into its strength. In the case of acetate ions in water, we need to calculate \(K_b\) to understand how it affects the solution's pH. To find \(K_b\), we can use the following relationship:
The ion-product constant for water \(K_w\) is used, which is \(1.0 \times 10^{-14}\) at 25°C. The relation between \(K_a\) and \(K_b\) is:

• \(K_b = \frac{K_w}{K_a}\)
• \(K_w = 1.0 \times 10^{-14}\)
• For acetic acid, \(K_a = 1.8 \times 10^{-5}\)
• The computed \(K_b\) for acetate ion becomes \(5.56 \times 10^{-10}\).

This calculated \(K_b\) helps determine how acetate ions interact with water and informs on the basic character of the resulting solution.

Acid-Base Reaction

In the context of sodium acetate, we focus on the reaction of acetate ions with water, which is an acid-base reaction. Here, acetate ions \( ext{CH}_3 ext{COO}^-\), considered as a base, can accept a proton from water, leading to the formation of acetic acid and hydroxide ions:
\[ \text{CH}_3 ext{COO}^- + \text{H}_2 ext{O} \rightleftharpoons \text{CH}_3 ext{COOH} + \text{OH}^- \]
This reaction illustrates the transitional state of reactants moving to products and vice versa, reaching a dynamic equilibrium where both forward and reverse reaction rates are equal. The acetate ions lead to the formation of hydroxide ions \(\text{OH}^-\), which is critical for evaluating the basicity of the solution:

• Acetate acts like a base, effectively raising the pH.
• The reaction is reversible and relies on the equilibrium constant \(K_b\) to understand the extent.
• The formation of \(\text{OH}^-\) from this reaction is a key factor in the basic nature of solutions with sodium acetate.

The calculated \(K_b\) value from the previous section is essential in quantifying this transformation's magnitude and ensuring that the system adheres to equilibrium principles.

ICE Table Method

The ICE table method is a systematic approach used to calculate the changes in concentrations of species at equilibrium in acid-base reactions. ICE stands for Initial, Change, and Equilibrium, representing different stages of the reaction process. This method helps in setting up the problem to solve for equilibrium concentrations effectively.
When applying the ICE table method to our sodium acetate solution, it helps visualize how the acetate ions reach equilibrium:

• **Initial**: Here, you start with an initial acetate ion concentration (0.36 M in this case) and 0 for acetic acid and \(\text{OH}^-\).
• **Change**: The concentration of acetate ions decreases by \(x\), while acetic acid and \(\text{OH}^-\) increase by \(x\).
• **Equilibrium**: At equilibrium, the concentrations are \(0.36 - x\) for acetate ions and \(x\) for both acetic acid and \(\text{OH}^-\).

With these expressions, substitute into the reaction's equilibrium constant equation. This allows you to solve for \(x\), which in this exercise represents the concentration of \(\text{OH}^-\).
By determining \(x\), the ICE table approach provides a clear path to calculating specific concentrations at equilibrium stages and directly aids in the computation of pH from \(\text{pOH}\), streamlining the process to solve complex chemical equilibria.