The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution ?

To determine the pH of a 0.1 M solution of chloroacetic acid and its corresponding sodium salt solution, we need to consider the acid dissociation and the properties of the salt. Chloroacetic acid (ClCH2COOH) is a weak acid, and its ionization constant (Ka) is given as 1.35 × 10–3. Let's break this down step by step.

Calculating the pH of Chloroacetic Acid Solution

First, we will find the pH of the 0.1 M chloroacetic acid solution. The dissociation of chloroacetic acid in water can be represented as:

ClCH2COOH ⇌ ClCH2COO⁻ + H⁺

The ionization constant (Ka) expression for this reaction is:

Ka = [ClCH2COO⁻][H⁺] / [ClCH2COOH]

Let’s denote the concentration of H⁺ ions produced as x. Since we start with 0.1 M of chloroacetic acid, at equilibrium, the concentrations will be:

• [ClCH2COOH] = 0.1 - x
• [ClCH2COO⁻] = x
• [H⁺] = x

Substituting these values into the Ka expression gives us:

1.35 × 10–3 = (x)(x) / (0.1 - x)

Assuming x is small compared to 0.1, we can simplify this to:

1.35 × 10–3 = (x²) / 0.1

Now, solving for x:

x² = 1.35 × 10–3 × 0.1

x² = 1.35 × 10–4

x = √(1.35 × 10–4) ≈ 0.0116 M

This x value represents the concentration of H⁺ ions. To find the pH, we use the formula:

pH = -log[H⁺]

pH = -log(0.0116) ≈ 1.93

Determining the pH of the Sodium Salt Solution

Next, let's calculate the pH of a 0.1 M solution of sodium chloroacetate (the sodium salt of chloroacetic acid). Sodium chloroacetate is a salt that will dissociate completely in solution:

ClCH2COONa → ClCH2COO⁻ + Na⁺

In this case, the chloroacetate ion (ClCH2COO⁻) can act as a weak base. It will hydrolyze in water as follows:

ClCH2COO⁻ + H₂O ⇌ ClCH2COOH + OH⁻

The base dissociation constant (Kb) can be calculated using the relationship between Ka and Kb:

Kw = Ka × Kb

Where Kw (the ion product of water) is 1.0 × 10⁻¹⁴ at 25°C. Thus:

Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.35 × 10⁻³) ≈ 7.41 × 10⁻¹²

Now, we can set up the Kb expression for the hydrolysis reaction:

Kb = [ClCH2COOH][OH⁻] / [ClCH2COO⁻]

Let y be the concentration of OH⁻ produced. At equilibrium, we have:

• [ClCH2COO⁻] = 0.1 - y
• [ClCH2COOH] = y
• [OH⁻] = y

Substituting into the Kb expression gives:

7.41 × 10⁻¹² = (y)(y) / (0.1 - y)

Assuming y is small compared to 0.1, we simplify to:

7.41 × 10⁻¹² = y² / 0.1

y² = 7.41 × 10⁻¹² × 0.1

y² = 7.41 × 10⁻¹³

y = √(7.41 × 10⁻¹³) ≈ 8.61 × 10⁻⁷ M

This y value represents the concentration of OH⁻ ions. To find the pOH, we use:

pOH = -log[OH⁻]

pOH = -log(8.61 × 10⁻⁷) ≈ 6.06

Finally, we can convert pOH to pH using the relationship:

pH + pOH = 14

pH = 14 - 6.06 ≈ 7.94

Summary of Results

In summary, the pH of a 0.1 M solution of chloroacetic acid is approximately 1.93, while the pH of a 0.1 M solution of sodium chloroacetate is around 7.94. This illustrates how the properties of acids and their salts can significantly influence the pH of solutions.