`K_(a)` for butyric acid is `2.0xx10^(-5)`. Calculate pH and hydroxyl ion concentration in `0.2M` aqueous solution of sodium butyrate.

Answer

To solve the problem of calculating the pH and hydroxyl ion concentration in a 0.2 M aqueous solution of sodium butyrate, we will follow these steps:

### Step 1: Understand the Hydrolysis of Butyrate Ion
Sodium butyrate (C3H7COONa) dissociates in water to give butyrate ions (C3H7COO⁻) and sodium ions (Na⁺). The butyrate ion can hydrolyze in water:
\[ \text{C}_3\text{H}_7\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_3\text{H}_7\text{COOH} + \text{OH}^- \]

### Step 2: Set Up the Initial Concentrations
Initially, we have:
- Concentration of C3H7COO⁻ = 0.2 M
- Concentration of C3H7COOH = 0 M
- Concentration of OH⁻ = 0 M

### Step 3: Define Changes at Equilibrium
Let \( x \) be the change in concentration at equilibrium. Thus, at equilibrium:
- Concentration of C3H7COO⁻ = \( 0.2 - x \)
- Concentration of C3H7COOH = \( x \)
- Concentration of OH⁻ = \( x \)

### Step 4: Write the Expression for Kb
The base dissociation constant (Kb) for the reaction can be expressed as:
\[ K_b = \frac{[\text{C}_3\text{H}_7\text{COOH}][\text{OH}^-]}{[\text{C}_3\text{H}_7\text{COO}^-]} \]
Substituting the equilibrium concentrations:
\[ K_b = \frac{x \cdot x}{0.2 - x} = \frac{x^2}{0.2 - x} \]

### Step 5: Relate Kb to Ka
We know that:
\[ K_w = K_a \cdot K_b \]
Where \( K_w \) (the ionization constant of water) is \( 1.0 \times 10^{-14} \) and \( K_a \) for butyric acid is given as \( 2.0 \times 10^{-5} \). Therefore:
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-5}} = 5.0 \times 10^{-10} \]

### Step 6: Substitute Kb into the Equation
Now we can substitute \( K_b \) into the previous equation:
\[ 5.0 \times 10^{-10} = \frac{x^2}{0.2 - x} \]

### Step 7: Neglect x in the Denominator
Since \( x \) is expected to be small compared to 0.2 M, we can simplify:
\[ 5.0 \times 10^{-10} = \frac{x^2}{0.2} \]
Thus:
\[ x^2 = 5.0 \times 10^{-10} \times 0.2 = 1.0 \times 10^{-10} \]

### Step 8: Solve for x
Taking the square root:
\[ x = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \]

### Step 9: Determine Hydroxyl Ion Concentration
The concentration of hydroxyl ions \( [OH^-] \) is equal to \( x \):
\[ [OH^-] = 1.0 \times 10^{-5} \, \text{M} \]

### Step 10: Calculate pOH and pH
To find pOH:
\[ pOH = -\log[OH^-] = -\log(1.0 \times 10^{-5}) = 5 \]
Now, using the relation \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answers
- **pH** = 9
- **Hydroxyl Ion Concentration** = \( 1.0 \times 10^{-5} \, \text{M} \)