What is the `OH^(-)` concentration of a 0.08 M solution of `[K_(a)(CH_(3)COOH)=1.8 xx 10^(-5)]`

Answer

To find the concentration of \( OH^- \) in a 0.08 M solution of sodium acetate, we can follow these steps:

### Step 1: Understand the Reaction
Sodium acetate (\( CH_3COONa \)) dissociates in water to produce acetate ions (\( CH_3COO^- \)) and sodium ions (\( Na^+ \)). The acetate ions can react with water to produce acetic acid (\( CH_3COOH \)) and hydroxide ions (\( OH^- \)):
\[ CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^- \]

### Step 2: Write the Expression for \( K_b \)
The base dissociation constant (\( K_b \)) for this reaction can be expressed as:
\[ K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} \]
Where:
- \( [CH_3COOH] \) is the concentration of acetic acid,
- \( [OH^-] \) is the concentration of hydroxide ions,
- \( [CH_3COO^-] \) is the concentration of acetate ions.

### Step 3: Relate \( K_b \) and \( K_a \)
We know that:
\[ K_a \cdot K_b = K_w \]
Where \( K_w = 1.0 \times 10^{-14} \) at 25°C. Given \( K_a \) for acetic acid is \( 1.8 \times 10^{-5} \), we can find \( K_b \):
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.55 \times 10^{-10} \]

### Step 4: Set Up the Equation
Let \( x \) be the concentration of \( OH^- \) produced. Initially, the concentration of acetate ions is 0.08 M. At equilibrium:
- \( [CH_3COOH] = x \)
- \( [OH^-] = x \)
- \( [CH_3COO^-] = 0.08 - x \approx 0.08 \) (since \( x \) will be very small)

Substituting into the \( K_b \) expression:
\[ K_b = \frac{x \cdot x}{0.08} = \frac{x^2}{0.08} \]
Setting this equal to \( K_b \):
\[ 5.55 \times 10^{-10} = \frac{x^2}{0.08} \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x^2 = 5.55 \times 10^{-10} \cdot 0.08 \]
\[ x^2 = 4.44 \times 10^{-11} \]
Taking the square root:
\[ x = \sqrt{4.44 \times 10^{-11}} \approx 6.67 \times 10^{-6} \, M \]

### Step 6: Conclusion
The concentration of \( OH^- \) in the solution is approximately:
\[ [OH^-] \approx 6.67 \times 10^{-6} \, M \]