What is the `[OH^(-)]` concentration of a 0.04 M solution of `CH_(3)COONa?`
`[K_(a)" of "CH_(3)COOH=2xx10^(-5),log2=20]`

Options

A. `5xx10^(-6)`
B. `6xx10^(-6)`
C. `2xx10^(-9)`
D. `3xx10^(-9)`

Answer

To find the concentration of hydroxide ions \([OH^-]\) in a 0.04 M solution of sodium acetate \((CH_3COONa)\), we will follow these steps:

### Step 1: Understand the dissociation of sodium acetate
Sodium acetate dissociates completely in water to give acetate ions \((CH_3COO^-)\) and sodium ions \((Na^+)\):
\[ CH_3COONa \rightarrow CH_3COO^- + Na^+ \]
The concentration of acetate ions \([CH_3COO^-]\) will be 0.04 M.

### Step 2: Determine the \(K_b\) for acetate ion
The acetate ion can react with water to produce acetic acid and hydroxide ions:
\[ CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^- \]
To find the \(K_b\) for acetate, we can use the relationship between \(K_a\) (the acid dissociation constant for acetic acid) and \(K_w\) (the ion product of water):
\[ K_b = \frac{K_w}{K_a} \]
Given:
- \(K_a\) for acetic acid \((CH_3COOH) = 2 \times 10^{-5}\)
- \(K_w = 1 \times 10^{-14}\)

Calculating \(K_b\):
\[ K_b = \frac{1 \times 10^{-14}}{2 \times 10^{-5}} = 5 \times 10^{-10} \]

### Step 3: Set up the equilibrium expression
Let \(x\) be the concentration of \([OH^-]\) produced at equilibrium. The equilibrium concentrations will be:
- \([CH_3COO^-] = 0.04 - x \approx 0.04\) (since \(x\) will be very small)
- \([CH_3COOH] = x\)
- \([OH^-] = x\)

The equilibrium expression for \(K_b\) is:
\[ K_b = \frac{[CH_3COOH][OH^-]}{[CH_3COO^-]} = \frac{x \cdot x}{0.04} = \frac{x^2}{0.04} \]

### Step 4: Solve for \(x\)
Setting the expression equal to \(K_b\):
\[ 5 \times 10^{-10} = \frac{x^2}{0.04} \]
Multiplying both sides by 0.04:
\[ x^2 = 5 \times 10^{-10} \times 0.04 = 2 \times 10^{-11} \]
Taking the square root:
\[ x = \sqrt{2 \times 10^{-11}} \approx 4.47 \times 10^{-6} \, \text{M} \]

### Step 5: Conclusion
The concentration of hydroxide ions \([OH^-]\) in the solution is approximately:
\[ [OH^-] \approx 4.47 \times 10^{-6} \, \text{M} \]