What is `[H^(+)]` in mol /L of a solution that is 0.20 M in ` CH_3COONa and 0.10 ` M in ` CH_3COOH? (Ka = 1.8 xx 10^(-5) `)
` 3.5 xx 10^(-4)`
` 1.1 xx 10^(-5)`
` 1.8 xx 10^(-5)`
` 9.0 xx 10 ^(-6)`

Options

A. ` 3.5 xx 10^(-4)`
B. ` 1.1 xx 10^(-5)`
C. ` 1.8 xx 10^(-5)`
D. ` 9.0 xx 10 ^(-6)`

Answer

To find the concentration of hydrogen ions \([H^+]\) in a solution that is 0.20 M in sodium acetate \((CH_3COONa)\) and 0.10 M in acetic acid \((CH_3COOH)\), we can use the concept of acid dissociation constant \(K_a\) and the common ion effect.

### Step-by-Step Solution:

1. **Identify the Components**:
- Sodium acetate \((CH_3COONa)\) dissociates in water to give acetate ions \((CH_3COO^-)\) and sodium ions \((Na^+)\).
- Acetic acid \((CH_3COOH)\) can dissociate to give acetate ions \((CH_3COO^-)\) and hydrogen ions \((H^+)\).

2. **Write the Dissociation Reaction**:
The dissociation of acetic acid can be represented as:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

3. **Set Up the Expression for \(K_a\)**:
The expression for the acid dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
Given that \(K_a = 1.8 \times 10^{-5}\).

4. **Substituting Known Concentrations**:
From the problem, we know:
- \([CH_3COO^-] = 0.20 \, M\) (from sodium acetate)
- \([CH_3COOH] = 0.10 \, M\) (from acetic acid)

We can substitute these values into the \(K_a\) expression:
\[ 1.8 \times 10^{-5} = \frac{(0.20)([H^+])}{0.10} \]

5. **Solving for \([H^+]\)**:
Rearranging the equation to solve for \([H^+]\):
\[ [H^+] = \frac{1.8 \times 10^{-5} \times 0.10}{0.20} \]
\[ [H^+] = \frac{1.8 \times 10^{-6}}{0.20} \]
\[ [H^+] = 9.0 \times 10^{-6} \, M \]

6. **Conclusion**:
The concentration of hydrogen ions \([H^+]\) in the solution is \(9.0 \times 10^{-6} \, M\).

### Final Answer:
\[ [H^+] = 9.0 \times 10^{-6} \, \text{mol/L} \]