What is `[H^(+)]` in `mol//L` of a solution that is `0.20 M` in `CH_(3)COONa` and `0.1 M` in `CH_(3)COOH`? `K_(a)` for `CH_(3)COOH` is `1.8xx10^(-5)`?

Options

A. `3.5xx10^(-4)`
B. `1.1xx10^(-5)`
C. `1.8xx10^(-5)`
D. `9.0xx10^(-6)`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Components**:
- Sodium acetate (\(CH_3COONa\)) dissociates completely in solution to give \(CH_3COO^-\) (acetate ion) and \(Na^+\) (sodium ion).
- Acetic acid (\(CH_3COOH\)) is a weak acid that partially dissociates to give \(H^+\) (hydrogen ion) and \(CH_3COO^-\).

2. **Write the Equilibrium Expression**:
The equilibrium expression for the dissociation of acetic acid is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
where \(K_a\) for acetic acid is given as \(1.8 \times 10^{-5}\).

3. **Determine the Concentrations**:
- The concentration of acetate ion \([CH_3COO^-]\) from sodium acetate is \(0.20 \, M\).
- The concentration of acetic acid \([CH_3COOH]\) is \(0.10 \, M\).
- Let \([H^+]\) be the concentration of hydrogen ions we want to find.

4. **Substituting Values into the Equilibrium Expression**:
Substitute the known values into the \(K_a\) expression:
\[ 1.8 \times 10^{-5} = \frac{(0.20)([H^+])}{0.10} \]

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

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