Caculate `[H^(+)]` in a solution that is 0.1 M both in `CH_(3)COOHandCH_(3)COONa`
`(pK_(a)CH_(3)COOH=4.7)`:-

Options

A. `1.2xx10^(9)`
B. `7xx10^(-5)`
C. `5xx10^(-5)`
D. `2xx10^(-5)`

Answer

To calculate the concentration of hydrogen ions \([H^+]\) in a solution that is 0.1 M in both acetic acid \((CH_3COOH)\) and sodium acetate \((CH_3COONa)\), we can use the Henderson-Hasselbalch equation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Concentration of acetic acid \([CH_3COOH] = 0.1 \, M\)
- Concentration of sodium acetate \([CH_3COONa] = 0.1 \, M\)
- \(pK_a\) of acetic acid = 4.7

2. **Write the Henderson-Hasselbalch Equation:**
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
where \([A^-]\) is the concentration of the conjugate base (sodium acetate) and \([HA]\) is the concentration of the weak acid (acetic acid).

3. **Substitute the Known Values into the Equation:**
\[ pH = 4.7 + \log\left(\frac{0.1}{0.1}\right) \]

4. **Calculate the Logarithm:**
Since \(\frac{0.1}{0.1} = 1\), we have:
\[ \log(1) = 0 \]
Therefore, the equation simplifies to:
\[ pH = 4.7 + 0 = 4.7 \]

5. **Convert pH to \([H^+]\):**
The concentration of hydrogen ions can be calculated using the formula:
\[ [H^+] = 10^{-pH} \]
Substituting the calculated pH:
\[ [H^+] = 10^{-4.7} \]

6. **Calculate \([H^+]\):**
Using a calculator or logarithm tables:
\[ [H^+] \approx 2.0 \times 10^{-5} \, M \]

### Final Answer:
\[ [H^+] \approx 2.0 \times 10^{-5} \, M \]