`0.1M` `CH_(3)COOH(aq.)` Is mixed with `0.1M` `NaOH(aq.).` The pH of solution on mixing will be (`K_(a)` of `CH_(3)COOH` is `1.8xx10^(-5)`

Options

A. `4.74`
B. `8.70`
C. `9.75`
D. None of the above

Explanation

To find the pH of the solution when `0.1M` `CH₃COOH` (acetic acid) is mixed with `0.1M` `NaOH`, we can follow these steps:

### Step 1: Identify the Reaction
When acetic acid (a weak acid) reacts with sodium hydroxide (a strong base), they undergo a neutralization reaction to form sodium acetate (the salt) and water. The reaction can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 2: Determine the Concentration of the Salt
Since both the acetic acid and sodium hydroxide are in equal concentrations and volumes, they will completely neutralize each other. The concentration of the resulting sodium acetate will be:
- Initial concentration of acetic acid = `0.1M`
- Initial concentration of sodium hydroxide = `0.1M`
- After mixing, the concentration of sodium acetate will be `0.1M` (since they react in a 1:1 ratio).

### Step 3: Calculate the pKa of Acetic Acid
The dissociation constant \( K_a \) of acetic acid is given as \( 1.8 \times 10^{-5} \). We can calculate \( pK_a \) using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1.8 \times 10^{-5}) \]
Calculating this gives:
\[ pK_a \approx 4.74 \]

### Step 4: Use the Henderson-Hasselbalch Equation
Since we have a salt formed from a weak acid and a strong base, we can use the Henderson-Hasselbalch equation to find the pH:
\[ pH = pK_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
In our case, since all the acetic acid has been converted to sodium acetate, we can consider:
- \([\text{A}^-] = 0.1M\) (sodium acetate)
- \([\text{HA}] = 0\) (no acetic acid left)

However, since we mixed equal volumes of both solutions, the concentration of the salt will be halved:
\[ [\text{A}^-] = \frac{0.1M}{2} = 0.05M \]

### Step 5: Calculate the pH
Now substituting into the Henderson-Hasselbalch equation:
\[ pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log(0.05) \]
Substituting \( pK_a \):
\[ pH = 7 + \frac{1}{2} \times 4.74 + \frac{1}{2} \log(0.05) \]
Calculating \( \log(0.05) \):
\[ \log(0.05) = -1.301 \]
Now substituting this value:
\[ pH = 7 + 2.37 - 0.6505 \]
\[ pH \approx 8.7195 \]

### Final Result
Thus, the pH of the solution after mixing is approximately **8.70**.