Calculate the pH of a solution containing 0.1 M `CH_(3)COOH`
and 0.15 M `CH_(3)COO^(-)`.
`(K_(a) "of" CH_(3)COOH=1.8xx10^(-5))`

Options

A. 9.1
B. 3.9
C. 10 .
D. 4 .91

Explanation

To calculate the pH of a solution containing 0.1 M acetic acid (CH₃COOH) and 0.15 M acetate ion (CH₃COO⁻), we can use the Henderson-Hasselbalch equation for an acidic buffer solution. Here’s the step-by-step solution:

### Step 1: Identify the components of the buffer solution
We have:
- Weak acid: CH₃COOH (acetic acid)
- Conjugate base: CH₃COO⁻ (acetate ion)

### Step 2: Write the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where:
- \([\text{A}^-]\) is the concentration of the conjugate base (CH₃COO⁻)
- \([\text{HA}]\) is the concentration of the weak acid (CH₃COOH)

### Step 3: Calculate pKₐ from Kₐ
Given:
- \(K_a\) of CH₃COOH = \(1.8 \times 10^{-5}\)

To find pKₐ:
\[ \text{pK}_a = -\log(K_a) \]
\[ \text{pK}_a = -\log(1.8 \times 10^{-5}) \]

### Step 4: Calculate the logarithm
Using a calculator:
\[ \text{pK}_a \approx 4.74 \]

### Step 5: Substitute the values into the Henderson-Hasselbalch equation
Now we can substitute the values into the equation:
- \([\text{A}^-] = 0.15 \, \text{M}\) (concentration of acetate)
- \([\text{HA}] = 0.1 \, \text{M}\) (concentration of acetic acid)

So, we have:
\[ \text{pH} = 4.74 + \log\left(\frac{0.15}{0.1}\right) \]

### Step 6: Calculate the log term
Calculate the ratio:
\[ \frac{0.15}{0.1} = 1.5 \]

Now calculate the logarithm:
\[ \log(1.5) \approx 0.176 \]

### Step 7: Final calculation of pH
Now substitute back into the equation:
\[ \text{pH} = 4.74 + 0.176 \]
\[ \text{pH} \approx 4.916 \]

### Step 8: Round the pH value
Rounding to two decimal places:
\[ \text{pH} \approx 4.92 \]

### Final Answer
The pH of the solution is approximately **4.92**.

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