A buffer solution is made up of acetic acid `[pK_(a)=5]` having conc.=1.5M and sodium acetate having conc.=0.15 M. What is the number `OH^(-)` ions present in 1 litre solution?
a. `10^(-10) N_(A)`
b. `10^(-4) N_(A)`
c. `10^(-3) N_(A)`
d. `10^(-6) N_(A)`

Options

A. `10^(-10) N_(A)`
B. `10^(-4) N_(A)`
C. `10^(-3) N_(A)`
D. `10^(-6) N_(A)`

Answer

To solve the problem, we need to determine the concentration of hydroxide ions \([OH^-]\) in a buffer solution made of acetic acid and sodium acetate. Here’s the step-by-step solution:

### Step 1: Identify the components of the buffer solution
The buffer solution consists of:
- Acetic acid \((CH_3COOH)\) with a concentration of \(1.5 \, M\) and \(pK_a = 5\).
- Sodium acetate \((CH_3COONa)\) with a concentration of \(0.15 \, M\).

### Step 2: Use the Henderson-Hasselbalch equation
The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
where:
- \([A^-]\) is the concentration of the base (sodium acetate).
- \([HA]\) is the concentration of the acid (acetic acid).

### Step 3: Substitute the values into the equation
Substituting the values into the equation:
\[ pH = 5 + \log\left(\frac{0.15}{1.5}\right) \]
Calculating the ratio:
\[ \frac{0.15}{1.5} = 0.1 \]
Now, substituting this back into the equation:
\[ pH = 5 + \log(0.1) \]
Since \(\log(0.1) = -1\):
\[ pH = 5 - 1 = 4 \]

### Step 4: Calculate pOH
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
We can find pOH:
\[ pOH = 14 - pH = 14 - 4 = 10 \]

### Step 5: Calculate the concentration of hydroxide ions
The concentration of hydroxide ions can be calculated using the formula:
\[ pOH = -\log[OH^-] \]
Rearranging gives:
\[ [OH^-] = 10^{-pOH} = 10^{-10} \]

### Step 6: Calculate the number of hydroxide ions in 1 liter
To find the number of hydroxide ions in 1 liter of solution, we multiply the concentration of hydroxide ions by Avogadro's number \((N_A)\):
\[ \text{Number of } OH^- = [OH^-] \times N_A = 10^{-10} \times N_A \]

### Conclusion
Thus, the number of hydroxide ions present in 1 liter of the solution is:
\[ 10^{-10} N_A \]
The correct answer is option **a. \(10^{-10} N_A\)**.

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