`K_a" for "RC-=CH" is "1xx10^-5`. Hence `pK_b` of conjugate base is…..........

Answer

To find the \( pK_b \) of the conjugate base of the given compound \( RC \equiv CH \) with a \( K_a \) value of \( 1 \times 10^{-5} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given value**:
We are given \( K_a = 1 \times 10^{-5} \).

2. **Calculate \( pK_a \)**:
The formula to calculate \( pK_a \) from \( K_a \) is:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1 \times 10^{-5}) \]

3. **Simplify the logarithmic expression**:
Using the property of logarithms, we can simplify:
\[ pK_a = -(-5 \log(10)) = 5 \cdot 1 = 5 \]
(since \( \log(10) = 1 \))

4. **Use the relationship between \( pK_a \) and \( pK_b \)**:
The relationship between \( pK_a \) and \( pK_b \) is given by:
\[ pK_a + pK_b = 14 \]
Now, substituting the value of \( pK_a \):
\[ 5 + pK_b = 14 \]

5. **Solve for \( pK_b \)**:
Rearranging the equation gives:
\[ pK_b = 14 - 5 = 9 \]

6. **Final answer**:
Therefore, the \( pK_b \) of the conjugate base is:
\[ pK_b = 9 \]