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

Answer

To solve the problem of finding the \( pK_b \) of the conjugate base given that \( K_a \) for \( RC \equiv CH \) is \( 1 \times 10^{-5} \), we can follow these steps:

### Step-by-Step Solution:

1. **Given Information**: We know that \( K_a = 1 \times 10^{-5} \).

2. **Calculate \( pK_a \)**: The relationship between \( K_a \) and \( pK_a \) is given by the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1 \times 10^{-5}) \]

3. **Using Logarithmic Properties**: We can break this down using the properties of logarithms:
\[ pK_a = -\log(1) - \log(10^{-5}) \]
Since \( \log(1) = 0 \) and \( \log(10^{-5}) = -5 \):
\[ pK_a = 0 - (-5) = 5 \]

4. **Relate \( pK_a \) to \( pK_b \)**: We use the relationship between \( pK_a \) and \( pK_b \):
\[ pK_a + pK_b = 14 \]
Now substituting the value of \( pK_a \):
\[ 5 + pK_b = 14 \]

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

### Final Answer:
Thus, the \( pK_b \) of the conjugate base is \( 9 \).

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