If the Kb of a certain aromatic amine is 2.24 × 10–9, then what is the value of pKb?
options: a) 9.22 b) −9.22 c) −8.65 d) 8.65
The correct answer and explanation is:
The correct answer is d) 8.65.
Explanation:
The pKb is calculated from the Kb of a base using the formula: pKb=−log(Kb)pKb = -\log(Kb)
Where:
- Kb is the base dissociation constant.
- pKb is the negative logarithm (base 10) of the Kb.
Given that the Kb of the aromatic amine is 2.24×10−92.24 \times 10^{-9}, we can calculate the pKb as follows: pKb=−log(2.24×10−9)pKb = -\log(2.24 \times 10^{-9})
To break this down:
- First, apply the logarithmic properties to simplify the expression: log(2.24×10−9)=log(2.24)+log(10−9)\log(2.24 \times 10^{-9}) = \log(2.24) + \log(10^{-9})
- The logarithm of 10−910^{-9} is: log(10−9)=−9\log(10^{-9}) = -9
- Now calculate log(2.24)\log(2.24). Using a calculator: log(2.24)≈0.350\log(2.24) \approx 0.350
- Adding these together: log(2.24×10−9)=0.350+(−9)=−8.65\log(2.24 \times 10^{-9}) = 0.350 + (-9) = -8.65
- Therefore: pKb=−(−8.65)=8.65pKb = -(-8.65) = 8.65
Thus, the pKb of the aromatic amine is 8.65.
Concept:
The pKb is a measure of the basicity of a compound. The lower the pKb, the stronger the base. In this case, since the Kb is quite small (2.24 × 10–9), the pKb value indicates that the aromatic amine is a weak base.