To determine the value of [tex]\( K_b \)[/tex], given the acid dissociation constant [tex]\( K_a \)[/tex] and the relationship between [tex]\( K_a \)[/tex] and [tex]\( K_b \)[/tex], we use the ion product of water [tex]\( K_w \)[/tex].
The relationship between [tex]\( K_a \)[/tex], [tex]\( K_b \)[/tex], and [tex]\( K_w \)[/tex] is described by the equation:
[tex]\[ K_a \cdot K_b = K_w \][/tex]
We are provided with:
- [tex]\( K_a = 1.3 \times 10^{-7} \)[/tex]
- [tex]\( K_w = 1.0 \times 10^{-14} \)[/tex] (a constant at 25°C)
We need to find [tex]\( K_b \)[/tex]. Rearranging the equation to solve for [tex]\( K_b \)[/tex]:
[tex]\[ K_b = \frac{K_w}{K_a} \][/tex]
Substitute the given values into the equation:
[tex]\[ K_b = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-7}} \][/tex]
Performing the division:
[tex]\[ K_b = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-7}} = 7.692307692307692 \times 10^{-8} \][/tex]
Upon simplification, this value is approximately:
[tex]\[ K_b \approx 7.7 \times 10^{-8} \][/tex]
Therefore, the value of [tex]\( K_b \)[/tex] is:
[tex]\[ \boxed{7.7 \times 10^{-8}} \][/tex]
