Answer:
Kb = [tex]4.45\times10^{-7}\ mol/L[/tex]
[tex]p^{Kb}=6.35[/tex]
Explanation:
For a weak organic base, the formula to find [tex]p^{OH}[/tex] is given by:
[tex]p^{OH}=p^{K_b}+\log c[/tex]
where c is the concentration of base.
Here c= [tex]5\times10^{-3}\ M[/tex]
[tex]p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05[/tex]
Substituting the above values in the formula,we get:
[tex]p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L[/tex]
Hence:
Kb = [tex]4.45\times10^{-7}\ mol/L[/tex]
[tex]p^{Kb}=6.35[/tex]
