Answer:
6.25 X10^{-9} = Ka
[tex]Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}[/tex]
Explanation:
The ionic equation for the hydrolysis of the cation of the given salt will be:
[tex]M^{+} + H_{2}O ---> MOH + H^{+}[/tex]
The expression for Ka will be:
Ka = [tex]\frac{[H^{+}][MOH]}{[M^{+}]}[/tex]
As given that the concentration of the salt is 0.1 M and pH of solution = 4.7, we can determine the concentration of Hydrogen ions from the pH
pH = -log [H⁺]
[H⁺] = antilog(-pH) = antilog (-4.7) = 2 X 10⁻⁵ M = [MOH]
Let us calculate Ka from this,
Ka = [tex]{2.5X10^{-5}X2.5X10^{-5}}{0.1} = 6.25 X10^{-9}[/tex]
The relation between Ka an Kb is
KaXKb =10⁻¹⁴
[tex]Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}[/tex]
