Question

The PH of a 0.1 M MCl (M is an unknown cation) was found to be 4.7. Write the net ionic equation for the hydrolysis of M and its corresponding equilibrium expression Kb. Calculate the value of Kb

Answer 1

Answer:

6.25 X10^{-9} = Ka

$Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}$

Explanation:

The ionic equation for the hydrolysis of the cation of the given salt will be:

$M^{+} + H_{2}O ---> MOH + H^{+}$

The expression for Ka will be:

Ka = $\frac{[H^{+}][MOH]}{[M^{+}]}$

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 = ${2.5X10^{-5}X2.5X10^{-5}}{0.1} = 6.25 X10^{-9}$

The relation between Ka an Kb is

KaXKb =10⁻¹⁴

$Kb= \frac{10^{-14}}{6.25 X10^{-9}}= 1.6X10^{-6}$