At 298k the base dissociating constant Kb of CH3NH2 is 4.5×10mol/dm
Write an expression for the kb of CH3NH2
I think you omitted the exponent for the Kb but that isn't needed in order to write the Kb expression.
CH3NH2 + HOH ==> CH3NH3^+ + OH^-
Kb = (CH3NH3^+)(OH^-)/(CH3NH2)
The expression for the dissociation constant Kb of CH3NH2 can be written as:
Kb = [CH3NH2+][OH-] / [CH3NH2]
Where:
- Kb is the dissociation constant of CH3NH2.
- [CH3NH2+] represents the concentration of the CH3NH2 cation.
- [OH-] represents the concentration of the hydroxide ion.
- [CH3NH2] represents the concentration of the CH3NH2 base.
To write the expression for Kb of CH3NH2, we need to understand the dissociation reaction that occurs at 298K.
The dissociation reaction of CH3NH2 (methylamine) in water can be represented as follows:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
In this reaction, CH3NH2 acts as a base and accepts a proton (H+) from water, forming CH3NH3+ (methylammonium ion) and OH- (hydroxide ion).
The Kb expression for CH3NH2 can be written using the concentrations of the products and reactants:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Here, [CH3NH3+] represents the concentration of methylammonium ion, [OH-] represents the concentration of hydroxide ion, and [CH3NH2] represents the concentration of methylamine.
To calculate the exact value of Kb for CH3NH2 at 298K, you need to know the concentrations of CH3NH3+ and OH-, which can be determined experimentally.