Question

n the laboratory, a general chemistry student measured the pH of a 0.342 M aqueous solution of ethylamine, C2H5NH2 to be 12.067. Use the information she obtained to determine the Kb for this base.

Answer 1

The value of
`K_b` of the an ethylamine is
`4.121* 10^(-4)`.

Step-by-step explanation:

The pH of the solution = 12.067

The pOH of the solution = 14 - pH =14-12.607 =1.933

`pOH=-\log[OH^-]`

`1.933=-\log[OH^-]`

`[OH^-]=0.0117 M`

`C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^(+)+OH^-`

Initially

0.342 M 0 0

At equilibrium

(0.342-x) x x

The value of x =
`[OH^-]=0.0117 M`

The expression of
`K_b`is given as:

`K_b=([C_2H_5NH_3^(+)][OH^-])/([C_2H_5NH_2])`

`K_b=(x^2)/((0.342-x))`

`K_b=((0.0117 )^2)/((0.342-0.0117))=4.121* 10^(-4)`

The value of
`K_b` of the an ethylamine is
`4.121* 10^(-4)`.