Question

calculate the pKa of acetic acid and the pKb of methylamine (CH3NH2). Acetic acid’s Ka is 1.8 x 10-5. Methylamine’s Ka is 4.2 x 10-4.

Answer 1

The
`pK_a` of acetic acid is 4.7.

The
`pK_b` of methylamine is 3.4.

Step-by-step explanation:

1) The dissociation constant of acetic acid =
`K_a=1.8* 10^(-5)`

The
`pK_a` is negative logarithm of dissociation constant.

`pK_a=-\log[K_a]`

`pK_a=-\log[1.8* 10^(-5)]=4.7`

The
`pK_a` of acetic acid is 4.7.

2) The dissociation constant of methylamine =
`K_b=4.2* 10^(-4)`

The
`pK_b` is negative logarithm of dissociation constant.

`pK_b=-\log[K_b]`

`pK_b=-\log[4.2* 10^(-4)]=3.4`

The
`pK_b` of methylamine is 3.4.