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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.

Respuesta :

Answer:

The [tex]pK_a[/tex] of acetic acid is 4.7.

The [tex]pK_b[/tex] of methylamine is 3.4.

Explanation:

1) The dissociation constant of acetic acid = [tex]K_a=1.8\times 10^{-5}[/tex]

The [tex]pK_a[/tex] is negative logarithm of dissociation constant.

[tex]pK_a=-\log[K_a][/tex]

[tex]pK_a=-\log[1.8\times 10^{-5}]=4.7[/tex]

The [tex]pK_a[/tex] of acetic acid is 4.7.

2) The dissociation constant of methylamine  = [tex]K_b=4.2\times 10^{-4}[/tex]

The [tex]pK_b[/tex] is negative logarithm of dissociation constant.

[tex]pK_b=-\log[K_b][/tex]

[tex]pK_b=-\log[4.2\times 10^{-4}]=3.4[/tex]

The [tex]pK_b[/tex] of methylamine is 3.4.