Question

Find the dissociation constant (k_a) of a 0.04 M weak acid HX whose pH is 4.8

Answer 1

given :

[HX]= 0.04 M

pH = 4.8

Calculation of [H3O+] from pH

[H3O+] = antilog (-pH)

= Antilog (-4.8)

= 1.585E-5 M

[H3O+] = 1.585 E-5 M

Dissociation reaction of HX

            HX(aq) + H2O (l) --- > X- (aq) +    H3O+ (aq)

I       0.04                                             0                      0

C         -x                                             +x                    +x

E     (0.04-x)                                         x                      x

Lets find equilibrium concentrations

[HA]= 0.04-x = 0.04 – 1.585 E-5 = 3.998 E-2 M

[H3O+] = 1.585 E-5

[X-]= 1.585 E-5

Ka = [X-][H3O+] / [HX]

Ka = (1.585E-5)² / [3.998E-2]

= 6.28 E-9

Ka (dissociation constant ) = 6.28 E-9