What is Kb and Ka in chemistry?

1 Answer
Jun 27, 2016

Answer:

#K_b# and #K_a# are the base and acid dissociation constants respectively. They are simply equilibrium constants.

Explanation:

For acid dissociation:

#HA+H_2OrightleftharpoonsH_3O^+ + A^-#

#K_a=([H_3O^+][A^-])/([HA])#

For base hydrolysis:

#A^(-)+H_2OrightleftharpoonsHO^(-) + HA#

#K_b=([HO^-][HA])/([A^-])#.

As with any mathematical expression, they can be manipulated, and as water itself undergoes self protonolysis, we can write:

#K_axxK_b=([H_3O^+]cancel([A^-]))/cancel([HA])xx([HO^-]cancel([HA]))/cancel([A^-])#

#=# #[HO^-][H_3O^+]# #=# #K_w#.

Of course, in water, at #298*K#, #K_w=10^(-14)#.

And if we take logarithms to the base 10, and mulitply each side by #-1#.

#14=pK_a + pK_b#

There is a lot of unsolicited info here so use it for reference. The defining equation for #K_a# is necessary to remember.