If #pK_a=7.4#, what is #pK_b#?

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Answer 1 · 2016-10-28T07:42:20

#K_axxK_b# #=# #K_w#, where #K_w=10^-14# under standard conditions.

Thus #K_b=K_w/K_a# #=# #10^-14/(4xx10^-8)#.

We can make it a bit more simply if we take logarithms to the base 10 of each side:

#log_10K_b=log_10K_w-log_10K_a#

OR

#-log_10K_w=-log_10K_a-log_10K_a#

But #-log_10K_w# #=# #-log_(10)10^-14# #=# #-(-14)#, and #-log_10K_a=pK_a#, and #-log_10K_p=pK_b#

And thus, finally, #pK_w=14=pK_a+pK_b#

Since #pK_a=-log_10(4.0xx10^-8)=-(-7.40)=7.4#.

#pK_b=14-7.40=6.60#.

And (finally!) #K_b=10^(-6.60)# #=# #??#