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!) $K_b=10^(-6.60)$ $=$ $??$

If pK_a=7.4pK_a=7.4, what is pK_bpK_b?