What is the weak base ionization constant (K_b) for HS^- equal to?

Jul 12, 2017

Well $p {K}_{b} = {10}^{14 - p {K}_{a}}$............; and thus ${K}_{b} = 1.25 \times {10}^{-} 7$............

Explanation:

Hydrogen sulfide dissociates according to the reaction......

${H}_{2} S \left(a q\right) + {H}_{2} O r i g h t \le f t h a r p \infty n s H {S}^{-} + {H}_{3} {O}^{+}$

Now ${K}_{a} \left({H}_{2} S\right) = {10}^{- 6.9} = 1.25 \times {10}^{-} 7$, and of course $p {K}_{a} = 6.9$.

In aqueous solution, we know that $p {K}_{a} + p {K}_{b} = 14$

And thus $p {K}_{b} = 14 - p {K}_{a} = 14 - 6.9 = 7.1$

${K}_{b} = {10}^{- 7.1}$...........=??

Note that the question SHOULD have quoted at least $p {K}_{a 1}$ for ${H}_{2} S$. ${K}_{a}$ and ${K}_{b}$ are established by measurement; you could not possibly be expected to remember them.