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How can you calculate the $\mathrm{pH}$ if you know for example the $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{b}$ value of a weak acid/base and how much acid/base can dilute in water. Let's take an example:

Quinoline has $\mathrm{p}K_\mathrm{a} = 4.5$ and $\pu{0.6 g}/\pu{100 ml}$ can dilute in water. I can easily calculate that the concentration of the saturated solution is $\pu{0.0465 M}$. So if the reaction is

$$ \ce{C9H7N + H2O -> C9H7NH+ + OH-}, $$

why $[\ce{OH-}]\neq \pu{0.0465 M}$? I know that I can get the correct answer by making an ICE-table and use that given $\mathrm{p}K_\mathrm{a}$ value but why is it necessary? It seems weird as after making the ICE-table it seems that all quinoline isn't diluted…

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    $\begingroup$ Quinoline with pKb=14-pKa= 9.5 is a very weak base. Why there should be almost 0.05 M OH- like if it had been strong base ? pH =~ 14 - 0.5(pKb - log c)=9.25 + 0.5.log c. $\endgroup$ – Poutnik Oct 22 '19 at 14:21
  • $\begingroup$ Yeah I know that it's obviously wrong but if Quinoline dilutes to water, shouldn't there be OH-? Or if not, what are there if Quinoline is diluted? $\endgroup$ – jte Oct 23 '19 at 15:23
  • $\begingroup$ There is OH-. It is always there, if there is water. But there are 2 chemical equilibriums. Dissociation of water and protonization of quinoline. Together with mass and charge balance, there is the set of equations for the set of variables. $\endgroup$ – Poutnik Oct 23 '19 at 15:46