# Calculating pH of a saturated weak acid/base solution [closed]

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…

• 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. Oct 22, 2019 at 14:21
• 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?
– jte
Oct 23, 2019 at 15:23
• 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. Oct 23, 2019 at 15:46