# Predicting Ka/Kb values using values for solubility in water

Given that $K_\mathrm{a}$ values are determined by the position of equilibrium for a species such as $\ce{HCl}$ where $K_\mathrm{a} = \ce{\frac{{[H^{+}]}{[Cl^{-}]}}{[HCl]}}$, I was wondering whether it would be possible to work out $K_\mathrm{a}$ values using 'Solubility in Water' values.

For example: $\ce{HCl}$ has a solubility of $\pu{720g/L}$ in water at 20 degrees Celsius. Is this possible or is there no direct correlation between solubility in water and $K_\mathrm{a}$?

• I'm no expert, but I found this link when googling your question. – JavaScriptCoder Mar 19 '18 at 21:55
• This question doesn't ask us to do OP's homework for them, so please don't VTC it even if you don't like it. – M.A.R. ಠ_ಠ Mar 19 '18 at 22:49
• There's no correlation. – Mithoron Mar 20 '18 at 0:34

Your example, $\ce {HCl}$, is considered to be a strong acid. Thus, it it dissociate $100\%$ in water. Thus, the molarity of the given solution (if it exists):
$$\frac{\pu{720 g}}{\pu{1.0 L}}\times\frac{\pu{1.0 mol}}{\pu{36.5 g}} = \pu{19.7 mol L^-1}$$ That means, since $\ce {HCl (aq) -> H+(aq) + Cl-(aq)}$, $[\ce {H+}]$ of the solution is $19.7$, and $[\ce {Cl-}]=19.7$ as well so you can find $\mathrm{pH}$ of the solution. But, since no $\ce {HCl}$ remains in the solution, $K_\mathrm{a}$ has no meaning.
• Hey nice answer! But, isn't $K_\mathrm{a}$ defined for $\ce{HCl}$. I mean, it's of the order $10^6$ (very high), but it is defined nonetheless, isn't it? So, why do you say that "$K_\mathrm{a}$ has no meaning"? – Gaurang Tandon Mar 21 '18 at 1:06
• By definition, $K_\mathrm{a} = \ce{\frac{{[H^{+}]}{[Cl^{-}]}}{[HCl]}}$ for $\ce {HCl}$ as you asked. But there is no $\ce {[HCl]}$ in solution because all dissociate in water. Thus, $K_\mathrm{a}$ for $\ce {HCl}$ has no meaning even though $\ce{[H^{+}]}$ has a finite value. – Mathew Mahindaratne Mar 21 '18 at 5:28
• A simple Google search gives you the approximate "pKa HCl" value, which is given as $-7$, meaning $K_a=~10^7$. That would give you the $\ce {HCl}$ concentration of said solution the value of $3.88\times10^{-5}$. I think I made my point clear now. So long! – Mathew Mahindaratne Mar 21 '18 at 6:47