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A weak acid in water dissociates reversibly as follows:

$$\ce{XH <=> X- + H+}$$

A strong base is one which dissociates completely.

Now, let's assume we're doing acid-base titrations, adding a weak acid to a strong base, trying to arrive at the titration curve.

My question is mainly why the $\mathrm{pH}$ at the equivalence point isn't $7$. Intuitively, it makes sense, one dissociates fully while the other doesn't. But if you dig into it a bit deeper you realise that that answer is incomplete.

Bear with me. You add the weak acid to the strong base until you reach the equivalence point. The strong base dissociates completely while the acid establishes an equilibrium as described above. However, thinking about it you realise that at no point should this be possible. An $\ce{OH-}$ ion from the base quickly reacts with any H+ ions produced meaning that $[\ce{H+}]$ is always near zero, meaning that more and more of the weak acid needs to dissociate to establish equilibrium. However, the $\ce{H+}$ concentration will always be 0 as any hydrogen ions are snapped up meaning that the acid will completely dissociate before it has a chance to establish equilibrium, at which point all of the $\ce{OH-}$ ions will also have been neutralised. The $\mathrm{pH}$ of this resulting solution should in other words at that point be $7$, because the acid, even though it's weak, acts as a strong one in an alkaline environment.

This however I know isn't correct. What's the issue with my logic above?


marked as duplicate by M.A.R. ಠ_ಠ, Mithoron, Tyberius, Todd Minehardt, airhuff Feb 22 '18 at 0:23

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While titrating weak acid with strong base, the titration does occur like a normal acid, because the base added quickly takes the dissociated $\ce{H^+}$ ion of the weak acid and thus the equilibrium of the acid is distorted. To maintain constant $\ce{K_c}$, the weak acid also produces $\ce{H^+}$ instantaneously, and thus its behaviour almost becomes like a strong acid if base is added fast(which is normal in titration) to maintain equilibium.But the $\ce{pH}$ at end point is not $7$ not due to the acid but the salt formed as a product of the titration.
Suppose the weak acid($\ce{HA}$) is being titrated with a strong base(say for example,$\ce{NaOH}$), and both are taken in same equivalents. Then the reaction does go until all the acid (though weak) is exhausted.$$\ce{HA + NaOH -> NaA + H_2O}$$At t= o, say both $\ce{HA }$ and $\ce{NaOH }$ has $c$ equivalents each. So, at the end of the reaction, both $\ce{HA }$ and $\ce{NaOH}$ will be totally exhausted and $\ce{NaA }$ will be $c$ equivalents.
Now, in the medium only $\ce{NaA }$ is present which will control the $\ce{pH}.$ But $\ce{NaA}$ is actually "SALT OF A WEAK ACID AND STRONG BASE". So it hydrolyses as,$$\ce{NaA + H_2O <=> HA + Na^+ + OH^-}$$So you can see as $\ce{HA }$ is a weak acid ,it's conjugate base (stronger) $\ce{A^- }$ takes the proton of water and leaves only $\ce{OH^- }$, which in turn is the reason for incresing the alkalinity of the medium.That's why generally, $\ce{pH }$ is above 7 at the end point.
By doing mathematical calculation, it can be easily shown that if the weak acid's $\ce{pK_a}$ is given and the final concentration of the salt is say, $\ce{ c' M}$. Then the $\ce{pH }$ of the medium can be calculated as ,$$\ce{pH = 1/2 pK_w + 1/2 pK_a + 1/2logc'}$$. Assuming the titration is being carried at $25$ degrees, then$$\ce{pH= 7 + 1/2pK_a +1/2logc' > 7 (generally) }$$ if your acid is tending to strong i.e. $\ce{pK_a}$ is very less positive, and $\ce{c'}$ is close to $1 $, the $\ce{pH}$ can be close to $7$, but $\ce{pK_a}$'s are normally of the order of $3$ or $4$. So, generally $\ce{pH}$ exceeds $7$.


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