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If you have a weak base XOH titrated with a strong monotonic acid, where do you start with that? Would you first have the base react with water and break up and determine how much of it does? Or just it breaking up alone, not reacting with water? Reacting just with the acid right away? I'm not sure where to start for this problem type...
I need to find the pH at the equivalence point. I have the concentrations of the acid and base, which are both equal, no volume given.
Equivalence point means equal numbers of moles of acid and base have been added. If $x$ moles of the weak base $\ce{B}$ and the strong acid $\ce{HA}$ have been added, then the $\ce{H^+}$ ions will react away $\ce{OH^-}$ until all the weak base has associated into $\ce{BH^+}$ and $\ce{OH^-}$ (which of course has been reacted with $\ce{H^+}$ forming water). This leaves us with water and $x$ moles of the salt $\ce{BHA}$. This salt is ionic and so soluble in water, so it completely ionises to form $\ce{BH^+}$ and $\ce{A^-}$ ions in solution.
As $\ce{BH^+}$ is the conjugate acid of the weak base we had originally, it can react with water in an equilibrium reaction as follows:
$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$
As $\ce{B}$ is a weak base, the above equilibrium will lie to the right, and hence there will be a considerable concentration of $\ce{H3O^+}$ in the solution at equivalence, which leads to an acidic pH.
Calculating the pH depends on the fact that $K_b K_a = K_w$ for a base and its conjugate acid. This is derived here.
For the equilibrium
$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$
the acid dissociation constant has the form
$K_a = \dfrac{[\ce{B}][\ce{H3O^+}]}{[\ce{BH^+}]}$
As we know $K_b K_a = K_w$, we can rewrite this equilibrium expression as
$[\ce{H3O^+}]=\sqrt{[\ce{BH^+}]\dfrac{K_w}{K_b}}$
To find pH you can $-\log$ this. (Note $[\ce{B}]=[\ce{H3O^+}]$ as one molecule of each is produced for each molecule of $\ce{BH^+}$ that reacts with a water molecule).
So in order to find the pH at the equivalence point, you need to know the $K_b$ of the base you're using. And as everything is in the same volume, all the volumes will cancel from the concentration terms.
I hope this helps, and please correct me if I've made any errors.
