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.

  • 1
    $\begingroup$ You asked where to start, but you didn't ask what you're looking for. Do you want to calculate pHs for every addition? The base concentration? Or simply describe the experiment? $\endgroup$
    – Molx
    Mar 27, 2015 at 2:53
  • $\begingroup$ Sorry, need to find pH. I have the concentrations of the acid and base, which are both equal, no volume given. $\endgroup$ Mar 27, 2015 at 2:55
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    $\begingroup$ If you have no volumes given, did they give you a ratio or do you have to draw a graph? If you do not have the ratio, it is impossible to determine the pH. $\endgroup$ Mar 27, 2015 at 3:38
  • $\begingroup$ I just have that they're both 0.10M solutions of each. No graph need, just need the pH at the equivalence point. $\endgroup$ Mar 27, 2015 at 3:41

1 Answer 1


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


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.

  • $\begingroup$ So for the "then the H+ ions will react away OH− until all the weak base has associated into BH+ and " part, if I have the base NH2OH, does that mean it first does this: NH2OH + H -> NH3 + OH ? $\endgroup$ Mar 28, 2015 at 7:32
  • $\begingroup$ And then NH3 + H2O -> NH2 + H3O? And then that's it? $\endgroup$ Mar 28, 2015 at 7:36
  • $\begingroup$ So we don't worry about the NH2OH reacting with water first because it's so weak and it won't do much, but in the second equation, NH3 would react with water significantly because it's strong? $\endgroup$ Mar 28, 2015 at 7:44
  • $\begingroup$ So if in a strong base weak acid problem you need to use Kb instead of Ka when at equilibrium, does that mean in this case you'd need to use Ka at the end instead of Kb? No wait, yea, we would need Ka, cause there H3O in the final equation. $\endgroup$ Mar 28, 2015 at 7:47
  • $\begingroup$ I'm not familiar with hydroxylamine so I couldn't say how it reacts specifically. The answer I wrote basically deals with a salt solution of a weak base, which would be produced at the equivalence point. Re $K_b$ or $K_a$ you have to remember that above I was considering $\ce{BH^+}$, which is the conjugate acid of $\ce{B}$, and so an acid dissociation constant was appropriate. In an example with a strong base and a weak acid, you would do everything exactly the same, except you would consider the association of the conjugate base. $\endgroup$
    – Ivan
    Mar 28, 2015 at 17:02

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