Why does the pH rapidly increase near the equivalence point during--why does the slope of the graph of a pH curve sharply increase around this area?


1 Answer 1


I'll try to answer your question firstly qualitatively, then quantitatively (using mathematical equation):

Let's consider the case of the titration of strong acid (volume $V_a$ and concentration$C_a$: unknown) with a strong base (volume $V_b$ and concentration$C_b$), and we'll denote $V_b(eq)$ the volume of base at the equivalence:

  1. The pH increases slowly at first because the pH scale is logarithmic, which means that a pH of 1 will have 10 times the hydronium ion concentration than a pH of 2. Thus, as the hydronium ion is initally removed, it takes a lot of base to change its concentration by a factor of 10, but as more and more hydronium ion is removed, less base is required to change its concentration by a factor of 10. Near the equivalence point, a change of a factor of 10 occurs very quickly, which is why the graph is extremely steep at this point. As the hydronium ion concentration becomes very low, it will again take a lot of base to increase the hydroxide ion concentration by 10 fold to change the pH significantly.
  2. Let's find the equation of the titration curve in the the region $0<V_b<V_b(eq)$:

The number of moles of $\ce{H3O+}$ before the titration is $C_aV_a$

The number of moles of $\ce{H3O+}$ after adding the volume $V_b$ of the base is $C_aV_a-C_bV_b$

Then, the concentration of $\ce{H3O+}$ is $$[\ce{H3O+}]=\frac{C_aV_a-C_bV_b}{V_a+V_b}$$ So, the titration curve in the region $0<V_b<V_b(eq)$ $$p\mathrm{H}=-\log\frac{C_aV_a-C_bV_b}{V_a+V_b}$$ If you trace this mathematical function $p\mathrm{H}=f(V_b)$, you will have a smooth slope at the beginning and as $V_b \rightarrow V_b(eq)$, the slope of the graph sharply increases as the function becomes not defined: $C_aV_a=C_bV_b(eq)$.

  • 2
    $\begingroup$ A while back I made a more formal derivation in a similar question, at the expense of making it significantly harder to read. Perhaps the asker can also browse the answers there. $\endgroup$ Commented Feb 12, 2015 at 17:46
  • $\begingroup$ You're right, the answer given there is really nice. Thank you $\endgroup$ Commented Feb 12, 2015 at 18:32

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