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I have a graph of a titration of a weak polyprotic acid with a strong base.
I graphed $\mathrm {pH}$ versus amount of strong base added.
Now since the acid is polyprotic, how do I determine which equivalence point is the final one?
Also not sure how to find the initial concentration of the weak acid? I know the volume I put in, as well as the volume of strong base at any given time.
A simple way to determine whether an equivalence point in a titration is the last one is to figure out the volume of titrant added for the first titration point, then realize all subsequent equivalence points will be at exact integer multiples of this volume, under the assumption that the analyte contains only one acid (a mixture of acids won't behave the same). When an integer multiple of the first equivalence point volume fails to be an equivalence point itself, then there will be no further equivalence points.
For example, let us imagine a generic polyprotic acid with formula $\ce{H_{k}A}$, being titrated with a strong monoprotic base such as $\ce{NaOH}$. Starting from the fully protonated acid, let us neutralize each ionizable hydrogen in steps. Let the acid $\ce{H_{k}A}$ being titrated have an initial concentration $c_{acid}$ in a solution of volume $V_{acid}$, and let the titrant have a concentration $c_{base}$ and a total volume added $V_{base}$.
By definition, to reach the first equivalence point ($V_{base}=V_{eq\ 1}$), we consider the reaction which represents the neutralization of only the first proton:
$$\ce{H_{k}A + OH- -> H_{k-1}A- + H2O}$$
Each one mole of acid $\ce{H_{k}A}$ must surrender exactly one mole of hydrogen ions $\ce{H+}$, each of which will react with exactly one mole of hydroxide ions $\ce{OH-}$. Thus:
$$n_{\mathrm{released}\ \ce{H+}}=n_\mathrm {acid}=c_\mathrm {acid} \times V_\mathrm {acid}$$ $$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{base}=c_\mathrm {base} \times V_\mathrm {eq\ 1} $$ $$n_{\mathrm{released}\ \ce{H^+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\ c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 1}$$ $$V_\mathrm {eq\ 1} = \frac{c_\mathrm {acid} \times V_\mathrm{acid}}{c_\mathrm {base}}$$
Now, to reach the second equivalence point ($V_{base}=V_{eq\ 2}$), we consider the reaction which represents the neutralization of the first and second protons:
$$\ce{H_{k}A + 2 OH- -> H_{k-2}A^{2-} + 2 H2O}$$
Each one mole of $\ce{H_{k}A}$ must surrender exactly two moles of $\ce{H+}$, each of which will react with exactly one mole of $\ce{OH-}$. The relevant equations are now:
$$n_{\mathrm{released}\ \ce{H+}}=2\times n_\mathrm {acid}=2\times c_\mathrm {acid} \times V_\mathrm {acid}$$ $$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{eq\ 2} $$ $$n_{\mathrm{released}\ \ce{H+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\ 2\times c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 2}$$ $$V_\mathrm {eq\ 2} = 2\times \frac{c_\mathrm {acid} \times V_\mathrm {acid}}{c_\mathrm{base}} = 2\times V_\mathrm {eq\ 1} $$
Notice how the added volume to reach the second equivalence point is exactly twice that added to reach the first equivalence point. This pattern repeats, i.e. the third equivalence point will happen at $V_\mathrm {base}=3\times V_\mathrm {eq\ 1}$, and the $k$-th equivalence point will happen at $V_\mathrm {base}=k\times V_\mathrm {eq\ 1}$. The equivalence points are equally spaced.
In short, if the titration of a pure polyprotic acid shows the first sudden pH swing after adding $\pu{15 mL}$ of the base titrant, then $V_\mathrm {eq\ 1}$ equals 15 mL, and the only possible next equivalence points are at $\pu{30 mL}$, $\pu{45 mL}$, $\pu{60 mL}$, etc. If during the titration another sudden pH swing is observed at $\pu{30 mL}$, but not at $\pu{45 mL}$, then you can deduce that there is no third equivalence point, and hence the acid is diprotic.
