Given a mixture of $\ce{HCl}$ and $\ce{MCl3}$ and the following dissociation constants for $\ce{M(OH)3}$, how can the concentrations of $\ce{HCl}$ and $\ce{MCl3}$ be determined separately by titrating this solution with a standard strong base (say, $0.1\:\mathrm{M}$ $\ce{NaOH}$) ? I want to be able to sketch the approximate titration curve for this titration too.

\begin{align} \ce{ M(OH)3 &~<=> M(OH)^{+}_{2} + OH^{-} } & \mathrm{p}K_{\mathrm{b}1} &= 0.5 \\ \ce{ M(OH)^{+}_{2} &~<=> M(OH)^{2+} + OH^{-} } & \mathrm{p}K_{\mathrm{b}2} &= 0.7 \\ \ce{ M(OH)^{2+} &~<=> M^{3+} + OH^{-} } & \mathrm{p}K_{\mathrm{b}3} &= 9.0 \end{align}

All of the species in above equilibria are water soluble.

Progress so far:

  1. $\ce{HCl}$ in the medium will first react with the base and give rise to an end point.
  2. As $\ce{MCl3}$ is a salt of weak base and a strong acid (i.e. $\ce{M(OH)3}$ and $ \ce{HCl} $), a solution of $\ce{MCl3}$ is acidic due to the following equilibria (hydrolysis reactions).

\begin{align} \ce{ M^{3+} + {H2O} &~<=> M(OH)^{2+} + H+ } & \mathrm{p}K_{\mathrm{a}1} &= 14.0 - 9.0 = 5.0 \tag{a} \\ \ce{ M(OH)^{2+} + {H2O} &~<=> M(OH)^{+}_{2} + H+ } & \mathrm{p}K_{\mathrm{a}2} &= 14.0 - 0.7 = 13.3 \tag{b}\\ \ce{ M(OH)^{+}_{2} + {H2O} &~<=> M(OH)_3 + H+ } & \mathrm{p}K_{\mathrm{a}3} &= 14.0 - 0.5 = 13.5 \tag{c} \end{align}

  1. Therefore, if further $\ce{NaOH}$ is added, $\ce{H+}$ generated by above equilibria will react with that $\ce{NaOH}$.
  2. Looking at the $\mathrm{p}K_\mathrm{a}$ values of $\mathrm{(b)}$ and $\mathrm{(c)}$ it can be assumed that reaction $\mathrm{(a)}$ is the dominant of the three. And it will give rise to a separate end point.

Now the problem is how to put all these information together and deduce a method to analyze the mixture.

Any help on how to sketch the titration curve, choose indicators and practically carry out the analysis would be greatly appreciated.

  • $\begingroup$ You mention both the titration curve and indicators. Are you planning to monitor the reaction with a pH meter? If so, an indicator might not be necessary. $\endgroup$
    – Molx
    Commented Jun 12, 2015 at 20:25
  • $\begingroup$ @Molx No. No pH meters. Just want to sketch the curve before carrying out the titration. $\endgroup$
    – user16096
    Commented Jun 12, 2015 at 23:56
  • $\begingroup$ What is the approximate concentration of the metal oxide? and of the hydrochloric acid? $\endgroup$
    – LDC3
    Commented Jun 13, 2015 at 17:33
  • $\begingroup$ @LDC3 No other information is given. $\endgroup$
    – user16096
    Commented Jun 14, 2015 at 5:32
  • $\begingroup$ When you use the align environment, you can use & as an anchor: use odd appearances to align at that point, use even appearances to separate equations in the same line. The \tag{...} will produce whatever you give as argument in text mode just left of the right margin. $\endgroup$ Commented Jun 15, 2015 at 5:53

1 Answer 1



The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the following in detail:

  • Bliefert, Claus; Linek, Alfred; Morawietz, Gerd (1978): PH-Wert-Berechnungen. Weinheim: Verlag Chemie.

There might also be some imprecise terms because I don't know many of the english counterparts to the german terms I am familiar with.


Let's think a bit about your problem first to simplify it from scratch:

  1. As you mentioned, reaction (a) will take place right after all protons from hydrochloric acid have reacted and before reaction (b) and (c) would start.
  2. The pK$_a$ of (b) and (c) are so close to another that they are unable to be separeted by titration and
  3. the pK$_a$ of (b) and (c) are so high that they are unable to be determined by titration in diluted aqueous solutions.

If we take a look at the titration curve of phosphoric acid, we can see what is meant by "too high to be determined". You can clearly identify the two inflection points for the first two protons but there is no third inflection point that could be determined. What is likewise to be seen, is that one can neglect this third deprotonation step in titration curves of diluted solutions (yellow vs. blue titration curve).

titration curve of phosphoric acid with a sodium hydroxide solution

Right now we can reduce your problem to a mixture of hydrochloric acid and M$^{3+}$. To continue we need to establish a proton condition (Protonen-Bedingung):

$$c(H_3O^+) = c(OH^-) + c(A^-) + c(X^-) - c(HB^+) \tag{1}$$

It means that protons come from the autoprotolysis of water (c(OH$^-$)), from hydrochloric acid (c(A$^-$)), from your metal ion (c(X$^-$)) and that the protons react with the used base (c(HB$^+$)).

To define an equation for the acid residue anions a small equation system (where small c's denote actual concentrations in solution and the big C denotes the origin concentration) $$\begin{align}c(HA) \cdot k_a - c(H_3O^+) \cdot c(A^-) =&~0\\c(HA)+c(A^-) =&~C(HA)\end{align} \tag{2.1}$$ needs to be solved to give $$c(A^-) = \frac{C(HA) \cdot k_a}{c(H_3O^+)+k_a}. \tag{2.2}$$

This can be done analogously for the other acid and also for the base.

What remains is the concentration of the hydroxide ions from the autoprotolysis of water that is simply determined by rearranging the equation for the autoprotolysis itself:

$$c(H_3O^+) \cdot c(OH^-) = k_W \Leftrightarrow c(OH^-) = \frac{k_W}{c(H_3O^+)} \tag{3}$$

Combining everything so far we get the following proton condition: $$x = \frac{k_W}{x} + \frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}} - \frac{x \cdot C(B) \cdot k_b}{x \cdot k_b + k_W} \tag{4}$$

where $c(H_3O^+)$ has been replaced by $x$ for the sake of simplicity.

The plot of titration curves usually shows the pH over the base volume (see eq. 8) or the degree of neutralisation ($\tau$ or $\gamma$) that is defined as $$\tau = \frac{c(B)}{C(HA)} \tag{5.1}$$

In the case of your mixture of two acids - and as always neglecting the increase in volume - it can be written as $$\tau = \frac{c(B)}{C(HA)+C(HX)} \tag{5.2}$$

To sketch the titration curve let's say that we know the two concentrations of the acids. To calculate (5.2) we also need to know c(B) that can easily be obtained by rearranging (4) to get: $$c(B) = \left(1+\frac{k_w}{x \cdot k_b}\right)\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right) \tag{6}$$

If one decides to include the increasing volume the concentrations of the acids and the base have to be extended to give $$\begin{align}C(HA) \rightarrow&~\frac{C(HA) \cdot V(HA)}{V(HA)+V(B)}\\C(B) \rightarrow&~\frac{C(B) \cdot V(B)}{V(HA)+V(B)}\end{align} \tag{7}$$

Inserting (7) in (4) and rearranging for V(B) we finally get a neat equation

$$V(B)=\frac{V(HA)~\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)} \tag{8}$$

As the pH is defined as $pH = -log_{10}(c(H_3O^+))$ and $c(H_3O^+)=x$ you can replace every $x$ in equation 8 with $10^{-x}$.

Now you are able to calculate with e.g. Excel enough (V(B),pH)-pairs to draw a titration curve and get some information about your system of two acids during the titration with your desired base.

The following figure shows some examples of your mixture using equation 8.

titration of a mixture of two acids with a sodium hydroxide solution

Please let me know if someone is interested in the full equation including also the second and third deprotonation step with pK$_a$ values 13.3 and 13.5 and I will add it here.

  • 1
    $\begingroup$ (+1) This is great! Glad I finally found it and it confirms the equation I have been using (your equation 8) for a long time. $\endgroup$
    – Ed V
    Commented Nov 16, 2021 at 22:29

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