Titrations

Titration is the procedure of identifying the quantity (moles or concentration) of a compound (which is called the analyte) using a well known reaction. In order to titrate a substance, there must be a reaction it participates in, and the end point of that reaction must be identifiable. Also, the quantity of the other reactants (the titrants) must be known. Acid-base reactions and redox reactions are most often used for titration.

Acid-base titrations are commonly used because they are simple and their end points can be easily detected using an indicator or a \(\text{pH}\) meter. For example, suppose we have a \(50 \text{ mL}\) hydrochloric acid solution whose concentration we do not know of. We want to know its molar concentration (let it be \(x\)). So, we prepare a \(0.1\text{ M}\) sodium hydroxide solution, and slowly mix it into the acid solution: \[\text{HCl}+\text{NaOH}\rightarrow\text{NaCl}+\text{H}_2\text{O}.\] The acid-base reaction equation above implies that the equivalence point is when the number of moles of \(\text{NaOH}\) added equals the number of moles of \(\text{HCl}.\) In this case the end point (or equivalence point) would be the point at which \(\text{pH}=7,\) so we slowly add \(\text{NaOH}\) until the \(\text{pH}\) meter reads \(7.\) If the \(\text{pH}\) is \(7\) when \(50 \text{ mL}\) of \(\text{NaOH}\) has been added, then we have \[\begin{align} (\text{# of moles of NaOH added})&=50\text{ mL}\times0.1\text{ M}=5\text{ mmol} &\qquad(1)\\ (\text{# of moles of HCl})&=50\text{ mL}\times x\text{ M}=50x\text{ mmol}, &\qquad(2) \end{align}\] where \((1)\) must equal \((2).\) Thus now we know that the initial molar concentration \(x\) of the hydrochloric acid solution was \(x=0.1\text{ M}.\)

\(\hspace{5cm}\) Imgur

Titration can also be done by using a well known redox reaction. In some cases the equivalence point can be detected using the bare eye for reactions involving color changes. For reactions that do not, a redox indicator or potentiometer may be used.

The oxidation of \(\text{Fe}^{2+}\) using potassium permanganate is a well known redox reaction. Below is the reaction equation: \[5\text{Fe}^{2+}(aq)+\text{MnO}_4^{-}(aq)+8\text{H}^+(aq)\rightarrow5\text{Fe}^{3+}(aq)+\text{Mn}^{2+}(aq)+4\text{H}_2\text{O}(l).\] Note that \(5\) moles of \(\text{Fe}^{2+}\) is oxidized to become \(\text{Fe}^{3+}\) while one mole of permanganate is reduced to form \(\text{Mn}^{2+}.\) Thus the ferrous ion and permanganate ion react with a \(5:1\) ratio. The use of permanaganate is very handy, due to the fact that its solution is bright purple, making it a good indicator. Suppose we have \(10 \text{ mL}\) of \(0.1\text{ M}\) potassium permanganate solution, and we want to know the concentration \(x\) of \(\text{Fe}^{2+}\) in another solution. We slowly add the ferrous solution (which is colorless) to the permanganate solution, until the purple color disappears. If the color disappeared when \(5 \text{ mL}\) of ferrous solution was added, then we have \[\begin{align} (\text{# of moles of Fe}^{2+}\text{ added})&=5\text{ mL}\times x\text{ M}=5x\text{ mmol} &\qquad(1)\\ (\text{# of moles of permanganate})&=10\text{ mL}\times0.1\text{ M}=1\text{ mmol}. &\qquad(2) \end{align}\] Since the ratio \((1):(2)\) must equal \(5:1,\) it must be true that \(x=0.1\text{ M}.\)