EDTA titrations involve the use of EDTA (ethylenediaminetetraacetic acid) as a titrant to determine the concentration of metal ions in a solution. In these titrations, the metal ion, represented as \( \text{M}^{n+} \), reacts with EDTA, which predominantly exists in its basic form with a charge of -4. The resulting complex can be expressed as \( \text{MY}^{n-4} \). For example, when barium ions (\( \text{Ba}^{2+} \)) react with EDTA, the complex formed is \( \text{BaY}^{2-} \), where the net charge is -2.
These titrations are conducted at a specific pH, which influences the fraction of EDTA in its basic form. The conditional formation constant, denoted as \( K'_{f} \), is crucial in this process and is calculated using the formula:
\[ K'_{f} = \alpha \cdot K_{f} \]
Here, \( \alpha \) represents the fraction of EDTA in its basic form, and \( K_{f} \) is the stability constant of the metal-EDTA complex. To determine \( \alpha \), one can refer to a pH chart or use a formula for non-integer pH values. As pH increases, the proportion of EDTA in its basic form also increases.
The formation constant can be derived from a log chart, where the inverse log gives the value of \( K_{f} \). The equilibrium expression for the conditional formation constant is:
\[ K'_{f} = \frac{[\text{MY}^{n-4}]}{[\text{M}^{n+}][\text{EDTA}]} \]
During the titration, as EDTA is added, the concentration of the metal ion gradually decreases, while the concentration of the metal-EDTA complex increases. The equivalence point is reached when the moles of metal ion equal the moles of EDTA, not necessarily their concentrations. Before reaching this point, there is an excess of metal ions, while after the equivalence point, there is an excess of EDTA.
To find the equivalence volume of the titrant, the relationship between the molarity and volume of the metal ion and EDTA is used:
\[ M_{\text{M}} \cdot V_{\text{M}} = M_{\text{EDTA}} \cdot V_{\text{eq}} \]
For instance, if you have 50 mL of 0.100 M barium ion and are titrating with 0.050 M EDTA, the calculation for the equivalence volume would be:
\[ 0.100 \, \text{M} \cdot 50 \, \text{mL} = 0.050 \, \text{M} \cdot V_{\text{eq}} \]
Solving for \( V_{\text{eq}} \) gives 100 mL, indicating that before reaching this volume, calculations will pertain to the conditions prior to the equivalence point.