EDTA binds metal ions to form complex ions (see Problem), so it is used to determine the concentrations of metal ions in solution:

${\text{M}}^{\text{n +}}{\text{(aq) + EDTA}}^{\text{4 -}}\text{(aq)}\to {\text{MEDTA}}^{\text{n - 4}}\text{(aq)}$

A $\text{50.0 - mL}$sample of ${\text{0.048 M Co}}^{\text{2 +}}$is titrated with ${\text{0.050 M EDTA}}^{\text{4 -}}$.Find ${\text{[Co}}^{\text{2 +}}\text{]}$and ${\text{[EDTA}}^{\text{4 -}}\text{]}$after.

1. $\text{25.0 mL}$and
2. $\text{75.0 mL}$of${\text{EDTA}}^{\text{4 -}}$are added (${\text{logK}}_{\text{f}}$of ${\text{CoEDTA}}^{\text{2 -}}\text{= 16.31}$).
   

Titration (also known as titrimetry and volumetric analysis) is a typical quantitative chemical analysis method used in laboratories to quantify the concentration of an analyte (a substance to be analysed).

The volume of ${\text{Co}}^{\text{2 +}}$is: $\text{50 mL =}\text{0.05 L}$

Number of moles of ${\text{Co}}^{\text{2 +}}$is: $\text{0.048 M}$

Number of moles of ${\text{EDTA}}^{\text{4 -}}$: $\text{0.05 M}$

First, write the reaction between ${\text{Co}}^{\text{2 +}}$and ${\text{EDTA}}^{\text{4 -}}$-

$$\begin{gathered} {\text{Co}}^{\text{2 +}}{\text{+ EDTA}}^{\text{4 -}}\rightleftharpoons {\text{CoEDTA}}^{\text{2 -}} \\ \text{Kf =}\frac{{\displaystyle {\text{[CoEDTA}}^{\text{2 -}}\text{]}}}{{\displaystyle {\text{[Co}}^{\text{2 +}}{\text{][EDTA}}^{\text{4 -}}\text{]}}}{\text{=10}}^{\text{16.31}} \end{gathered}$$

Here calculate ${\text{[Co}}^{\text{2 +}}]$after adding $\text{25 mL EDTA}$

$\begin{array}{rcl}\left[{\text{Co}}^{\text{2 +}}\right]& =& \frac{{\displaystyle \text{0.05L×0.048 M - 0.025 L×0.05 M}}}{{\displaystyle \text{0.05 L + 0.025 L}}}\\ & =& \text{0.0153 M}\end{array}$

Do the same calculation for ${\text{[CoEDTA}}^{\text{2 -}}\text{]}$-

$\begin{array}{rcl}\left[{\text{CoEDTA}}^{\text{2 -}}\right]& =& \frac{{\displaystyle \text{0.025 L×0.05 M}}}{{\displaystyle \text{0.05 L + 0.025 L}}}\\ & =& \text{0.0167 M}\end{array}$

Now determine ${\text{[EDTA}}^{\text{4 -}}\text{]}$from the $\text{Kf}$expression

$\begin{array}{rcl}\left[{\text{EDTA}}^{\text{4 -}}\right]& =& \frac{{\displaystyle \text{0.0167 M}}}{{\displaystyle {\text{0.0153 M×10}}^{\text{16.31}}}}\\ & =& {\text{5.32×10}}^{\text{- 17}}\text{M}\end{array}$

Therefore, the value for concentration is obtained as$\left[{\text{EDTA}}^{\text{4 -}}\right]\text{=}{\text{5.32×10}}^{\text{- 17}}\text{M}$and$\left[{\text{Co}}^{\text{2 +}}\right]\text{= 0.0153 M}$.

(b)

Because the volume is post equivalence point, the calculations will be –

$\begin{array}{rcl}\left[{\text{CoEDTA}}^{\text{2 -}}\right]& =& \frac{{\displaystyle \text{0.05 L×0.048 M}}}{{\displaystyle \text{0.025 L}}}\\ & =& \text{0.0192 M}\\ \left[{\text{EDTA}}^{\text{4 -}}\right]& =& \frac{{\displaystyle \text{0.075 L×0.05 M - 0.05 L×0.048 M}}}{{\displaystyle \text{0.0125 L}}}\\ & =& \text{0.0108 M}\end{array}$

Now determine $\left[{\text{Co}}^{\text{2 +}}\right]$from the $\text{Kf}$expression-

$\begin{array}{rcl}\left[{\text{Co}}^{\text{2 +}}\right]& =& \frac{{\displaystyle \text{0.0192 M}}}{{\displaystyle {\text{0.0108 M + 10}}^{\text{16.31}}}}\\ & =& {\text{8.71×10}}^{\text{- 17}}\text{M}\end{array}$

Therefore, the value for concentration is obtained as $\left[{\text{EDTA}}^{\text{4 -}}\right]\text{= 0.0108 M}$ and$\left[{\text{Co}}^{\text{2 +}}\right]\text{=}{\text{8.71×10}}^{\text{- 17}}\text{M}$.