Question

Find [Ca²⁺] in 0.10 M CaY^2- at pH 8.00

Short answer

The concentration of [Ca²⁺] is 2.3×10^-5Min 0.10 M CaY^2- at pH 8.00.

Step-by-step solution

Introduction

The fraction of all free EDTA (in Y^4- format) is expressed in terms of ${\alpha }_{Y^{4-}}$. It can be expressed as

${\alpha }_{Y^{4-}}=\frac{\left[Y^{4-}\right]}{\left[H_6{Y}^{2+}\right]+\left[H_5{Y}^{+}\right]+\left[H_{4}Y\right]+\left[H_3{Y}^{-}\right]+\left[H_2{Y}^{2-}\right]+\left[H{Y}^{3-}\right]+\left[Y^{4-}\right]}$

From table 12-1 it was found that for, ${\alpha }_{Y^{4-}}=4.2\times {10}^{-3}$

From table 12-2 it was found that for CaY^2-,$K_f={10}^{10.65}$

Determine the conditional formation constant

$\begin{array}{l}{K}_f^{\text{'}}={\alpha }_{Y^{4-}}{K}_f\\ K_f^{\text{'}}=4.2\times {10}^{-3}\times {10}^{10.65}\\ =4.2\times {10}^{7.65}\end{array}$

Determine concentration of [Ca2+]

$$\begin{gathered} \frac{C{a}^{2+}+EDTA\leftrightharpoons Ca{Y}^{2-}}{InitialConcentration000.1} \\ FinalConcentrationxx0.1-x \end{gathered}$$

From the above ICE table we can write

$$\begin{gathered} \frac{\left[Ca{Y}^{2-}\right]}{\left[C{a}^{2+}\right]\left[EDTA\right]}=K_f^{\text{'}} \\ \frac{0.1-x}{x^2}=4.2\times {10}^{7.65} \\ x=2.3\times {10}^{-5} \end{gathered}$$

Therefore, the concentration of [Ca²⁺] is 2.3×10^-5M