Question

For a given value of ${\left[\mathrm{HL}\right]}_{\mathbf{org}}$in Equation 23-13, over what pH range (how many pH units) will D change from if n = 2?

Short answer

$∆\mathrm{pH}=2$

Step-by-step solution

Given Information:

Distribution change pH value for D =0.01 to 100 as n = 2

To Find the[HL]org :

For distribution of metal-chelate complex between phases is

$\mathrm{D}=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{aq}}^{\mathrm{n}}}$

${\mathrm{ML}}_{\mathrm{n}}$is the form extracted into organic solvent

For distribution coefficients as localid="1654772730303" ${\mathrm{D}}_1=0.01$

$$\begin{gathered} 0.01=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{aq}}^{\mathrm{n}}} \\ \mathrm{As}{\mathrm{D}}_2=100 \\ 100=\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_{\mathrm{aq}}^{\mathrm{n}}} \\ \frac{{\mathrm{D}}_2}{{\mathrm{D}}_1}=\frac{\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_2^{\mathrm{n}}}}{\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_1^{\mathrm{n}}}} \end{gathered}$$

For n = 2 in the equation and to eliminate the values to simplify the equation

$$\begin{gathered} \frac{{\mathrm{D}}_2}{{\mathrm{D}}_1}=\frac{\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_2^2}}{\frac{{\mathrm{K}}_{\mathrm{M}}{\mathrm{\beta K}}_{\mathrm{a}}^{\mathrm{n}}}{{\mathrm{K}}_{\mathrm{L}}^{\mathrm{n}}}\frac{{\left[\mathrm{HL}\right]}_{\mathrm{org}}^{\mathrm{n}}}{{\left[{\mathrm{H}}^{+}\right]}_1^2}} \\ \frac{{\mathrm{D}}_2}{{\mathrm{D}}_1}={\left(\frac{{\left[{\mathrm{H}}^{+}\right]}_1}{{\left[{\mathrm{H}}^{+}\right]}_2}\right)}^2 \\ \frac{100}{0.01}={\left(\frac{{\left[{\mathrm{H}}^{+}\right]}_1}{{\left[{\mathrm{H}}^{+}\right]}_2}\right)}^2 \\ \sqrt{{10}^4}=\left(\frac{{\left[{\mathrm{H}}^{+}\right]}_1}{{\left[{\mathrm{H}}^{+}\right]}_2}\right) \\ {10}^2=\left(\frac{{\left[{\mathrm{H}}^{+}\right]}_1}{{\left[{\mathrm{H}}^{+}\right]}_2}\right) \end{gathered}$$

To calculate △pH

$$\begin{gathered} △\mathrm{pH}=-\log \left(\frac{{\left[{\mathrm{H}}^{+}\right]}_1}{{\left[{\mathrm{H}}^{+}\right]}_2}\right) \\ △\mathrm{pH}=-\log \left({10}^2\right) \\ △\mathrm{pH}=2 \end{gathered}$$