Q11P Consider the extraction of Mnt聽... [FREE SOLUTION]

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Chapter 23: Q11P (page 629)

Consider the extraction of $M^{nt}$ from aqueous solution into organic solution by reaction with protonated ligand HL : $D = M^{n +} (aq) + nHL (org) 鈻� {ML}_{n} (org) + {nH}^{+} (aq) K_{extraction} = \frac{{[{ML}_{n}]}_{org} {[H^{+}]}_{aq}^{n}}{{[M^{n +}]}_{aq} {[HL]}_{org}^{n}}$ .Rewrite Equation 23 - 13 in terms of role="math" localid="1654863844402" $K_{e x t r a c t i o n}$ and express $K_{extraction}$ in terms of the constants in Equation 23 - 13 . Give a physical reason why each constant increases or decreases $K_{extraction}$

Short Answer

Expert verified

In this task let us rewrite the equation 23 - 13 in terms of $K_{extraction}$ and express it in terms of the constants.
Further, we shall give a physical reason why each constant increases/decreases the $K_{extraction}$

Step by step solution

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Let us write the Equation 23 - 13 for distribution of metal-chelate complex

between phases as $D = \frac{K_{M} {尾碍}_{a}^{n}}{K_{L}^{n}} \frac{[{HL}_{org}^{n}}{{[H^{+}]}_{aq}^{n}}$ .

Here, ${ML}_{n}$ is the form extracted into organic solvent.

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Here, let us further consider the following term for distribution coefficient (D) .

$D = \frac{[{ML}_{n}]}{[M^{n +}]}$

Now, we shall take the formula from the task and rearrange it in the way so that we make it equal to the term above

K_{extraction} = \frac{[{ML}_{n}] {[H^{+}]}^{n}}{[M^{n +}] {[HL]}^{n}} \frac{{[HL]}^{n}}{[H^{+}]} 脳 K_{extraction} = \frac{[{ML}_{n}]}{[M^{n +}]}

Then, let us substitute it in order to combine the resulting equation with the

one from 23 - 13

$D = \frac{[{ML}_{n}]}{[M^{n +}]} D = \frac{{[HL]}^{n}}{{[H^{+}]}^{n}} 脳 K_{extraction}$

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Let us combine the resulting equation with Equation :

D = \frac{K_{M} {尾碍}_{a}^{n}}{K_{L}^{n}} \frac{{[HL]}_{org}^{n}}{{[H^{+}]}_{aq}^{n}}

$\frac{{[HL]}^{n}}{{[H^{+}]}^{n}} 脳 K_{extraction} = \frac{K_{M} {尾碍}_{a}^{n}}{K_{L}^{n}} \frac{{[HL]}_{org}^{n}}{{[H^{+}]}_{aq}^{n}}$

Let us eliminate it further to make it as a simplified one. So we have

$K_{extraction} = \frac{K_{M} {尾碍}_{a}^{n}}{K_{L}^{n}}$

Result:

$K_{M}$ increases due to ${ML}_{n}$ being more soluble in organic phase
$尾$ - increases because ${ML}_{n}$ is more soluble in organic form and ligand in this

case binds the metal more tightly

$K_{a}$ increases because ligand loses $H^{+}$ easier and therefore increases the

formation of ${ML}_{n}$

$K_{L}$ decreases as HL is more soluble in organic phase and then $L^{-}$ cannot react

with ${MN}^{+}$ in order to form ${ML}_{n}$ .

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