Question

$EDTA,$ often abbreviated as $H_{4}Y$, forms very stable complexes with almost all metal ions. Calculate the fraction of EDTA in the fully protonated form, $H_{4}Y$ in a solution obtained by dissolving $0.1$ mol ${Na}_{4}Y$ in $1$ litre.

Given, the acid dissociation constants of $H_{4}Y$ are as follows:
$k_1=1.02\times {10}^{-2}$, $k_2=2.13\times {10}^{-3}$, $k_3=6.92\times {10}^{-7}$; $k_4=5.50\times {10}^{-11}$

• A. $3.82\times {10}^{-26}$
• B. $3.82\times {10}^{-6}$
• C. $3.82\times {10}^{-20}$
• D. $3.82\times {10}^{-30}$

Answer 1

The correct option is A $3.82\times {10}^{-26}$

${Na}_{4}Y+H_{2}O\rightleftharpoons {Na}_{3}HY+NaOH\frac{k_w}{k_4}=1.818\times {10}^{-4}$
$0.1$ Initial
$0.1-x$ $x$ $x$ Final
$\frac{x^2}{0.1-x}=1.818\times {10}^{-4}$ $\Rightarrow$ $5500.55x^2+x-0.1=0$
$x=4.17\times {10}^{-3}$
${Na}_{3}HY+H_{2}O\rightleftharpoons {Na}_2{H}_{2}Y+NaOH\frac{k_w}{k_3}=1.445\times {10}^{-8}$
$x$
$x-y$ $y$ $y+x$
Since $y<<x$

$x-y\simeq x$

$x+y\simeq x$
$1.445\times {10}^{-8}=\frac{y\times x}{x}=y$
${Na}_2{H}_{2}Y+H_{2}O\rightleftharpoons Na{H}_{3}Y+NaOH\frac{k_w}{k_2}=4.7\times {10}^{-12}$
$y$
$y-z$ $z$ $z+x$
$y-z\simeq y$, $z+x\simeq x$
$4.7\times {10}^{-12}=\frac{z\times x}{y}$
$z=1.628\times {10}^{-17}$
$Na{H}_{3}Y+H_{2}O\rightleftharpoons H_{4}Y+NaOH$ $\frac{k_w}{k_1}=9.8\times {10}^{-13}$
$z$
$z-t$ $t$ $t+x$
$z-t\simeq z$

$t+x\simeq x$
$9.8\times {10}^{-13}=\frac{t.x}{z}$

$t=3.82\times {10}^{-26}$