Question

Calculate the pZn of a solution prepared by mixing 25.0 mL of 0.0100 M EDTA with 50.0 mL of 0.00500 M Zn2 . Assume that both the Zn2 and EDTA solutions are buffered with 0.100 M NH3 and 0.176 M NH4Cl.

Answer 1

Answer:

$\mathbf{pZn ^{2+} =8.8569 }$

Explanation:

Using the approach of Henderson-HasselBalch equation, we have :

$pH = pKa[NH^+_4] + log \dfrac{[NH_3]}{[NH_4^+]}$

where;

the pKa of $NH^+_4$ = 9.26

concentration of $NH_3$ = 0.100 M

concentration of $NH_4Cl$ = 0.176 M

∴

the pH of the buffered solution is :

$pH = 9.26 + log \dfrac{[0.100]}{[0.176]}$

$pH = 9.26 + log (0.5682)$

$pH = 9.26 +(-0.2455)$

$pH =9.02$

The Chemical equation for the reaction of $Zn ^{2+}$ and EDTA is :

$Zn^{2+}_{(aq)} + Y^{4-}_{(aq)} \iff ZnY^{2-} _{(aq)}$

Here;

$Y^{4-}_{(aq)}$ denotes the fully deprotonated form of the EDTA

The formation constant $K_f$ of the equation for the reaction can be represented as:

$K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}$      ----- (1)

The logarithm of the formation constant of Zn - EDTA complex = 16.5

$K_f$  = $10^{16.5}$

$K_f$  = $3.16 \times 10^{16}$

Since the formation constant in the above equation signifies that the EDTA is present in  $Y^{4-}$,

Then:

$\alpha _{Y^{4-} }= \dfrac{Y^{4-}}{C_{EDTA}}$

${Y^{4-}}= \alpha_ {Y^{4-}} \times {C_{EDTA}}$

From (1)

$K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}$  

$K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ \ \alpha_ {Y^{4-}} \times {C_{EDTA}}}$

∴

$K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }$

where;

$K_f'$ = conditional formation constant

$\alpha _Y{^4-}$ = the fraction of EDTA that exit in the form of the presences of the 4 charges .

So at equivalence point :

all the $Zn^{2+}$ initially in titrand  is now present in $ZnY^{2-}$

$K_f' = K_f \times \alpha _Y{^4-}$

Obtaining the data for the value of $\alpha _Y{^4-}$ at the reference table:

$\alpha _Y{^4-}$  =  $5.4 \times 10^{-12}$

∴

$K_f' = 3.16 \times 10^{16} \times 5.4 \times 10^{-2}$

$K_f' = 1.7064 \times 10^{15}$

To calculate the moles of  EDTA ,$Zn^{2+}$  , $ZnY^{2-}$ ; we have:

moles of  EDTA = 0.0100 M × 0.025 L

moles of  EDTA = $2.5 \times 10^{-4} \ mole$

moles of $Zn^{2+}$ = 0.00500 M  × 0.050 L

moles of $Zn^{2+}$ = $2.5 \times 10^{-4} \ mole$

moles of  $ZnY^{2-}$  =  $\dfrac{initial \ mole}{total \ volume}$

moles of  $ZnY^{2-}$  = $\dfrac{2.5 \times 10^{-4}}{ 0.025 + 0.050 }$

moles of  $ZnY^{2-}$  = $\dfrac{2.5 \times 10^{-4}}{ 0.075 }$

moles of  $ZnY^{2-}$  = 0.0033333 M

Recall that:

$K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }$

$K_f' = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }$

Assume Q² is the amount of complex dissociated in $ZnY^{2-}$

$ZnY^{2-} \iff Zn^{2+} + C_{EDTA}$  

i.e $Q^2 = Zn^{2+} + C_{EDTA}$

$1.707 \times 10^{15}= \dfrac{0.0033333}{Q}$

$Q= \dfrac{0.0033333}{1.707 \times 10^{15}}$

$Q^2= \dfrac{0.0033333}{1.707 \times 10^{15}}$

$Q^2= 1.9527 \times 10^{-18}$

$Q= \sqrt{1.9527 \times 10^{-18}}$

Q = $1.397 \times 10^{-9}$ M

$[Zn^{2+}]= 1.39 \times 10^{-9} \ M$

∴

$pZn ^{2+} =- log [Zn^{2+}]$

$pZn ^{2+} = - log (1.39 \times 10^{-9} ) \ M$

$\mathbf{pZn ^{2+} =8.8569 }$