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

`\mathbf{pZn ^(2+) =8.8569 }`

Step-by-step explanation:

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

`pH = pKa[NH^+_4] + log ([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 ([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 = ([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 * 10^(16)`

Since the formation constant in the above equation signifies that the EDTA is present in
`Y^(4-)`,

Then:

`\alpha _(Y^(4-) )= (Y^(4-))/(C_(EDTA))`

`{Y^(4-)}= \alpha_ {Y^(4-)} * {C_(EDTA)}`

From (1)

`K_f = ([ZnY^(2-)])/([Zn^(2+) ][Y^(4-)])`

`K_f = \frac{[ZnY^(2-)]}{[Zn^(2+) ] \ \ \alpha_ {Y^(4-)} * {C_(EDTA)}}`

∴

`K_f' = K_f * \alpha _Y{^4-} = ([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 * \alpha _Y{^4-}`

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

`\alpha _Y{^4-}` =
`5.4 * 10^(-12)`

∴

`K_f' = 3.16 * 10^(16) * 5.4 * 10^(-2)`

`K_f' = 1.7064 * 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 * 10^(-4) \ mole`

moles of
`Zn^(2+)` = 0.00500 M × 0.050 L

moles of
`Zn^(2+)` =
`2.5 * 10^(-4) \ mole`

moles of
`ZnY^(2-)` =
`(initial \ mole)/(total \ volume)`

moles of
`ZnY^(2-)` =
`(2.5 * 10^(-4))/( 0.025 + 0.050 )`

moles of
`ZnY^(2-)` =
`(2.5 * 10^(-4))/( 0.075 )`

moles of
`ZnY^(2-)` = 0.0033333 M

Recall that:

`K_f' = K_f * \alpha _Y{^4-} = ([ZnY^(2-)])/([Zn^(2+) ] \ C_(EDTA) )`

`K_f' = ([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 * 10^(15)= (0.0033333)/(Q)`

`Q= (0.0033333)/(1.707 * 10^(15))`

`Q^2= (0.0033333)/(1.707 * 10^(15))`

`Q^2= 1.9527 * 10^(-18)`

`Q= \sqrt{1.9527 * 10^(-18)}`

Q =
`1.397 * 10^(-9)` M

`[Zn^(2+)]= 1.39 * 10^(-9) \ M`

∴

`pZn ^(2+) =- log [Zn^(2+)]`

`pZn ^(2+) = - log (1.39 * 10^(-9) ) \ M`

`\mathbf{pZn ^(2+) =8.8569 }`