Which is the value of the conditional formation constants, Ki of the Cd^2x-EDAT

Question

Which is the value of the conditional formation constants, Ki of the Cd^2x-EDAT complex at pH-9. The Cd^2x-EDTA complex has a formation constant Ki=16.5 0.985 0.041 0.676 16.5 0.028 Find the pH of a solution prepared by dissolving 1.00 of potassium hydrogen phihate and 1.50 g of disodium phihate in 50 0 mL. of water. M(C3HSO4K)=204.221. M(C3HSO4Na_2)=210.094 Calculate the pH when 50.00 mL of 0.100 M HNO_3 is mixed with 40.00 ml. of 0.0750 M NaOH. Calculate the pH during the titration of 50.00 ml. of a 0.1500 M methylamine with 0.1000 M HCl. K_b, = 4.42 times 10^-4. When the volume added is V=0 mL. V=50 mL V=75 mL V=80 mL. Calculate |Mn^2+| when 50.00 mL of 0.1000 M Mn^2+ is titrated with 17.00 mL of 0.2000 M EDTA. The titration is buffered to pH 11 for which a_r4 =0.81. K_f = 7.76 times 10^13

Explanation / Answer

7)

Answer c) 0.676

The formation of a metal–EDTA complex, let’s consider the reaction between Cd2+ and EDTA.

EdTA is a tetradentate ligand so let us represent it as Y4

Cd2+(aq)+Y4(aq) CdY2(aq)

The reaction’s formation constant

Kf=[CdY2] / [Cd2+][Y4] = 16.5

Kf´ is a pH-dependent conditional formation constant .

Kf = Kf × Y4

Where Y4–isthe fraction of EDTA present as Y4–.

Kf = Kf × Y4 = 16.5 x 0.041 = 0.676

Section B

1)

According to Henderson Hassel balch equation

pH = pKa +log [salt] / [Acid]

Here salt = Na2P (Disodiumphthalate)

Number of moles of Na2P = 1.50 / 210.094 = 0.00714 = 7.14 x 10-3

Number of moles of KHP = 1.00 / 204.221 = 0.00490 = 4.90 x 10-3

This much moles in 50.0 ml So molarity is

molarity of Na2P = 7.14 x 10-3 /.0500L = 142.8  x 10-3

molarity of KHP = 4.90 x 10-3 /.0500L = 98 x 10-3

pKa of KHP = 5.4

pH = pKa +log [salt] / [Acid]

pH = 5.4 + log 142.8  x 10-3/ 98 x 10-3

= 5.4 + 0.16 = 5.56 = 5.6

2)

Both are strong

M1 = 50.0 ml of 0.1000 M HNO3 = 0.1000x50.0 /1000 = 0.005 M

pH = - log(-0.005) = 2.3

40.0 ml of 0.0750 M NaOH =0.0750x40.0 /1000 = 0.003 M

pH = 14 -{- log(-0.003)} = 14 - 2.5 = 11.5

M2 = 10-11.5 = 6.3 x10-12

M1 V1 +M2V2 = Mnew(V1+V2)

0.005 x 50.0 +6.3 x10-12 x 40.0 = Mnew x 90.0

0.25 +3.0 x 10-10 = 0.25

Mnew = 0.25/90 = 0.0027

pH = - log (0.0027) = 2.6

3)50.0 ml of 0.1500 methyl amine = 0.0075

pH of as weak base = 14 -1/2{pKb - logC} = 12.9

C of HCI = 0.005

pH of HCl = 2.3

M1 V1 +M2V2 = Mnew(V1+V2)

10-12.9 V1 +10-12.9V2 = Mnew(100)

pH after mixing = 2.6

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