Question

Volumes and molarities for titration of CdC2O4 solutions.

PART A Data

Initial reading of buret (KMnO4) (mL) 0.50

Final Reading of buret (KMnO4) (mL) 14.00

Molarity of KMnO4 solution (M) 0.000969

PART B DATA

Initial Reading of buret (NH3) (mL) 1.00

Final reading reading of buret (NH3) mL 22.50

Molarity of NH3 solution (M) 5.00

FOR PART A Find:

Moles of MnO4- used for titration (mol)

Moles of C2O42- in 100.0mL of solution (mol)

Molarity of C2O42-(M)

Molarity of Cd2+ (M)

Ksp of CdC2O4

For PART B Find:

Total Moles of C2O42-

Molarity of C2O42- (M)

Total moles of Cd2+ (mol)

Moles of [Cd(NH3)4 2+] (mol)

Molarity of [Cd(NH3)4 2+] (M)

Moles of NH3 added by titration (mol)

Moles of NH3 that did not react with Cd2+ (mol)

Molarity of NH3 that did not react with Cd2+ (M)

Kf for [Cd(NH3)4 2+]

Answer 1

Part A) has been solved here with all sub-parts. kindly post other part separately.

FOR PART A

Note that 1 mole KMnO4 contain 1 mole of K+ cation and 1 mole of MnO4- anion. So KMnO4 or MnO4- can be used interchangeably for stoichiometric calculations.

i) Moles of MnO4- used for titration (mol) = Volume of KMnO4 used x Molarity of KMnO4 solution

                                                                = (Final buret reading -Initial buret reading) x ( 0.000969 ) M

                                                                = (14.00 - 0.50) mL x ( 0.000969 ) M

                                                               = 1.31x 10-5

Concept: KMnO4 is an oxidizing agent. It oxidizes Oxalate (C2O42-) ( C in +3 oxidation state) present in the solution to CO2 ( C in +4 oxidation state). And in the process, Mn (+7) in MnO4- is reduced to Mn (+2) . We need balanced redox reaction for this change to calculate mole ratio of KMnO4 and oxalate. Assuming the reaction is taking place in acidic medium.

Reduction Half : MnO4– + 8H+ + 5e– → Mn2+ + 4H2O

Oxidation Half : C2O42- → 2CO2 + 2e–

Combining both the reactions, overall reaction: (overall electrons should be cancelled)

2MnO4– + 16H+ + 5C2O42- → 2Mn2+ + 10CO2 + 8H2O

So, 2 mole of MnO4– react with 5 mole of C2O42

Or, 1 mole of MnO4– react with 2.5 mol of C2O42

So, 1.31x10-5 mole of MnO4– reacted with 1.31x10-5 x 2.5 mol = 3.27x10-5 mol of C2O42

ii) Moles of C2O42- in 100.0mL of solution (mol) = 3.27x10-5 mol ( as calculated above)

iii) Molarity of C2O42-(M) : Volume of solution in L = 0.1 L

                                         Therefore, mol/litre (M) = 3.27x10-5 mol / 0.1 L = 3.27x10-4 M

Now the salt CdC2O4 conatins 1 mole Cd2+ for 1 mole of C2O42-

hence, for 1 M of C2O42-, 1 M of Cd2+ is present

So, 3.27x10-4 M of C2O42-, implies, 3.27x10-4 M of Cd2+ is present

iv) So, Molarity of Cd2+ (M) = 3.27x10-4 M

v) Ksp of CdC2O4

The salt is sparingly soluble and its saturated solution in water exists as:

CdC2O4 (s)    Cd2+ (aq) + C2O42- (aq)

1- s                         s                s

Where, s is the solubility of salt in M unit. and s = 3.27x10-4 M (as calculated above)

Ksp = [Cd2+] [C2O42-] = s 2

Putting the value of s =3.27x10-4 M ; we have

Ksp = s2 = 1.07 x 10-7