Volumes and molarities for titration of CdC2O4 solutions. Part A Initial reading

Question

Volumes and molarities for titration of CdC2O4 solutions.

Part A
Initial reading of buret (KMnO4) (mL) 24
Final Reading of buret (KMnO4) (mL) 38
Molarity of KMnO4 solution (M) 0.00117

Part B
Initial Reading of buret (NH3) (mL) 0
Final reading reading of buret (NH3) mL 12
Molarity of NH3 solution (M) 5.00

FOR PART A calculate:

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 calculate:

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+]

Explanation / Answer

Part A: Number of moles of KMnO4 = concentration of KMnO4 X Volume of KMnO4

= 0.00117 mol L-1 X (38-24) 10-3 L = 0.00117 X 0.014 mol = 1.638 X 10-5 mol

This titration is carried in the presence of sulfuric acid and the balanced equation is given as:

5CdC2O4 + 2KMnO4 + 8H2SO4 ------> 10CO2 +CdSO4 + K2SO4 + MnSO4 + H2O

From the above equation we know that 5 mol of CdC2O4 are consumed per 2 mol of KMnO4

Number of moles of CdC2O4 = 2/5 X Number of moles of KMnO4

= 0.4 X 1.638 X 10-5 mol = 6.552 X 10-6 mol

6.552 X 10-6 moles of CdC2O4 were added per 100 mL of water

Molarity of CdC2O4 = 6.552 X 10-6 mol / 0.1 L

= 6.552 X 10-5 mol L-1 = 6.552 X 10-5 M

CdC2O4 (s) <=> Cd2+(aq) + (C2O4)-2(aq)

CdC2O4 dissociates to give Cd2+(aq) and (C2O4)-2(aq) in 1:1 ratios

Therefore, concentration of Cd2+(aq) = concentration of (C2O4)-2(aq) = 6.552 X 10-5 M

Ksp = [Cd2+]e[(C2O4)-2]e

= 6.552 X 10-5 X 6.552 X 10-5

= 42.93 X 10-10

Ksp = 4.293 X 10-9

NB: Ksp does not have units

Part B:

When NH3 was added to the aqueous solution of CdC2O4, allCd2+(aq) ions convert toCd(NH3)42+(aq) ions.

Cd2+(aq) + (C2O4)-2(aq) + 4NH3(aq) ------> Cd(NH3)42+(aq) + (C2O4)-2(aq)

However, you have not provided the Volume of CdC2O4 solution. This is the amount of  CdC2O4 pipetted out. The molarity of  (C2O4)-2 and Cd2+ ions depends on this volume.

Number of moles of NH3 added = concentration of NH3 X Volume of NH3

= 5.0 mol L-1 X (12-0) 10-3 L = 0.06 mol

Since ammonia is used in excess, we can not directly correlate the Number of moles of NH3 added to the number of moles of Cd2+.  Hence volume of CdC2O4 solution is required to calculate the concentration of Cd2+ , NH3 used andCd(NH3)42+ ions.  Using these calculations we can calculate Kf, which is given by following equation.

Kf = [Cd(NH3)42+] / [Cd2+] [NH3]4

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