12.2.06 1 Nickel(II) Ethylenediamine Salt

Preparation and Analysis of a Complex Nickel Salt

Dr. Fred O. Garces Chemistry 201

Miramar College

OH2

OH2Ni

OH2

H2O

H2O

H2O

SO4 [Ni(en)x(H2O)6-2x]SO4 • Y H2O

12.2.06 2 Nickel(II) Ethylenediamine Salt

Objective

Two Nickel-ethylenediamine complexes will be synthesize

with the general formula, [Ni(en)x(H2O)6-2x]SO4 • Y H2O.

The analysis of the two nickel salts consist of (1)

determining the percent yield of the products and (2) the

analysis of the “x” and “y” values in the formula in order

to determine the the nickel and ethylenediamine content

in the salts. These information will provide the chemical

formula for the two salts.

12.2.06 3 Nickel(II) Ethylenediamine Salt

Introduction Nickel(II) is a good Lewis acid and prefers to bind six Lewis base donor atoms (atoms with lone pair of electrons). In pure water nickel(II) sulfate dissolves in water to form [Ni(OH2)6]2+, also known as Ni2+

(aq). When ethylenediamine is added, (abbreviated en; which contains 2 donor atoms per molecule) some or all of the water may be replaced in [Ni(OH2)6]2+ as shown in the scheme.

H2O

OH2Ni

N

N

N

N

OH2

OH2Ni

OH2

H2O

N

N

OH2

OH2Ni

OH2

H2O

H2O

H2O

NNi

N

N

N

N

N

SO4

SO4

SO4

SO4

en

en

en

12.2.06 4 Nickel(II) Ethylenediamine Salt

Introduction In the analysis of the salts, note that the nickel(II) and SO4

2- are the only

charged species present in the product and they have equal but opposite

charge which means they are present in equimolar amounts. The salt is

also hygroscopic which means that waters (hydrates) are trapped in the

lattice. The general formula is can be expected to have the formula:

[Ni(en)x(OH2)6-2x]SO4• yH2O

OH2

OH2Ni

OH2

H2O

H2O

H2O

OH2

OH2Ni

OH2

H2O

N

N

H2O

OH2Ni

N

N

N

N NNi

N

N

N

N

N

SO4 SO4 SO4 SO4

6 H2O Y H2O Y H2O Y H2O

x =1 x =2 x =3

12.2.06 5 Nickel(II) Ethylenediamine Salt

•

Synthesis

12.2.06 6 Nickel(II) Ethylenediamine Salt

Analysis The analysis of the ethylenediamine in the salt (C2H4N2) will be based on titrating the salt with NaOH. The chemical to be neutralize by the NaOH is ethylenediamine, which is like a double barrel ammonia, the structure is- :NH2-CH2-CH2-NH2. When this is release from from the metal complex salt, both nitrogen are protonated and forms- [NH3-CH2-CH2-NH3]+2, this can be neutralize by two OH- to form two water and one ethylenediamine. The amount of ethylenediamine is determine by the following equation: mol ethylenediamine = [ M(HCl) x V(HCl) - V(NaOH) x M(NaOH) ]/2 where M is molarity, and V is volume. The analysis of the salts will be based on the following formula: [Ni(en)x(H2O)6-2x]SO4•H2O, the analysis will be to determine the x and y in this formula. I will clarify complete analysis in lab Monday but at this point in the semester you all should have the expertise to understand the analysis based on the description of the procedure. You are expected to have the prelab ready before you begin the experiment. You should also answer the prelab question before you begin the experiment.

12.2.06 7 Nickel(II) Ethylenediamine Salt

Analysis of [Ni(en)x(OH2)6-2x]SO4• yH2O The analysis of the nickel salt A & B will consist of two parts. Analysis of Nickel (250 mg): UV Spectroscopy Standard concentrations of Nickel: Conc of Nickel in Salts Conc. Nickel, moles of Nickel, mass of Nickel, % Nickel compound mol nickel, mole of SO4

2-, mass of SO42-, % SO4

2-

Analysis of ethylenediamie 50 mg sample: Titration Addition of HCl to salt, determines ethylenediamine moles moles of en, mass of en, % of en Composition of salt % Ni, % SO4

2-, % en g % H2O moles Ni, moles SO4

2-, moles en, moles H2O g x, y [Ni(en)x(OH2)6-2x]SO4• yH2O

12.2.06 8 Nickel(II) Ethylenediamine Salt

Nickel Analysis The analysis of the nickel salt will consist of two parts. The first part will be to analyze the Nickel in both salt-A and salt-B. The Nickel will be analyze via abosorption spectroscopy, using Beer's Law. Recall that the absorbance, A is proprtional to the concentration of the absorbing specie via A = ebc, where, e is the molar absorptivity constant, b is the path lenght and c is the concentration of the Nickel. For this part, you will prepare a stock solution of Nickel via NiSO4. You will then take this stock solution and prepare two other diluted solution. Be sure to calculate these concentration before you use the spectrometer. Take both salt-A and salt-B and prepare a solution based on the instructions of the lab. You will measure the absorbance of the nickel of these salts and use Beer's Law to calculate the Nickel concentration in your salts.

12.2.06 9 Nickel(II) Ethylenediamine Salt

Absorption Spectra Nickel Salt

12.2.06 10 Nickel(II) Ethylenediamine Salt

Analysis Ni+2 and SO42-

Determination of Ni+2 and SO4-2 in Salts A & B

50 mL Stock solution Cs = 1.0 g Ni(H2O)6]SO4 + 10.0 mL 1M H2SO4, 40mL H2O

C10 = 10 ml Stock in 25 mL solution C15 = 15 mL Stock in 25 mL Solution Salts 0.25 g + 5.00 mL, 1M H2SO4, 25 mL solution

Concentration (M) Ni+2

Abs

orba

nce

12.2.06 11 Nickel(II) Ethylenediamine Salt

Analysis Ni+2 and SO42-

Determination of Ni+2 and SO4-2 in Salts A & B

Concentration of Ni+2 g moles of Ni+2 g mass of Ni+22 g % Ni+2

Concentration [Ni+2] , UV-Vis Analysis 1. Moles of Ni via volume of solution prepared 2. Mass of Ni via atomic Wt of Ni 3. % Ni in Salt via Mass Ni in sample / Wt (.25g Nickel salt) * 100 This is the % of Nickel for any size sample. Assume 100 g sample of salt, calculate mol Nickel for 100 g sample. mol Ni+2 = mol SO4

2-

12.2.06 12 Nickel(II) Ethylenediamine Salt

Analysis: Titration The analysis of the ethylenediamine in the salt (N2C2H4) will be based on titrating the salt with NaOH. The chemcial to be neutralize by the HCl is ethylenediamine, which is like a double tail ammonia, the sturcture is- :NH2-CH2-CH2-NH2: . When this is release from from the metal complex salt, both nitrogen are protonated and forms- [NH3-CH2-CH2-NH3]+2, this can be neutalize by two OH- to form two water and one ethylenediamine. The amount of ethylenediamine is determine by the following equation:

[Ni(en)x(OH2)6-2x]SO4• yH2O + 2HCl (excess) g [Ni(H2O)6]+2 + SO4

-2 + enH2+2 + HCl (unreacted)

Note that an excess amount of HCl is used in the reaction so there will be unreacted HCl that will cause the solution to be acidic

12.2.06 13 Nickel(II) Ethylenediamine Salt

Analysis: Titration en The moles of enH2

+2, can be determine by analyzing the excess unreacted HCl.

[Ni(en)x(OH2)6-2x]SO4• yH2O + 2HCl (excess) →

[Ni(H2O)6]+2 + SO4-2 + enH2

+2 + HCl (unreacted) Note from the reaction that the moles of 2 mol HCl reacted = 1 mol en, or 2 mol HCl (reacted) = 1 mol en to form 1mol enH2

2+ The excess is titrated with NaOH, and the reacted HCl can be calculated by:

mol en = 1/2 HCl (reacted) = 1/2 [HCl (excess) - HCl (unreacted)] = 1/2 [HCl (excess) - NaOH (to neutralize unreacted)] = 1/2 [MHCl • VHCl (excess) - MNaOH • VNaOH (to neutralize)] mol en = 1/2 [MHCl • VHCl (excess) - MNaOH • VNaOH (to neutralize)] Mass en = mol en * 60.0 g / mol

12.2.06 14 Nickel(II) Ethylenediamine Salt

Analysis: Titration Solution: 0.05 g Salt + 10mL DI + 10.0 ml 0.2N HCl + methyl Red 1. Titrate with NaOH 2. Percentage of en in the salts: 3. Mass en in salt = moles en * 60.0 g / mol % en in salt = (Mass en in salt / Mass of sample titrated) * 100

12.2.06 15 Nickel(II) Ethylenediamine Salt

Formula Analysis [Ni(en)x(OH2)6-2x]SO4• yH2O mol Ni = mol SO4

2-

mol en determines mol H2O via ((H2O)6-2x

Excess H2O assigned as hydrated H2O Composition of salt % Ni, % SO4

2-, % en g % H2O moles Ni, moles SO4

2-, moles en, moles H2O g x, y [Ni(en)x(OH2)6-2x]SO4• yH2O Calculate molar mass of Salt A and Salt B Calculate % Yield