8 ... Ni(SO3NH2)2*4H2O = { A mL x A Factor - B mL x B Factor x 0.247 } x 5.5

9 ... Ni(SO3NH2)2*4H2O = { A mL - B mL x [B Factor / A Factor] x 0.247 } x 5.5 x A Factor

The formula in step 9 is in the form used in the ChemTrak Library. Using the formula in step 9, sample size and titrant normality for the chloride test and the nickel test can be optimized for your desired bath concentration and analysis accuracy, and your choice will determine the A Factor and the B Factor which can be used to calculate the k1 and k2 in ChemTrak. However, you must be consistent when selecting either g/L or oz/gal as the measurement units for each test since the arithmetic used to compare the results does not include conversion of measurement units.

Explanation of Chloride Method

The method for determining nickel chloride is a complexometric titration involving color changes as nickel chromate (NiCrO4) is displaced by silver chromate (Ag2CrO4) as it reacts with silver nitrate (AgNO3). The yellowish color is due to the formation of NiCrO4 in solution. (Almost all chromate salts are yellow, like NiCrO4, the red color of Ag2CrO4 is an exception.)

In step 3, the amount must be sufficient to allow complete reaction: NiCl2 + K2CrO4 —> NiCrO4 + 2 KCl. This reaction shows a 1:1 ratio of NiCl2 to K2CrO4. The 5.0 mL bath sample has a maximum of 53 g/L x 5.00 mL = .265 grams of NiCl2 which requires (194.190/129.616) x .265 = .397 grams of K2CrO4. If you change the sample size, the amount of K2CrO4 should be changed in proportion.

The reaction in step 4 is: NiCrO4 + AgNO3 —> Ni(NO3)2 + Ag2CrO4. The endpoint is determined by the reddish-brown color of Ag2CrO4. The reaction shows a 2:1 ratio of the AgNO3 molecule in the titrant to the original NiCl2*6H2O in the solution. So, for each molecule of AgNO3 in the used titrant there is 0.5 x 237.706 g/L = 118.853 g/L of NiCl2*6H2O in the original solution. Conversion for the sample size and molarity of the titrant gives 0.1M x 118.853 g/L / 5 mL = 2.377 g/L per mL of titrant as shown in the preceding table.

Explanation of Nickel Method

The method for determining total nickel uses EDTA (ethylenediaminetetraacetic acid). EDTA removes free metal ions from solution as they are changed into complex ions. The pM (negative log of metal ion concentration) vs volume of titrant in EDTA titrations is similar to pH vs volume of titrant in acid-base titrations. At the end point of an EDTA titration, the pM rapidly increases due to the removal of metal ions from solution by EDTA. Any method which can measure disappearance of free metal ions can be used to detect the end point in EDTA titrations. A common indicator is eriochrome black T which undergoes a change in color from red, in the presence of metallic salts with the ions of Mg, Mn, Zn, Cd, Hg, Pb, Cu, Al, Fe, Ti Co, Pt, and Ni, to blue when the metal has been complexed. Murexide is another indicator which changes from violet to blue as Cu or Co salts are removed from solution. Murexide complexes Ni to form a yellow compound and the solution color changes to purple when the nickel is complexed with EDTA.

In step 3, ammonium hydroxide serves as a buffer in the pH range of 9 to 10. The pH of the solution is critical since H+ ions compete with the nickel ions to complex with the EDTA molecule.

In step 4, 100 grams of murexide indicator is a mix of 1 gram murexide and 99 grams NaCl. The murexide complexes with nickel ions to form the yellowish nickel-murexide complex.

At the endpoint, the nickel ions are complexed with the EDTA leaving the purple murexide free. Each EDTA molecule wraps up a Ni-2 ion, taking it out of solution. The total nickel per liter of 1.0M EDTA is then 58.71 grams - the molecular weight of Ni. Conversion to our sample size of 2 mL and titrant molarity gives 0.1 x 58.71 g/L / 2 mL = 2.935 g/L nickel per mL of 0.1M EDTA titrant as shown in the preceding table.

Reference

Volumetric Analysis of Metal Finishing Solutions

Andrew K. McFadyen, B.Sc., C.Chem., FRSC

Finishing Publications Ltd, 1998, Pg 76-77