Determination

Nickel metal and nickel chloride with calculation of nickel sulfamate in sulfamate nickel plating baths.

Application

Variations of this method can be used for sulfamate nickel or Watts nickel per BAC 5746, QQ-N-290, AMS 2403, AMS 2423, and AMS 2424. as well as for an electroless nickel bath used for plating per MIL-C-26074, ASTM B733 or AMS 2404. For electroless nickel baths, the concentration of total nickel is directly useful, since there is only one nickel measurement needed for the bath control. In the case of sulfamate and Watts baths, the total nickel is one step in the process because there are two nickel concentrations that must be known to control the bath. It is possible to test chloride and make additions of nickel chloride to true up chloride concentration, then after additions of nickel chloride, test total nickel and make additions of nickel sulfamate. However, this requires more work than calculation of both additions from a single sample.

The following table shows concentrations of nickel sulfamate and nickel chloride for a typical sulfamate nickel plating bath (per BAC 5746). If you prefer ounces and gallons, convert g/L (grams per liter) to oz/gal (ounces per gallon) by dividing by 7.5. The factor of 7.5 is (28.375 grams/ounce)/(3.7854 liters/gallon).

 min mid max Nickel Metal 75 g/L 94 g/L 113 g/L 10 oz/gal 12.5 oz/gal 15 oz/gal Ni(SO3NH2)2 323 g/L 405 g/L 480 g/L 43 oz/gal 53 oz/gal 64 oz/gal Cl 3.375 g/L 4.125 g/L 4.875 g/L 0.45 oz/gal 0.55 oz/gal 0.65 oz/gal NiCl2*6H2O 11.33 g/L 13.8 g/L 16.35 g/L 1.51 oz/gal 1.84 oz/gal 2.18 oz/gal

Method (NiCl2*6H2O and chloride ion)

The following titration can be used to determine the concentration of nickel chloride salts and/or chloride ion. This result can be used later to calculate nickel sulfamate concentraton from the total nickel concentration.

1 Pipette a 5.00 mL sample from bath.

2 Dilute to ~100 mL with DI water.

3 Add 1.0 mL of 5% K2CrO4 solution. The solution will turn yellow-green.

4 Titrate with 0.1M AgNO3 to a reddish-brown endpoint.

5 Record titrant used as B mL.

Calculation of NiCl2*6H2O and chloride ion.

B mL x B Factor = NiCl2*6H2O

 Sample Titrant B Factor NiCl2*6H2O (g/L) 5.0 mL 0.1M 2.377 NiCl2*6H2O (g/L) 10.0 mL 0.1M 1.189 Cl- (g/L) 5.0 mL 0.1M 0.355 NiCl2*6H2O (oz/gal) 5.0 mL 0.1M 0.317 NiCl2*6H2O (oz/gal) 10.0 mL 0.1M 0.159 Cl- (oz/gal) 5.0 mL 0.1M 0.024

Method (Ni Metal)

1 Pipette a 2.0 mL sample from bath.

2 Dilute to ~100 mL with DI water.

3 Add 10 mL of concentrated ammonium hydroxide.

4 Add 0.1 gram of murexide indicator to pale straw color.

5 Titrate with 0.1M EDTA to a blue-violet endpoint.

6 Record titrant used as A mL.

Calculation of Total Nickel

A mL x A Factor = Ni Metal

 Sample Titrant A Factor Ni (g/L) 2.0 mL 0.1M 2.935 Ni (g/L) 5.0 mL 0.1M 1.174 Ni (g/L) 10.0 mL 0.5M 2.935 Ni (oz/gal) 2.0 mL 0.1M 0.392 Ni (oz/gal) 5.0 mL 0.1M 0.156 Ni (oz/gal) 10.0 mL 0.5M 0.392

Calculation (Nickel Sulfamate)

This table shows the molecular weights used in the following calculations.

 MolecularWeight Nickel Metal 58.710 Chloride Ion 35.450 Potassium chromate (K2CrO4) 194.190 Silver Nitrate (AgNO3) 169.873 Nickel Chloride (NiCl2) 129.616 Nickel Chloride Hexahydrate(NiCl2*6H2O) 237.690 Nickel Sulfamate Anhydrous (Ni(SO3NH2)2 250.860 Nickel Sulfamate Tetrahydrate (Ni(SO3NH2)2*4H2O) 322.930

The concentration of nickel sulfamate can be calculated since we know that the total nickel comes from only from the nickel chloride and the nickel sulfamate. This is expressed in the first step below. Subsequent steps use the molecular weight factors to substitute the weight of the salts for the weight of the nickel in the salts. If you have trouble following along, just go to step 9 for the final answer which matches the formula from the reference source; although the reference source does not show the derivation or how to adjust the calculation for different sample sizes and titrant normalities.

1 ... Total Nickel = Nickel in Ni(SO3NH2)2*4H2O + Nickel in NiCl2*6H2O

2 ... Nickel in Ni(SO3NH2)2*4H2O = Total Nickel - Nickel in NiCl2*6H2O

3 ... Ni(SO3NH2)2*4H2O = { Total Nickel - Nickel in NiCl2*6H2O } x [322.93 / 58.71]

4 ... Total Nickel = A mL x A Factor

5 ... Ni(SO3NH2)2*4H2O = { A mL x A Factor - Nickel in NiCl2*6H2O } x [322.93 / 58.71]

6 ... Nickel in NiCl2*6H2O = [B mL x B Factor] x [58.71 / 237.69]

7 ... Ni(SO3NH2)2*4H2O = { A mL x A Factor - [B mL x B Factor] x [58.71 / 237.69] } x [322.93 / 58.71]

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