Testing/analyzing Baths for Passivation of Stainless Steel

A. Here is a procedure for the Dichromate Determination:

SCOPE: To determine the concentration of Sodium Dichromate in a solution.

EQUIPMENT:

PROCEDURE: Using the pipet, transfer 5 mL of the sample into the Erlenmeyer flask. Using the graduated cylinder, add 100 mL deionized water to the flask. Using the graduated cylinder, add 5 mL sulfuric acid and 10 mL potassium iodide solution to the flask. Titrate with 0.1N sodium thiosulphate until the solutions turns to a straw color. Add 2 mL starch indicator to the flask and continue to titrate until the blue-black color disappears. Record the mL used as V1. Place 500 mL of sample in the graduated cylinder. Measure the density of the solution with the hydrometer. Record this value as "P".

CALCULATION:

REPLENISHMENT:

A. There are several books that can be ordered through ASM International and AESF that provide procedures for analysis of metal finishing solutions. Here is a procedure for the determination of Nitric Acid:

Ira Donovan, M.S.F.
Kansas City, Missouri

"The Canning Handbook of Surface Finishing Technology"

on AbeBooks

or eBay or

Amazon

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A. Working through the math of the nitric acid procedure, V1 mL of N1 NaOH can neutralize V1*N1 mmol of HNO3, which is equal to V1*N1*(63.01) mg of HNO3.
Dividing by the number of mL in the sample produces a concentration (for a mL sample) of V1*N1*12.6 mg/mL, which produces a value of V1*N1*1.26 %w/v, not %v/v.

The Canning Handbook (pg. 772 of the 21st edition) supports this correction, and provides correct factors to find %v/v terms of concentrated solutions of nitric acid (1.46 for SG 1.38 and 1.30 for SG 1.41).

A. Peter,
The F1 constant in the titration equations is just hiding all of the other math involved. If you run the numbers from scratch and trace through all the stoichiometry, dilutions, etc., the number given for F1 should appear. In procedures such as this, the math is typically "precooked" since the intended user is only interested in the final answer and might not have been a chem major.

I have to confess, I don't know where 2.483 comes from either. I'm getting a totally different number.

I managed to find something that explains the reactions involved at
http://www.meduniv.lviv.ua/files/kafedry/tokshim/English/Analytical/Manuals_Analytchem/Iodometry.pdf

Essentially, adding the Potassium Iodide and Sulfuric Acid generates 3 moles of iodine (I2) for every mole of dichromate in solution. The iodine is then titrated by the Thiosulphate with an equivalence of 2 moles Thiosulphate per mole of iodine.

The normality of the thiosulphate solution would seem to be equal to molarity (mol/L) here.

So you have your mL of titrant (V1), divide by 1000 to convert it to liters, multiply by the molarity (N1, though assigned as 0.1 in the procedure) to convert to moles of thiosulphate titrant, divide by 6 to convert moles thiosulphate to moles dichromate, and multiply by the molar mass of sodium dichromate (261.97 g/mol) to get the grams of sodium dichromate present in the original 5 mL sample. Now for %w/w, you need to divide that by the total mass of that original sample, which is its density times 5 (for the 5 mL).

Stringing all that together, the F1 in %w/w = V1*F1*N1/density
would presumably be 261.97/(1000*6*5) = 0.0087323

I rather suspect I'm missing something here.