Sulfuric acid and dissolved aluminum in anodizing baths.
Acid-base (or neutralization) titrations depend upon the neutralization of an unknown quantity of acid with a known quantity of base or vice versa. These titrations are used to measure acid concentrations in acid etch tanks, anodizing tanks, and plating tanks, and to measure base concentration in caustic etch tanks. As an example, we will explain the process of testing the sulfuric acid and aluminum concentration in a bath used for MIL-A-8625 Type III hard coat anodizing. A similar titration may be used for a MIL-A-8625 Type II anodizing bath; although the concentrations of sulfuric and aluminum are normally different. The following table shows concentrations of acid and aluminum for a hard coat anodizing bath set up per BAC 5821 Solution 2. 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 | |
| Sulfuric acid | |||
| (66°Be H2SO4) | 210 g/L | 240 g/L | 270 g/L |
| Aluminum | |||
| (dissolved) | 0 g/L | 3.75 g/L | 7.5 g/L |
1 Pipette a 5.0 mL sample from bath into each of two 250 mL flasks.
Label one flask A, and label the other flask B.
2 Dilute each flask with 25 mL of DI water.
3 Add 10 mL 33% KF (potassium fluoride) to flask A and mix.
4 Add a few drops of phenolphthalein to each flask.
5 Titrate flask A with A mL of 1.0N NaOH and
flask B with B mL of 1.0N NaOH to a pink endpoint.
A mL x Factor = H2SO4
| Sample | Titrant | Factor | |
| H2SO4 (g/L) | 5.0 mL | 1.0N | 9.8 |
| H2SO4 (oz/gal) | 5.0 mL | 1.0N | 1.31 |
| H2SO4 (g/L) | 10.0 mL | 1.0N | 4.9 |
| H2SO4 (oz/gal) | 10.0 mL | 1.0N | 0.65 |
(B mL - A mL) x Factor = dissolved aluminum
| Sample | Titrant | Factor | |
| Al (g/L) | 5.0 mL | 1.0N | 1.8 |
| Al (oz/gal) | 5.0 mL | 1.0N | 0.24 |
| Al (g/L) | 10.0 mL | 1.0N | 0.9 |
| Al (oz/gal) | 10.0 mL | 1.0N | 0.12 |
In this analysis, we will determine the concentration of sulfuric acid in a bath sample by titrating with a sodium hydroxide reagent of known normality after removing dissolved aluminum by adding potassium fluoride to the sample. We will also determine the amount of dissolved aluminum using a second sample without the addition of potassium fluoride and comparing the results of the two titrations.
The KF removes the dissolved aluminum from flask A as AlF3. With the dissolved aluminum removed, the free acid can be accurately determined. You must add enough KF to remove ALL of the dissolved aluminum. The BAC 5821 hard coat bath contains a maximum 7.5 g/L of dissolved aluminum, so a 5 mL sample could contain as much as 0.0375 gram of aluminum, and you should add at least 0.08 gram of KF to insure complete removal. If the dissolved aluminum can reach 20 g/L, a 5 mL sample could contain as much as 0.1 gram of aluminum, and you should add at least 0.22 grams of KF. 10 mL of 33% KF solution contains 1.917 grams of KF, so this amount specified by Boeing should be adequate for up to 59 grams of dissolved aluminum. (KF molecular weight is 58.1 amu; Al molecular weight is 26.98 amu; 0.1 gram of Al x 58.1/26.98 = 0.215 gram of KF).
After addition of KF, the choice of indicators is less important because methyl orange or phenolphthalein will both change color at almost the same equivalence point; however, phenolphthalein is the better choice because it insures the total dissociation of the acid.
As the sodium hydroxide is added, sodium sulfate and water is formed. The reaction is:
H2SO4 + 2 NaOH —> Na2SO4 + 2 H2O.
From the reaction equation: two moles of NaOH have reacted with one mole of H2SO4, when the solution begins to transition from acid to base (pH of 3.6) so: (98.073 grams) x (1 mole H2SO4/2 moles NaOH) = 49.04 grams of H2SO4 is neutralized by one mole (1 liter of 1.0N) of NaOH. Since the bath sample is 5.0 mL, 49.04/5.0 = 9.8 grams per mL titrant.
For every 3 molecules of NaOH, one molecule of Al(OH)3 is formed. The reaction is:
Al2(SO4)3 + 6 NaOH —> 3Na2SO4 + 2Al(OH)3
From the reaction equation: six moles of NaOH have reacted with two moles of Al when the solution begins to transition from neutral to base (pH of 8.6) so: 2 x (26.982 grams) x (2 moles Al/6 moles NaOH) = 8.985 grams of Al reacts with one mole (1 liter of 1.0N) of NaOH. Since the bath sample is 5.0 mL, 8.985/5.0 = 1.8 grams per mL of additional titrant, where the additional titrant is A mL - B mL.
We have documented a slightly less accurate analysis using a single bath sample that is normally used by commercial suppliers as Anodize Method 1. This Anodize Method 2 analyzes acid and aluminum using two bath samples. The total removal of the aluminum and independent testing of the two samples is the more accurate method of analysis. The first cited reference estimates an accuracy of ±3% when the method is carefully performed.
Process Control Analytical Procedures
Volumetric Analysis of Metal Finishing Solutions
Chemical Analysis of Plating Solutions