| Acid-base titrations 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, let’s go through the process of testing the sulfuric acid and aluminum concentration in a bath used for MIL-A-8625 Type II anodizing. A similar titration may be used for a MIL-A-8625 Type III anodizing bath; although the concentrations of sulfuric and aluminum are normally different. The target range of acid concentration for a typical (Type II) sulfuric anodizing tank is 180 to 200 g/L, and dissolved aluminum in normally held between 4 and 12 g/L. If you prefer ounces and gallons, convert g/L (grams per liter) to oz/g (ounces per gallon) by dividing by 7.5. The factor of 7.5 is (28.375 grams/ounce)/(3.7854 liters/gallon). | |||||
| minimum | mid-range | maximum | |||
| Sulfuric acid (66°Be H2SO4) | 180 g/L | 190 g/L | 200 g/L | ||
| Alumimum (dissolved solid) | 4 g/L | 8 g/L | 12 g/L | ||
| The following information is taken from the table of Chemical Data, and the periodic table. The Chemical Data table shows data for other commercially available sulfuric acid concentrations, and these may be substituted for the 66°Be H2SO4 as long as the different acid concentration is taken into account on make ups, additions and titrations. A titration that computes concentration in %v/v will assume the acid concentration. If the titration computes concentration in g/L or oz/g, the acid concentration will not affect the computation; however, any computed addition must take the acid concentration into account. For example, an addition of 1 liter of 66°Be H2SO4 will add 1715 grams of H2SO4 to your tank, but this is not the case for 31°Be H2SO4. | |||||
| molecular weight |
solute weight |
moles /liter |
|||
| Sulfuric acid (66°Be H2SO4) | 98.073 | 1715 g/L | 17.49 | ||
| Aluminum | 26.982 | ||||
| This titration will compute both acid concentration and the concentration of dissolved aluminum in the bath. The titration steps and explanation follow. | |||||
| Pipette a 3.0 mL sample from bath. | This amount is critical since we will measure the number of molecules in this sample and use the sample size to compute the weight of those molecules (grams) per unit of volume (liter). Any error in the sample size will cause an error in the computation of both acid and dissolved metal concentration. | ||||
| Dilute with 50 mL of DI water. | This measurement is not critical because the critical number of molecules of each measured substance is unaffected by the dilution. | ||||
| Add a few drops of methyl orange. | The dissolved aluminum in solution binds up a few OH- ions, releasing a few H+ ions to create a weak acid. This makes the selection of indicator more important than for most simple acid-base titrations. Methyl orange transitions from orange color to yellow color at pH of 3.6: this color change occurs when the NaOH neutralizes the H2SO4, before there is sufficient NaOH to convert the remaining Al2(SO4)3 to salt and Al(OH)3. A different approach to analyzing for dissolved aluminum is to add enough NaOH to drive the pH to trigger a phenolphthalein color change at pH 8.6, and then add a bit of KF (potassium fluoride) or NH4HF2 (ammonium bifluoride) to the solution to bind the Al into AlF3, subsequently measuring how much NaOH is required to get the phenolphthalein color change again. If you are using KF or NH4HF2 as a binder, you should try the Anoplex approach since you can get sulfuric and aluminum tested in one sample without using the fluoride. | ||||
| Titrate with B mL of 1.0N NaOH from orange to a yellow endpoint. |
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 = 9.807 grams per mL titrant. This is the number we need for the first part of the titration: B mL of 1.0N NaOH x 9.807 = grams/liter of H2SO4. | ||||
| Add a few drops of phenolphthalein indicator. | At this point in the titration the H2SO4 has been neutralized by the NaOH, but there is insufficient NaOH in solution to remove all of the dissolved aluminum as aluminum hydroxide. Phenolphthalein changes from clear to a pink color at pH of 8.6. This pH indicates near completion of the reaction when the solution becomes highly alkaline due to an abundance of free OH- ions. | ||||
| Continue titration to total of A mL of 1.0N NaOH from colorless to pink endpoint. | As the sodium hydroxide continues to be added, sodium sulfate and aluminum hydroxide 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 = 1.797 grams per mL of additional titrant, where the additional titrant is A mL - B mL. This is the number we need for the second part of the titration: A mL - B mL of 1.0N NaOH x 1.797 = grams/liter of dissolved Al. | ||||
| This method analyzes acid and aluminum using one bath sample, and the anodizing baths in the ChemTrak template library calculate the concentrations of both components when the two titrant amounts are entered. A similar technique is used to speed up titrations for caustic etch tanks requiring analysis of NaOH (or free caustic) and dissolved aluminum. | |||||