I have a 2 part question concerning the use of H2SO4 to adjust the pH of a solution:
1. How many grams per liter of H2SO4 is required to adjust the pH of a solution from 3.8 to 2.8 assuming the sulfuric acid fully ionizes?
2. If in reality however, the actual sulfuric acid ionization/dissociation is a 2 step process and not fully ionized such that:
If, K1 = [H+][HSO4-]/[H2SO4] is complete and is equal to 100 mol/L,
But if the second reaction is not complete such that:
K2 = [H+][HSO4--]/[HSO4-] is incomplete and equal to .012 mol/L,
then how many grams/liter of sulfuric acid in this case would be required to change the pH of the solution from 3.8 to 2.8?
Note: For a formatted solution, please see the attachment.
1. How many grams per litre of H2SO4 are required to adjust the pH of a solution from 3.8 to 2.8 assuming the sulphuric acid fully ionizes?
Solution: [H+] = 10-pH
Increase in [H+] due to increase in pH = 10-2.8 - 10-3.8 = 0.001426 moles/litre
Each mole of sulphuric acid yields 2 moles of hydrogen ions.
Therefore, increase in moles of sulphuric acid = 0.001426/2 = 0.000713 moles/litre
Required mass ...
The solution calculates the mass of sulfuric acid to affect pH change.