Calculating pH of methanoic/formic acid (HCOOH) When calculating the pH of methanoic/formic acid (HCOOH) why do we not multiply the given molarity of the acid by 2 as there are 2 moles of H+ ions, as seen in chemical formula HCOOH?
However when it comes to calculating the pH of sulfuric acid (H₂SO₄), we do multiply the given molarity by 2 as there are 2 moles of H+ ions.
The reason for this is that the hydrogen atom attached to the carbon will not be lost. Only the hydrogen attached to the oxygen will be ionized. In fact, in a solution of formic acid, not all molecules will be ionized anyways. Only very strong acids will fully ionize. Even in the case of sulfuric acid, not all molecules will fully ionize and lose the second proton. It is actually a simplification.