To calculate the acid dissociation constant (Ka) for HCrO4(-) and CrO4(2-) from a pH of 3.50 in a 0.0025 mol/L KHCrO4 solution, the concentration of [H+] is determined to be 3.16 x 10^(-4) mol/L. The equilibrium expression for Ka is established as Ka = [H+][CrO4(2-)]/[HCrO4(-)], where [CrO4(2-)] is equal to [H+]. After substituting the values, the calculated Ka is approximately 4.58 x 10^(-5) mol/L, which is close to the textbook answer of 4.1 x 10^(-5) but slightly different. The discussion highlights the importance of stoichiometry and equilibrium in acid-base calculations, with participants sharing methods and confirming results. Clarification on the calculations and potential discrepancies with textbook answers is encouraged.