give the pH of a solution containing .972 g HBr in 628 mL of a soluion. I did .972/80.9=.012 mol. what do i do next...i'm stuck!
mols = g/molar mass which is what you have.
M = mols/L
pH = -log(H^+)
HBr is a strong acid, ionized 100% just like HCl).