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!

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  1. 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).

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