## Answers

Concentration of H = 9.3x10'M Equation representing the ionization of HI: HI(aq) →→ H(aq) + 1" (aq) Since 1 mol of HI forms 1 mol of H* upon dissociation [H*] = [HI] = 9.3x102 M Now, pH = -log[H] =-log(9.3x10-) = 2.03

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