Acid-Base Equilibria

34) (a) [HNO₃] = 0.0575 M

            HNO₃(aq) H⁺(aq) + NO₃^-(aq)

            pH = -log[H⁺] = -log(0.0575) = 1.240

      (b) m = 0.723 g HClO₄          V = 2.00 L

            HClO₄(aq) H⁺(aq) + ClO₄^-(aq)

            [HClO₄] = n/V = 0.723 g HClO₄ x 1 mol HClO₄/100.46 g HClO₄/2.00 L = 3.60 x 10^-3 M

            pH = -log[H⁺] = -log(3.60 x 10^-3) = 2.444

      (c) V_c = 5.00 mL              V_d = 0.750 L

            [HCl]_c = 1.00 M         [HCl]_d = ?

            HCl(aq) H⁺(aq) + Cl^-(aq)

            n_c = n_d

            [HCl]_c x V_c = [HCl]_d x V_d

            [HCl]_d = [HCl]_c x V_c/V_d = 1.00 M x 5.00 mL/(0.750 L x 10³mL/L) = 6.67 x 10^-3 M

            pH = -log[H⁺] = -log(6.67 x 10^-3) = 2.176

      (d) [HCl] = 0.020 M          [HI] = 0.010 M

           V_HCl = 50.0 mL            V_HI = 125 mL

           HCl(aq) H⁺(aq) + Cl^-(aq)

           HI(aq) H⁺(aq) + I ^-(aq)

           [HCl] = n/V

           n_HCl = [HCl] x V = 0.020 mol HCl/L x 50.0 mL x 1 L/10³ mL = 0.0010 mol HCl

           n_HI = [HI] x V = 0.010 mol HI/L x 125 mL x 1 L/10³ mL = 0.0012 mol HI

           [H⁺] = n/V = (0.0010 mol H⁺ + 0.0012 mol H⁺)/((125 mL + 50.0 mL) x 1 L/10³ mL)

           [H⁺] = 0.013 M

           pH = -log[H⁺] = -log(0.013) = 1.89