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) Vc = 5.00 mL              Vd = 0.750 L

            [HCl]c = 1.00 M         [HCl]d = ?

 

            HCl(aq) H+(aq) + Cl-(aq)

            nc = nd

            [HCl]c x Vc = [HCl]d x Vd

            [HCl]d = [HCl]c x Vc/Vd = 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

           VHCl = 50.0 mL            VHI = 125 mL

 

           HCl(aq) H+(aq) + Cl-(aq)

           HI(aq) H+(aq) + I -(aq)

         

           [HCl] = n/V

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

           nHI = [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