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KEY


                        GENERAL CHEMISTRY-II (1412)
                                 S.I. # 24


Calculate the pH of each of the following strong acid solutions:

1.   0.00835 M HNO3
              0.00835 M HNO3 = 0.0835 M H+; pH = -log(0.00835) = 2.08



2.   0.525 M HClO4
              0.525 g HClO4 x 1 mol HClO4 = 2.612 x 10-3 = 2.61 x 10-3 M HClO4
               2.00 L Soln     100.5 g HClO4

              [H+] = 2.61 x 10-3 M; pH = -log (2.612 x 10-3) = 2.583

3. 5.00 mL of 1 M HCl diluted to 0.500 L
             Mc x Vc = Md x Vd; 0.500 L = 500 mL

              1.00 M HCl x 5.00 mL HCl = Md HCl x 500 mL HCl

              Md HCl = 1.00 M x 5.00 mL = 1.00 x 10-2 M HCl = 1.00 x 10-2 M H+
                           500 mL

              pH = - log (1.00 x 10-2) = 2.000


4. a mixture formed by adding 50.0 mL of 0.020 M HCl to 150 mL of
              0.010 M HI.
              [H+] total = mol H+ from HCl + mol H+ from HI ; mol = M x L
                               total L solution

              [H+]total = (0.020 M HCl x 0.0500 L) + (0.010 M HI x 0.150 L)
                                             0.200 L

              [H+]total = 1.0 x 10-3 mol H+ + 1.50 x 10-3 mol H+ = 0.013 M
                                       0.200 L

              pH = -log (0.0125) = 1.90




5. A 0.100 M solution of brmoacetic acid (BrCH2COOH) is 13.2% ionized.
Calculate [H+] , [BrCH2COO-], and [BrCH2COOH].
KEY



          [H+] = 0.132 x [BrCH2COOH]initial = 0.0132 m

                 BrCH2COOH (aq)  H+ (aq) + BrCH2COO- (aq)
Initial           0.100 M         0             0
Equil.            0.087           0.0132 M      0.0132 M

6. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.200 L to form a
solution with a pH of 3.25?
       [H+] = 10-pH = [C2H3O2-] [HC2H3O2] = x – [H+].
Substitute into the Ka expression and solve for x. solve

[H+] = 10-pH =10-2.90 = 1.26x10-3 = 1.3x10-3 M

Ka = 1.8x10-5 = [H+][C2H3O2-] = (1.26 x 10-3)2
                 [HC2H3O2]       (x-1.26x10-3)

1.8x10-5 (x – 1.26x10-3) = (1.26x10-3)2;

1.8x10-5 x = 1.585x10-6 + 2.266x10-8 = 1.608x10-6;

x = 0.08931 = 0.089 M HC2H3O2


7. What are tow kinds of molecules or ions that commonly function as weak bases?

Organic amines and anions that are the conjugate bases of weak acids function as weak
bases.

8. If a substance is a Lewis acid, is it necessarily a Bronsted-Lowery acid? Is it
necessarily an Arrhenius acid? Explain.

No. If a substance is a lewis acid, it is not necessarily a Bronsted or an Arrhenius acid.
Lewis acid is an electron pair acceptor which is not necessarily an H+ donor.

9. Which member of each pair produces the more acidic aqueous solution:

          a. ZnBr2 or CdCl2    ZnBr2, smaller cation radius, same charge

          b. CuCl or Cu(NO3)2 Cu(NO3)2, higher cation charge

          c. Ca(NO3)2 or NiBr2 NiBr2, smaller cation radius, same charge

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#24 Key

  • 1. KEY GENERAL CHEMISTRY-II (1412) S.I. # 24 Calculate the pH of each of the following strong acid solutions: 1. 0.00835 M HNO3 0.00835 M HNO3 = 0.0835 M H+; pH = -log(0.00835) = 2.08 2. 0.525 M HClO4 0.525 g HClO4 x 1 mol HClO4 = 2.612 x 10-3 = 2.61 x 10-3 M HClO4 2.00 L Soln 100.5 g HClO4 [H+] = 2.61 x 10-3 M; pH = -log (2.612 x 10-3) = 2.583 3. 5.00 mL of 1 M HCl diluted to 0.500 L Mc x Vc = Md x Vd; 0.500 L = 500 mL 1.00 M HCl x 5.00 mL HCl = Md HCl x 500 mL HCl Md HCl = 1.00 M x 5.00 mL = 1.00 x 10-2 M HCl = 1.00 x 10-2 M H+ 500 mL pH = - log (1.00 x 10-2) = 2.000 4. a mixture formed by adding 50.0 mL of 0.020 M HCl to 150 mL of 0.010 M HI. [H+] total = mol H+ from HCl + mol H+ from HI ; mol = M x L total L solution [H+]total = (0.020 M HCl x 0.0500 L) + (0.010 M HI x 0.150 L) 0.200 L [H+]total = 1.0 x 10-3 mol H+ + 1.50 x 10-3 mol H+ = 0.013 M 0.200 L pH = -log (0.0125) = 1.90 5. A 0.100 M solution of brmoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+] , [BrCH2COO-], and [BrCH2COOH].
  • 2. KEY [H+] = 0.132 x [BrCH2COOH]initial = 0.0132 m BrCH2COOH (aq)  H+ (aq) + BrCH2COO- (aq) Initial 0.100 M 0 0 Equil. 0.087 0.0132 M 0.0132 M 6. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.200 L to form a solution with a pH of 3.25? [H+] = 10-pH = [C2H3O2-] [HC2H3O2] = x – [H+]. Substitute into the Ka expression and solve for x. solve [H+] = 10-pH =10-2.90 = 1.26x10-3 = 1.3x10-3 M Ka = 1.8x10-5 = [H+][C2H3O2-] = (1.26 x 10-3)2 [HC2H3O2] (x-1.26x10-3) 1.8x10-5 (x – 1.26x10-3) = (1.26x10-3)2; 1.8x10-5 x = 1.585x10-6 + 2.266x10-8 = 1.608x10-6; x = 0.08931 = 0.089 M HC2H3O2 7. What are tow kinds of molecules or ions that commonly function as weak bases? Organic amines and anions that are the conjugate bases of weak acids function as weak bases. 8. If a substance is a Lewis acid, is it necessarily a Bronsted-Lowery acid? Is it necessarily an Arrhenius acid? Explain. No. If a substance is a lewis acid, it is not necessarily a Bronsted or an Arrhenius acid. Lewis acid is an electron pair acceptor which is not necessarily an H+ donor. 9. Which member of each pair produces the more acidic aqueous solution: a. ZnBr2 or CdCl2 ZnBr2, smaller cation radius, same charge b. CuCl or Cu(NO3)2 Cu(NO3)2, higher cation charge c. Ca(NO3)2 or NiBr2 NiBr2, smaller cation radius, same charge