1. Construct table and provide CA/CB (if any): CrO₄^-2   HC₂O₄^-1   NH₄⁺¹   HClO₄    PCl₃     (10)
2. List theories (Arrenhius, Bronsted-Lowry, or Lewis) that are applicable: (10)
    (a) Br + Br Br₂              (b) HClO₃ + H₂O  H₃O⁺¹ + ClO₃^-1
    (c) Cl⁺¹ + Cl^-1 Cl₂          (d) H + Cl  HCl           (e) HNO₃ + H₂S  H₃S⁺¹ + NO₃^-1

3. (a) For HAc/Ac^-1       Prove K_w = K_a^. K_b     (12)
    (b) Explain how  HAc/NaAc works as buffer.
    (c) Using Ka/Kb, explain why NH₄CN is basic in water.

4. Calculate pH resulting from mixing:     (16)
    (a) 40 mL of 1.00 M HCl and 60 mL of 1.00 M NaOH.
    (b) 50 mL of 1.00 M HCN and 30 mL of 1.00 M NaOH.

5. Calculate pH:    (20)
   (a) 3.00 M H₃PO₄    (b) 10.0 M Na₃PO₄    (c) 0.100 M NaClO (d) 0.50M NaHCO₃

6. When 2.95 g of C₃H₉N diluted to exactly 1 liter, pH of 11.40 results.    Calculate K_b     (8)
7.  Calculate pH, [HF] & [F^-1] for 0.200 M HF.      (8)

8. Calculate Ka if 1.0M weak acid ionizes 3.0%. (8)

9.  1.00 L of buffer contains 1.00 mol HAc and 0.50 mol NaAc:     (12)
     (a) Calculate pH after adding 0.25 mol OH^- (assume volume remains 1.00L)
     (b) Calculate pH after adding 0.25 mol H⁺ (assume volume remains 1.00L)

Answers
1.           CrO₄^-2          HC₂O₄^-1      NH₄⁺¹       HClO₄      PCl₃
     CA  HCrO₄^-1      H₂C₂O₄     none             none         HPCl₃⁺¹
     CB      none              C₂O₄^-2       NH₃             ClO₄^-1       none

2.  (a) none    (b) Arrenhius/BL/Lewis    (c) Lewis     (d) none     (e) BL/Lewis

3. (a)     HAc  H⁺¹  +  Ac^-1      Ac-  HAc  +  OH^-1
           K_a = [H⁺¹][Ac^-1]/[HAc]K_b = [HAc][OH^-1]/[Ac^-1]K_a^. K_b  = [H⁺¹][OH^-1]  =  K_w
    (b)  If additional base added, H⁺¹ combines with OH^-1 and shift takes place to the RIGHT
          If additional acid added, Ac^-1 combines with H⁺¹  and shift takes place to the LEFT

    (c)  NH₄⁺¹ H⁺¹ + NH₃     K_a = 5.6x10^-10
          CN^-1  HCN + OH^-1    K_b = 1.6x10^-5      Solution basic because  K_b > K_a

4.   (a) In:  0.040 mol   0.060 mol
                 HCl     +    NaOH   ®NaCl   +   H₂O
         Eq:     0             0.020 mol
        NaOH in excess by 0.020 mol     [OH^-1] = 0.020mol/0.10L = 0.20M   pOH = 0.70   pH = 13.30

     (b) In:  0.050 mol   0.030 mol
                HCN     +    NaOH NaCN   +   H₂O
         Eq:   0.020 mol        0            0.030 mol

        HCN/NaCN buffer now exists
        Ka = [H⁺][CN^-]/[HCN] = [H⁺][0.030mol/0.080L]/[0.020mol/0.80L] = 6.2x10^-10
        [H⁺] = 4.1x10^-10 M          pH = 9.38

5.  (a)   H₃PO₄ H⁺¹  +  H₂PO₄^-1      K_a = 7.1x10^-3 = x²/3    x = [H⁺¹] = 0.15 M    pH = 0.84
              3-x             x              x
    (b)   PO₄^-3HPO₄^-2   +   OH^-1     K_b = 2.22x10^-2 = x²/10    x = [OH^-1] = 0.47 M     pOH = 0.33
           10-x            x                 x                                                                                  pH = 13.67

    (c)   ClO^-1 HClO   +   OH^-1         K_b = 3.3x10^-7 = x²/0.1     x = [OH^-1] = 1.8x10^-4    pOH = 3.74
            0.1-x         x               x                                                                                       pH = 10.26

 (d)   HCO₃^-1 H⁺¹  +  CO₃^-2              Ka = 4.7x10^-11
        HCO₃^-1   H₂CO₃  +  OH^-1          Kb = 2.2x10^-8     Since Kb > Ka, solution BASIC
        0.5-x ~ 0.5       x               x
                                                           x²/0.5 = 2.2x10^-8       x = [OH^-1] = 1.1x10^-4
                                                          pOH = 3.96            pH = 10.04

6.   mol B = 2.95/59.0 = 0.0500M    [B] = 0.0500 M     when pH = 11.40,  [OH^-1] = 0.0025 M
            B   =    BH⁺¹   +   OH^-1
        0.0475M    0.0025M      0.0025M
         K_b = [BH⁺¹][OH^-1]/[B] = (0.0025)²/0.0475 = 1.3x10^-4

7.     HF   H⁺¹  +  F^-1       Ka = 6.8x10^-4
      0.2-x          x          x

      x²/(0.2-x) = 6.8x10^-4 = x²/0.2
     x = 0.012      pH = 1.92    [F-1] = 0.012M     [HF] = 0.19M

8.   HA   H⁺¹  +  A^-1
      0.97         0.03    0.03      Ka = [H⁺¹][A^-1]/[HA] = (0.03)2/0.97 = 9.3x10^-4

9.   (a) Addition of 0.25M OH^-1 causes 0.25M HAc to shift RIGHT
           at equilibrium:  [HAc] = 0.75M      [Ac^-1] = 0.75M
           K_a = [H⁺¹][Ac^-1]/[HAc] = [H⁺¹][0.75]/[0.75] = 1.8x10^-5
           [H⁺¹] = 1.8x10^-5     pH = 4.74
    (b) Addition of 0.25M H⁺¹ causes 0.25M Ac^-1 to shift LEFT
           at equilibrium:  [HAc] = 1.25M      [Ac^-1] = 0.25M
           K_a = [H⁺¹][Ac^-1]/[HAc] = [H⁺¹][0.25]/[1.25] = 1.8x10^-5
           [H⁺¹] = 9.0x10^-5     pH = 4.05