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                            GENERAL CHEMISTRY-II (1412)
                                     S.I. # 20

1. Calculate the concentration of H+ (aq) in (a) a solution in which [OH-] is 0.010 M,
(b) a solution in which [OH-] is 1.8x10-9 M.
       a. using [H+][OH-] = 1.0 x 10-14
       [H+] = 1.0 x 10-14 = 1.0 x 10-14 = 1.0 x 10-12M
                 [OH-]        0.010

       The solution is basic because [OH-] > [H+]

       b. [H+] = 1.0 x 10-14 = 1.0 x 10-14 = 3.6 x 10-6M
                 [OH-]          1.8 x 10-9

       The solution is acidic because [H+] > [OH-]

2. Define pH and give the formula by which it is calculated.
        The molar concentration of H+ (aq) in an aqueous solution is usually very small.
For convenience, we therefore usually express [H+] in terms of pH, which is the negative
logarithm in base 10 of [H+].

                  pH = -log[H+]

3. Calculate the pH of a solution when the value of [H+] is 1.0 x 10-7M and when [H+]
is 1.0 x 10-3M.
                  pH = -log (1.0 x 10-7) = -(-7.00) = 7.00 = Neutral
                  pH = -log (1.0 x 10-3) = -(-3.00) = 3.00 = Acidic

4. Give the pH of a solution when the value of [OH-] is 2.0 x 10-3M.
                  [H+] = Kw = 1.0 x 10-14 = 5.0 x 10-12M
                        [OH-] 2.0 x 10-3

                  pH = -log (5.0 x 10-12) = 11.30 = Basic

5. Determine [H+] if the pH of a sample is 3.76.
       pH = -log[H+] = 3.76
       log[H+] = -3.76
       [H+] = antilog(-3.76) = 10-3.76 = 1.7 x 10-4M.
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6. Provide the equation for determining pOH and show how it is related to pH.

        pOH = -log[OH-]
take the negative log of both sides  -log[H+] + (-log[OH-]) = -log Kw
we obtain the following expression: pH + pOH = 14.00 at 25°C.

7. List the names and formulas for the 7 strong acids and the 7 strong bases.

          7 STRONG ACIDS                            7 STRONG BASES
 1.    HCl – hydrochloric acid                  NaOH – sodium hydroxide
 2.    HBr – hydrobromic acid                   KOH – potassium hydroxide
 3.    HI – hydroiodic acid                     Ba(OH)2 – barium hydroxide
 4.    HNO3 – nitric acid                       LiOH – lithium hydroxide
 5.    HClO3 – chloric acid                     RbOH – rubidium hydroxide
 6.    HClO4 – perchloric acid                  CsOH – cesium hydroxide
 7.    H2SO4 – hydrogen sulfate                 Sr(OH)2 – strontium hydroxide

8. List 3 main points of strong acids

       1.Completely ionized

       2. Only ions are present

       3. No equilibrium

List 3 main points of weak acids

       1. Not completely ionized

       2. Molecules and Ions present

       3. Equilibrium possible, Eq const.


9. What do weak acids and weak bases have in common?

       They are weak electrolytes that ionize only to a limited extent in water and
contain Ionized and non-ionized species.

10. Why is HCl- as strong acid and a weak base?

       Because Cl- will not react with H+. If it did, HCl(aq) {a strong acid} would form
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      which does not exist

#20 Key

  • 1.
    KEY GENERAL CHEMISTRY-II (1412) S.I. # 20 1. Calculate the concentration of H+ (aq) in (a) a solution in which [OH-] is 0.010 M, (b) a solution in which [OH-] is 1.8x10-9 M. a. using [H+][OH-] = 1.0 x 10-14 [H+] = 1.0 x 10-14 = 1.0 x 10-14 = 1.0 x 10-12M [OH-] 0.010 The solution is basic because [OH-] > [H+] b. [H+] = 1.0 x 10-14 = 1.0 x 10-14 = 3.6 x 10-6M [OH-] 1.8 x 10-9 The solution is acidic because [H+] > [OH-] 2. Define pH and give the formula by which it is calculated. The molar concentration of H+ (aq) in an aqueous solution is usually very small. For convenience, we therefore usually express [H+] in terms of pH, which is the negative logarithm in base 10 of [H+]. pH = -log[H+] 3. Calculate the pH of a solution when the value of [H+] is 1.0 x 10-7M and when [H+] is 1.0 x 10-3M. pH = -log (1.0 x 10-7) = -(-7.00) = 7.00 = Neutral pH = -log (1.0 x 10-3) = -(-3.00) = 3.00 = Acidic 4. Give the pH of a solution when the value of [OH-] is 2.0 x 10-3M. [H+] = Kw = 1.0 x 10-14 = 5.0 x 10-12M [OH-] 2.0 x 10-3 pH = -log (5.0 x 10-12) = 11.30 = Basic 5. Determine [H+] if the pH of a sample is 3.76. pH = -log[H+] = 3.76 log[H+] = -3.76 [H+] = antilog(-3.76) = 10-3.76 = 1.7 x 10-4M.
  • 2.
    KEY 6. Provide theequation for determining pOH and show how it is related to pH. pOH = -log[OH-] take the negative log of both sides  -log[H+] + (-log[OH-]) = -log Kw we obtain the following expression: pH + pOH = 14.00 at 25°C. 7. List the names and formulas for the 7 strong acids and the 7 strong bases. 7 STRONG ACIDS 7 STRONG BASES 1. HCl – hydrochloric acid NaOH – sodium hydroxide 2. HBr – hydrobromic acid KOH – potassium hydroxide 3. HI – hydroiodic acid Ba(OH)2 – barium hydroxide 4. HNO3 – nitric acid LiOH – lithium hydroxide 5. HClO3 – chloric acid RbOH – rubidium hydroxide 6. HClO4 – perchloric acid CsOH – cesium hydroxide 7. H2SO4 – hydrogen sulfate Sr(OH)2 – strontium hydroxide 8. List 3 main points of strong acids 1.Completely ionized 2. Only ions are present 3. No equilibrium List 3 main points of weak acids 1. Not completely ionized 2. Molecules and Ions present 3. Equilibrium possible, Eq const. 9. What do weak acids and weak bases have in common? They are weak electrolytes that ionize only to a limited extent in water and contain Ionized and non-ionized species. 10. Why is HCl- as strong acid and a weak base? Because Cl- will not react with H+. If it did, HCl(aq) {a strong acid} would form
  • 3.
    KEY which does not exist