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					Chapter 14. Acids and Bases
AP Chemistry 12
Reference:p. 653-662, Zumdahl


1. The Nature of Acids and Bases:




    How do we tell if a substance is an acid ?

               a) litmus paper



               b) react with magnesium to produce hydrogen



               c) pH scale




    How do we tell from a formula if a substance is an acid, base or salt ?

       HCl - hydrochloric acid                NaOH
       H2SO4 - sulfuric acid                  KOH
       HNO3 - nitric acid                     Sr(OH)2
       CH3COOH - acetic acid


2. Arrhenius Definition of an acid and base:

a) Acid: Any substance that increases the concentration of ____________________ions
in aqueous solution.


The structure of an H+ ion:

- often known as a PROTON.
- A proton in water undergoes the following equilibrium:




       The chemical species H3O+(aq) is referred to as a HYDRONIUM ion.



                                                                                  1
       In the above equilibrium, Keq =    ___________ =



    So... [H3O+] must _______________ the [H+]
    For this reason, we will use the terms proton, hydrogen ion and _______________
     ion interchangeably.



b) Base: Any substance that increases the concentration of hydroxide ions in aqueous
solution.
              KOH(s) --




But.... test a solution of Na2CO3 with litmus paper.




3. Bronsted-Lowry Definition

   o In order for an acid to work like an acid, there must also be a base acting like a base.


a) Bronsted-Lowry Definition of

       Acid:   substance which _______________a proton to another substance



       Base: substance which ______________ a proton from another substance



   o Getting back to our original demonstration, the Na2CO3 solution had to act as a base
      accepting a proton. Where did that proton come from ?




Example: ammonia dissolved in water:




                                                                                                2
b) Bronsted-Lowry Conjugate acid/base pairs

     As we said before, if a chemical is acting as an acid, it must lose a proton, either to
       water or to something else. When it loses its proton, it turns into a chemical
       species called CONJUGATE BASE.

       Examples :

       Bronsted Acid                                            Conjugate Base

               HCl
               HSO4-

               HPO42-


     Conversely, if a chemical is acting as a base, by gaining a proton, it will turn into its
       CONJUGATE ACID.

       Bronsted Base                                            Conjugate Acid


               CO32-

               CH3COO-
               HPO42-


   o Note in the above examples, HPO42- can either act as an acid or a base. It can either
     donate or accept a proton.
   o Such chemicals that can act as either an acid or a base are called
     AMPHIPROTIC (or amphoteric) in nature. The reasons as to how they will react are
     simple: it depends on what they are reacting with.



Definitions:
    Monoprotic Acid: acid which can supply 1 proton (HCl).
    Diprotic Acid: acid which can supply 2 protons (H2CO3).
    Triprotic Acid: acid which can supply 3 protons (H3PO4).
    Polyprotic Acid: a general definition of acids that can give away more than 1 proton.


It is all “relative” when it comes to Acid-Base Chemistry:

     As we said before, in order for an acid to act as an acid, there must be something
       willing to act as a base, that is, to accept the proton the acid will be donating. All
       acids have some 'degree of losing' their protons. This is known as the relative
       strength of the acid. We have a list, in order, of the relative strengths of
       Bronsted-Lowry Acids and bases.



                                                                                                  3
     In order for a chemical to act as an acid, it must be mixed with a chemical that is
       capable of accepting the proton (or acting as base).

     ―The Stronger Wins!!!‖ That is, if a substance is a stronger acid than another
       substance, then it will be the acid.

Strong Acids:
                 The top six acids (see table) are considered strong acids, that is, they can
                  only donate protons and they ionize in water to completion.
                 Equilibrium lies far on the right.



Weak Acids:
          Partial ionization! Establishes equilibrium with water....
          Equilibrium lies far on the left.
          The stronger the ―weak acid:, ____________________________!!!




   o Those chemicals that are amphiprotic will only donate a proton if it is mixed with a
     substance willing to accept it (weaker acid or a base).


                HCO3- + PO43- <====> CO32- + HPO42-

                              -
                   The HCO3 ion acts as an ________________ in this case because the
                    PO43- is capable of accepting the _______________.


So we can recognise an amphiprotic (or amphoteric) substance if
       1) __________________________________

       2) __________________________________

   o In every Bronsted-Lowry reaction there are two acid/base conjugate pairs.


       HCO3- +      PO43- <====>        CO32- + HPO42-



   o The actual position of the equilibrium will be determined by the relative strengths of
     the two acids in the equation. We will talk about that more later.

For a generic acid ―HA‖:          HA(aq) + H2O(l) < ==== > H3O+(aq) + A-(aq)

    In this case, HA is a stronger acid than H2O, or you can think of it as H2O has a
     greater affinity to H+.



                                                                                              4
              CHEMISTRY 12           BRONSTED ACIDS AND BASES
WS #1

1. Write the formula for the conjugate base for each of the following Bronsted acids.
              a) H2O                        f) HPO42-
               b) HI                         g) H2

               c) HNO2                       h) NH4+
               d) H3PO4                      i) NH3

               e) H2PO4-                     j) HCO3-

2. Write the formula for the conjugate acid for each of the following Bronsted bases.

               a) HO2-                       f) NH3

               b) SO42-                      g) H2PO4-

               c) PO43-                      h) HPO42-

               d) C2H3O2-                    i) N2H4

               e) NH2-                       j) C5H5N


3. Identify the conjugate acid-base pairs in the following equations :

   a) HSO4-(aq) + PO43-(aq) <===> SO42-(aq) + HPO42-(aq)


   b) C2H3O2-(aq) + H3O+(aq) <===> HC2H3O2 (aq) + H2O (l)


   c) HCO3-(aq) + H2PO4-(aq) <===> H2CO3 (aq) + HPO42-(aq)


   d) HNO3 (aq) + N2H4 (aq) <===> NO3-(aq) + N2H5+(aq)




   e) HIO3 (aq) + HC2O4-(aq) <===> H2C2O4 (aq) + IO3-(aq)



                                                                                        5
C. Lewis Definition of Acids and Bases:

       Acid: Acids are substances that accept __________________ pairs from other
       atoms, ions or molecules, forming coordinate covalent bond

       Bases: substances that donate electron pairs when forming coordinate covalent
       bond.

The Lewis Theory is mainly used to explain the formation of complexes.

Example:

       :NH3 + BCl3 --



The lone pair electrons from ammonia form a bond with the boron. Note that chlorine is
very electronegative, making boron slightly positive.

Example:
                     –
       CO2(g) + OH       (aq)   --- HCO3 – (aq)

       [H—O ] –          O==C==O --



D. Another View of Acids:
   Any compound that has one or more hydrogen atoms that are weakly bound from the
    rest of the molecule. When the compound is dissolved in water, the hydrogen ions ionize
    from the rest of the molecule.
   Most Acids are written with hydrogen as the first element.
   Organic acids are mainly due to the carboxylic acid functional group and are written as




Assignment: Exercises on p. 704 # 29, 31, 35, and 37




                                                                                          6
E. Acid strength can be explained using the concept of
electronegativity, bonding, and size of the ions.
Reference: p. 693-695
    Strength of acid is inversely proportional to the _______________ of the bond
       between the rest of the molecule and hydrogen.
    More electronegative anion tends to attract the electrons more, making the bond
       between the hydrogen and the anion _______________________.
               e.g. HBr is a ___________________ acid than H 2S




   As the anion increases in size, the bond between the hydrogen and the anion becomes
    weaker, thus increasing the strength of the acid:
        e.g. HBr is a stronger acid than HCl




   For oxoacids (acids that contain H, O, and another element), the strength depends upon
    the strength of the oxygen-hydrogen bond.

Example: sulphuric acid:




Example: a stronger acid is produced in the presence of more oxygen atoms (more
electronegative, causing a greater pull of electrons).
       Hyporchlorous acid               chloric acid               perchloric acid




Example: a more electronegative central atom results in a stronger acid.
       Telluric acid                selenic acid                    sulphuric acid




                                                                                          7
Example: a more electronegative central atom results in a stronger acid.

       Phosphoric acid        <       sulphuric acid          <       perchloric acid




Example: For organic acids, electronegative atoms such as F, Cl, Br, I, O, and S that are
near carbon atoms tends to withdraw electron density from the –O—H bond and increases
the strength of the acid.

Acetic acid    <       chloroacetic acid      <        trichloroacetic acid




Other Considerations:
1. Strong Bases.
All metal hydroxides are strong bases.
soluble hydroxides (group IA metals, Strontium, and barium) have a stronger effect.
Less soluble hydroxides do not provide too much hydroxide ions.




2. Weak Bases:

All bases related to ammonia are weak.
So, all amines and amides are weak base.



3. Base strength:

The presence of electronegative elements tends to decrease the strength of the organic
bases.




Assignment: Exercises on p. 707 # 113 and 115




                                                                                            8
                             Acid-Base Calculations

1. What is the [H+] in a 0.50 M HCl solution ?




2. What is the [H3O+] in a 15.0 M HNO3 solution ?




3. What is the [H+] in a 0.50 M CH3COOH solution ?




4. What is the [H3O+] in a 3.0 M solution of benzoic acid ?




                                                              9
5. What is the [H+] in water?




------------------------------------------------------------
     This idea of an equilibrium constant for water is a very useful tool. It is used so
       often, we give it its own symbol as well, kw. Because we usually dissolve acids (and for
       that matter bases) in water, if we can determine the [H+], then we can easily
      calculate the [OH-] by using the formula :




     This value for Kw 1.0 x 10-14 is only valid at 25o C. The Kw value changes due to a
      change in temperature.
     Autoionization of water




6. What is the concentration of Hydronium ion for 0.20 M solution of Nitrous acid?




Assignment:




                                                                                              10
                                   Calculation of Ka values

       If we are given the [H+] concentration and the concentration of the acid solution,
we can easily calculate the Ka value for the acid.

        Example #1 :


               An unknown weak acid, HW, is found to have a [H+] of 1.5 x 10-5 M when we
mix a 2.5 M solution of the acid. Calculate Ka for this acid.

        Answer #1 :


                 HW (aq) <====> H+(aq) + W-(aq)


               [H+][W-]              (1.5 x 10-5)(1.5 x 10-5)
        Ka = ___________          = ________________________
                   [HW]                (2.5 - (1.5 x 10-5))


                                      (1.5 x 10-5)2
                                   = _______________
                                         (2.5)



                                   = 9.0 x 10-11

------------------------------------------------------------
         Example #2 :


                 Another weak acid, HZ, has a 1.8 x 10 -8 M [OH-] in a 0.025 M solution of the
acid. Calculate the Ka of HZ.

        Answer #2 :


                 HZ (aq) <====> H+(aq) + Z-(aq)


        If [OH-] = 1.8 x 10-8 M the [H+] = 5.6 x 10-7 M


               [H+][Z-]           (5.6 x 10-7)(5.6 x 10-7)
        Ka = __________ =         ________________________
                   [HZ]             (0.025 - (5.6 x 10-7))


                                  (5.6 x 10-7)2



                                                                                            11
                         = ______________
                              (0.025)


                         =   1.2 x 10-11

- Assignment :

       - Ka worksheet question #2

       - Read sections 20-6, 20-7, 20-8




                                            12
Day 7.

- go over Ka worksheet question #2


                            Measuring [H+] concentrations

        We can measure the concentration of the hydronium ion a number of different ways:
by using a series of indicators such as phenolphthalein and methyl orange to give us an
approximate indication; by doing a titration with a standardized base solution and calculating
how much of it is needed to neutralize the acid; by using a mechanical device that actually
measures the concentration directly. in time we will use all of these methods to try and
determine the acidity of a certain solution. First though, let's do something to the physical
numbers we are using to 'clean them up' a little bit.
        As we saw the other day, a 0.50 M CH3COOH solution has a [H3O+] of 3.0 x 10-3 M
and a [OH-] of 3.3 x 10-12 M. Chemists, being a lazy breed, like to simplify things somewhat.
Instead of always having to write this power of 10 when dealing with [H3O+], they looked
for a shortcut. Along came a mathematician who suggested that they use a mathematical
device called a logarithm to express this power of 10. For example :


         the logarithm of 1 x 10-5 is -5.0
         the logarithm of 5 x 10-5 is -4.3 (because the number can be rewritten as 10 -4.3)

       Chemists thought that this was a pretty good idea but they were still stuck with a
negative sign. In order to eliminate this (and to give the impression that it was their idea),
they developed a system of writing the [H3O+] in terms of negative logarithms of the
concentration. Because the first digit(s) (those preceding the decimal) in the logarithm has
everything to do with the POWER of 10, chemists decided to call this conversion pH. (p for
power H for H+)
         pH is therefore defined as the following :


         pH = -log[H+]

         We can also define pOH :


         pOH = -log[OH-]




                                                                                                 13
       Example :


               [H3O+] = 3.0 x 10-3 M


               pH = -log(3.0 x 10-3) = 2.52

        NOTE : Because the numbers in front of the decimal in a pH or pOH indicate which
power of 10 we are talking about, they are NOT counted when considering significant
figures. Only the digits AFTER the decimal point are significant in a logarithmic number.

       More examples :


               Calculate pH for the following [H+]


       a) 5.4 x 10-12 M        pH = 11.27
       b) 5.4 x 10-7 M         pH = 6.27
       c) 15.0 M               pH = -1.176
       d) 1.9 x 10-17 M       pH = 16.72


There is a relationship between [H+] and [OH-] that we have seen before :
       Kw = [H+][OH-]
At 25oC
       1.0 x 10-14 = [H+][OH-]

Let's see if we can make a logarithmic relationship for these.
        If [H3O+] = 1.0 x 10-5 M, what is [OH-] ?
                [OH-] = 1.0 x 10-9 M
       If [H3O+] = 1.0 x 10-5 M, what is pH and pOH ?
              pH = -log(1.0 x 10-5) = 5.00
               pOH = -log(1.0 x 10-9) = 9.00

       Notice what happens when we ADD pH and pOH
We get 14.00. No matter what numbers we pick for [H3O+], at 25oC,
       pH + pOH = 14.00. WHY ???


       1.0 x 10-14 = [H+][OH-] if we LOG both sides of the equation, we get (trust me)
               14.00 = pH + pOH

In fact, at any temperature the following applies :

       pKw = pH + pOH




                                                                                         14
                                      pH calculations

      Given the same set of information as in previous examples, we can calculate the pH
and pOH of a certain concentration of a weak acid.

       Example :
           Calculate the pH of a 1.0 x 10-2 M solution of HCN.

       Solution :
                         HCN <===> H+ + CN-     Ka = 4.8 x 10-10


       I         1.0 x 10-2      0        0
       C                 -X     +X       +X
       E      1.0 x 10-2- X      X        X

       X is insignificant


                [H+][CN-]                               X2
       Ka =     ____________         4.8 x 10-10 =   _____________

                 [HCN]                                    1.0 x 10-2


                                     4.8 x 10-12 =   X2


                                     2.2 x 10-6 = X = [H+]


                pH = -log[H+] = -log(2.2 x 10-6) = 5.66



Assignment:

       Read section 20-10
       Ka worksheet questions #3, 4

- go over Ka worksheet questions #3, 4

pH worksheet # 1,2,3 odds.




                                                                                           15
Lesson!


Calculations Involving Weak Acids:
Example: Determining the pH.

The hypochlorite ion (OCl-) is a strong oxidizing agent often found in household bleaches and
disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine.
In addition to its oxidizing abilities, the hypochlorite ion has relatively high affinity for protons (it is a
much stronger base than Cl-, for example) and forms the weakly acidic hypochlorous acid (HOCl, K a= 3.5
x 10 –8). Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid. (pH=4.23)




                                                                                                            16
Example 2: pH of a mixture of acids.

Calculate the pH of a solution that contains 1.00 M HCN (Ka= 6.2 x 10-10) and 5.00 M HNO2 (Ka=4.0 x 10 –
4
  ). Also calculate the concentration of cyanide ion in this solution at equilibrium.




                                                                                                     17
Example 3: Percent dissociation (or percent ionization):

                           % dissociation = amount dissociated           X 100%
                                               initial concentration

Calculate the % dissociation of acetic acid (Ka= 1.8 x 10 –5) in each of the following solutions:
a) 1.00 M CH3COOH
b) 0.100 M CH3COOH




                                                                                                    18
Example 4:

Lactic Acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to
pain and a feeling of fatigue. In a 0.100M aqueous solution, lactic acid is 3.7% dissociated. Calculate the
Ka for this acid. (1.4 x 10 –4)




                                                                                                          19
AP Chemistry 12
Weak Acid problems
1. A solution is prepared by mixing 90.0 mL of 5.00M HCl, 30.0 mL 8.00M HNO3.
   Water is then added until the final volume is 1.00L. Calculate [H+], [OH-], and the pH of
    this solution. (1.9M, 5.3x 10-15M, -0.28)

2. Boric acid, H3BO3, is commonly used in eyewash solutions in chemistry laboratories to neutralize
    bases splashes in the eye. It acts as a monoprotic acid, but the dissociation reaction is slightly different
    from that of other acids:

        B(OH)3 + H2O < == > B(OH)4 - + H+         Ka= 5.8 x 10-10

 Calculate the pH of a 0.50 M solution of boric acid. (4.77)

3. Calculate the percent dissociation for a 0.22M solution of chlorous acid (HClO 2, Ka=1.2 x10-2). (21%)

4. Calculate the pH of a solution containing a mixture of 0.10 M HCl and 0.10 M HOCl.
    (1.00)




Additional Assignment:

                                          Read p. 682-693

Review questions:
p. 696 # 24, 26-28

Review Problems:

p. 696 # 49, 51, 59, 62




                                                                                                             20
Lesson!


                                        INDICATORS

       Indicators are weak organic acids or bases. These chemicals change colours in the
presence of certain chemicals such as acids or bases. Acid-base indicators are weak acids
that have different colours in their acid form and their conjugate base form.

We can abbreviate the acid as H In.

                                                        -
                    HIn +        H 2O    == In               +    H 3O +

Example:       Bromcresol green has the following equilibrium :


                       HBg + H2O <====> H3O+ + Bg-

       When bromcreosol green is added to an acid, it turns yellow. This is because the
extra H+ ions in the acid upset the equilibrium for the indicator, shifting it towards the
left. The HBg form of the indicator (the conjugate acid form) is favoured. Because the
solution turns yellow, the HBg form must be yellow in colour.


                       HBg + H2O => H3O+ + Bg-
                  Conj. Acid               Therefore, the indicator is in its conj. Acid if
                                           it is in acidic solution.

        When bromcreosol green is added to a base, the additional OH - ions react with the
H+ ions in the indicator equilibrium lowering the [H+]. This will shift the equilibrium right
producing more Bg- ions. Because the solution turns blue, this must mean that the basic
form of the indicator is blue.


                       HBg + H2O <= H3O+ + Bg-
                                           Conj. Base
                                                               An indicator is in its conj.
                                                               Base form in the presence of
                                                               A base.

        If we added a base slowly to the yellow acidic solution, there comes a time when the
solution goes through a transition point in colour. In this case, where the solution turns
green. At this point the yellow and blue forms are in equal numbers hence an intermediate
colour is seen. What exactly is happening here ?




                                                                                             21
        If we look at the Ka expression for bromcreosol green we have


                         [H+][Bg-]
                Ka =    ____________
                          [HBg]


        At the transition point, [Bg-] = [HBg]. Why?

Because if they did not equal each other, the solution would be either yellow or blue, and not
between the two. If this is the case, let's see what happens to the K a expression:


                         [H+][Bg-]
                Ka =    ____________                [H+]
                                                =
                          [HBg]

        So we can see that at the transition point, the [H+] is the same as the Ka value for
the indicator (this is only true for indicators!). All indicators have a different K a value,
therefore they all have different points at which they change colour.

Note that if we keep this in mind, we can come up with a relationship for indicators:

        pKa = pH

        If we look at the indicator table, we can see that there is a range at which the
indicator will change.

For example, methyl violet is yellow at a pH level less than 0.0 while it is blue at pH levels
greater than 1.6. During the interval 0.0 - 1.6, the indicator is in different shades of green.

 For our purposes, we will take the midpoint of the range (0.8) as the exact transition
point of the indicator.

        Example : What is the Ka of methyl violet ?
        Solution :
                Ka = [H+] = antilog (-0.8)
                                 = 0.2

Universal Indicators: made up of mixtures of several indicators. This solution changes
colour several times over a range of pH values.




                                                                                              22
Estimating the pH of a solution by using more than one indicator.

If we use two or more indicators, we can isolate the pH level of a certain solution between
certain limits.

Example :
       A solution turns phenolphthalein magenta and indigo carmine blue. What is the range
of pH that the solution can have ?

Solution :
Phenolphthalein will be magenta only if the pH > 10.0. Indigo carmine will be blue if pH <
11.4.

        Therefore the pH must be between 10.0 and 11.4



Colours of indicators:

If we can determine either the pH or the pOH of a solution, we can compare this to the
data table to determine what colour a specific indicator will be in this solution.

Example : What colour will Bromcresol green be in the following solutions :

               a) [H+] = 1.9 x 10-11    (pH = 10.72 - blue)
               b) pOH = 7.8             (pH = 6.2 - blue)
               c) pH = 4.0             (green)

Example : What colour would chlorophenol red be in a 0.050 M solution of HCN ?

Solution :
                        HCN is a weak   acid therefore this is a Ka
                        calculation.
               HCN <====>       H+ +     CN-                     Ka = 4.8 x 10-10
        I       0.050           0       0
        C      -X               +X      +X
        E      0.050-X           X       X
               X is insig

               [H+][CN-]                                          X2
                                             -10
        Ka =    ---------       = 4.8 x 10         =   -------
                 [HCN]                                           0.050

                                 2.4 x 10-11 = X2

                                 4.9 x 10-6 = X = [H+]



                                                                                              23
                      pH = -log[H+] = -log(4.9 x 10-6) = 5.31

               With a pH of 5.31, chlorophenol red would be
               ORANGE. (an intermediate colour)

Assignment :

      -   Preparation of Lab 20-B
      -   Complete the pH worksheet
      -    indicator worksheet




                                                                24
     CHEMISTRY 12                           INDICATOR WORKSHEET

1. What is the approximate Ka for the following indicators :

      a) Orange IV
      b) Alizarin Yellow
      c) Bromthymol blue

2.    What colour would thymol blue be in the following
      solutions ?

      a)    a solution of pH 7.0
      b)    a solution of pOH 5.0
      c)    0.050 M HClO4
      d)    6.3 x 10-6 M LiOH
      e)    2.0 M benzoic acid
      f)    1.0 x 10-2 M hydrogen sulfide
      g)    3.1 x 10-5 M Ca(OH)2

3.    Determine the approximate pH of the following
      solutions:

      a) A solution which turns orange in chlorophenol red
      b) A solution which turns red in methyl red and blue
         in methyl violet.
      c) A solution which turns yellow in alizarin yellow,
         blue in bromcreosol green and magenta in
         phenolphthalein.




ANSWERS :

1.    a)   2 x 10-2
      b)   4 x 10-11
      c)   5 x 10-7
2.    a)   yellow
      b)   green
      c)   orange
      d)   green
      e)   orange
      f)   yellow
      g)   blue
3.    a)   5.2 - 6.8
      b)   1.6 - 4.8
      c)   10.0 - 10.1




                                                                  25
day 10
                                         Lab 20-B

        Use acetate sheets instead of glass plates to do lab on
        Be consistent with colours. If you say that a certain indicator goes blue in HCl and
red in NaOH, use those colour descriptions for the remaining acids in your data table.

         After procedure discussion :

        Question #1 :
                  Draw another data table similar to the one in the lab. Compare the relative
strengths of the acids to the indicators. In order to do this, we must compare the colours
that the indicators changed with HCl and NaOH. Because HCl is a stronger acid than the
indicator, any other acid that is stronger than the indicator will go the same colour as the
HCl. Conversely, because the NaOH is a weaker acid than the indicator, any acid that went
the same colour as NaOH must be a weaker acid than the indicator.
        Fill in the new data table with > and < signs comparing the acid to the indicator.

       Question #2 :
               Deductive reasoning time. Assume for arguments sake the following table
was part of what you got in question #1 :


                        HIn1/In1-       HIn2/In2-      HIn3/In3-
HA1/A1-                    >               <               >
HA2/A2-                    >               <               <

        Now the logic begins :
        According to the data, HA1 > HIn1 but < HIn2. This means that it must be between
them (in strength). It must also be stronger than HIn3. So far we can deduct the following:

             HIn2
             HA1
             HIn1 HIn3 at this time, we can't tell these two apart.
       Now, HA2 , like HA1, is between HIn1 and HIn2 but HA2 is weaker than HIn3
whereas HA1 was stronger than HA3 . If this is the case, HA1 MUST be stronger than HA2
AND HIn3 MUST be stronger than HIn1. Rewriting the logic pattern we get :
             HIn2
             HA1
             HIn3
             HA2
             HIn1

       Continue this logic for the complete table. When that is completed, question 2 also
asks you to write the dissociation equations for all of these weak acids. For example :
                HIn1 <====> H+ + In1-



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         After all the equations have been completed, you must draw two arrows - one stating
increasing strength of acids and the other indicating increasing strength of bases. The final
result should look similar to your Ka table (except for the fact that you have no Ka values to
put in).

        Follow-up Question #1 :
                 instead of using the appendix in the lab manual to identify the indicators
used in this lab, give out a copy of the acid/base indicator table from government exam.

- Assignment :
        Lab 20-B Questions #1,2
                     Follow-ups #1,2,3
                             (due in 2 days)

- Quiz tomorrow on Ka and pH worksheets




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