Acids Bases and Salts

1. Electrolytes

 Electrolytes are compounds that ionize in water to produce aqueous solutions that conduct an electric current. Nonelectrolytes are substances that do not ionize, remain as molecules, and do not conduct an electric current.

Strength: Strong electrolytes are molecules that ionize 100% (or nearly so) and conduct an electric current well. Weak electrolytes barely or partially ionize; most molecules remaining un-ionized, and conduct an electric current poorly.

Examples: Nitric acid (HNO3) is a strong electrolyte. HNO3 => H+1 + NO3-1

If 1.00 mole of HNO3 is dissolved in water, it will ionize to produce 1.00 mole of H+1 and 1.00 mole of NO3-1 . There will not be any HNO3 left, unionized. By contrast, one mole of a weak electrolyte will produce much less than 1.00 mole of its constituent ions. One mole of acetic acid

HC2H3O2 <=> H+1 + C2H3O2-1

might produce .05 moles of H+1 and .05 moles of C2H3O2-1 and still contain .95 moles of the original acetic acid.

Some problems with detailed solutions:

1. Calculate the concentrations of ions in .020 M HCl solution. HCl is a strong electrolyte.

  HCl => H+1 + Cl-1
moles/liter before ionization .020    
moles/liter after ionization none .020 .020

2. Calculate the concentrations of ions in .015 M Ba(OH)2 solution. Ba(OH)2 is a strong electrolyte.

First, how will Ba(OH)2 ionize? Ba, in group 2A, should ionize by losing 2 electrons to form a +2 ion. [Recall that during the mass spectrometry/combining capacity unit you discovered that Ba would have a combining capacity of 2; and during the ionization potential unit, you discovered that this was because Ba had two outer electrons with very low ionization potentials.] Barium's electronegativity is 1.0 indicating a very weak hold on electrons. Oxygen's electronegativity is 3.5, meaning an electronegativity difference of 2.5, a condition for ionic bonding. What about the two OH groups? H's electronegativity is 2.1, so there is an electronegativity difference with oxygen of 1.4, suggesting a polar but NOT an ionic bond between O and H. Where will barium's two lost electrons go? One will go to each OH group.

  Ba(OH)2 => Ba+2 + 2OH-1
moles/liter before ionization .015    
moles/liter after ionization none .015 .030

3. Calculate the concentrations of ions in a solution that contains .92 g of magnesium bromide in 500 mL of solution.
Magnesium bromide, with its big electronegativity difference, is a strong electrolyte. The formula, molecular weight and ionic charges are easily obtained from the periodic table.

formula: MgBr2 molecular weight: 24.3 + 80.0 + 80.0 = 184.3
ionic charges: Mg+2 +2Br -1

Molarity = (.92/184.3) moles /.500 liter = .010 moles/liter

  MgBr2 => Mg+2 + 2Br -1
moles/liter before ionization .010    
moles/liter after ionization none .010 .020

More problems:

Problem Solution
What are the concentrations of the ions in .010 M NaCl? [Na+1] = [Cl -1] = .010 M
What are the concentrations of the ions in .015 M K2SO4 [K+1] = .030M, [SO4 -2] = .015M
3.63 g of KAl(SO4)2 are dissolved in .250 L of solution. What are the ionic concentrations? [K+1] = [Al+3] = .0562 M,
[SO4 -2] = .112 M
.269 g HNO3 are added to 36.3 mL of 1.18 M HNO3 solution. Assume no change in solution volume. What are the final concentrations of the ions? [H+1] = [NO3-1] = 1.30 M

 

 

 

Acid-Base 2

2. Arrhenius Theory

Acids and Bases Defined:

Acid

 An acid is something that contributes hydrogen ions, H+, to solution. For example: HCl in water ionizes to H+ + Cl-
                HCl   =>  H+ + Cl-
hence an aqueous solution of HCl is acidic.

Base

 A base is something that contributes hydroxyl ions, OH-, to solution. For example: NaOH in water ionizes to Na+ + OH-
             NaOH => Na+ + OH-
hence an aqueous solution of NaOH is basic.

 

Water

 When a water molecule ionizes, both hydrogen and hydroxyl ions are formed, hence water is considered neutral
H2O <=>  H+ + OH-
Kw The equilibrium constant for the reaction is given by the expression
                              [H+]   [OH-]
                  Keq = ----------------
                                  [H2O]
Most water molecules do not ionize.
   Water is 55.6M               (1000 g/ L = (1000/18) moles/L = 55.6 M
    [H+] = [OH-] = .0000001 M
   This means that only 1 in 556000000 water molecules ionize! The other 555999999 remain H2O!
So the concentration of water, 55.6M, is, for all practical purposes, constant. Since Keq is also constant, the equation may be rewritten
    Kw = Keq [H2O] =  [H+]   [OH-]  or more simply Kw =  [H+]   [OH-] 
    Experimentally, Kw is found to be 1.0 E -14
[H+], [OH-] If Kw =  [H+]  [OH-] = 1.0 E -14, and if [H+] = [OH-] = x, then x2  = 1.0 E -14, and x = 1.0 E-7.
From this, we can see that [H+] = [OH-] = .0000001, as stated above.

Strong Acids

A strong acid ionizes 100%
For example, 0.1 M HCl ionizes to 0.1 M  [H+] and 0.1 M [Cl-]
HCl  =>     H+   +    Cl-
0.1 M       none       none      before ionization
none         0.1 M    0.1 M     after ionization

Because HCl is a strong acid, the reaction goes to completion, leaving no HCl.
Examples of Strong Acids
Formula
HCl
HBr
HI
HNO3
HClO4
H2SO4		 Name
hydrochloric
hydrobromic
hydroiodic
nitric
perchloric
sulfuric

 

Determination of [H+] and [OH-] for a strong acid

Consider a .050M nitric acid solution
In 1 L of water we have		 H2O       <=>      H+     +      OH-
55.6 moles   1E-7 moles   1E-7 moles
to which we add .05 moles of HNO3


which ionizes 100% to

 HNO3    <=>    H+     +    NO3-
.050 moles

none           .050 moles    .050 moles
The total [H+] = .050 + 1E-7 = .050
This affects the water equilibrium

But Kw = 1 E-14 at equilibrium.
Clearly we are not at equilibrium.
[.050] [1E-7] is not 1 E -14!
Addition of the acid has disturbed the equilibrium.
Some H+ will combine with OH- to produce H2O		 H2O <=>          H+      +       OH-
55.6 moles    .050 moles     1E-7 moles


                     .050-x           1E-7 -x

Kw = 1 E-14 = (.050-x)(1E-7-x)
       x must be less than 1E-7
            so .050-x = .050
Kw = 1 E-14 = (.050)[OH-]
         [OH-] = 2E-13
We now know that in a .050M HNO3   solution
[H2O] = 55.6M
[ H+] = .050M
[OH-] = 2E-13
[NO3-] = .050		  

A simpler solution!

In a .050 M HNO3 solution, [ H+] = .050M; [NO3-] = .050; and [H2O] = 55.6M.
We need only calculate [OH-] from Kw. Kw = 1 E-14 = (.050)[OH-]; [OH-] = 2E-13

 

Strong Bases

A strong base ionizes 100%
For example, 0.30 M NaOH ionizes to 0.30 M  [Na+] and 0.30 M [OH-]
NaOH  =>  Na+   +  OH-
0.30 M       none       none      before ionization
none         0.30 M    0.30 M     after ionization

Because NaOH is a strong base, the reaction goes to completion, leaving no NaOH.

Examples of Strong Bases

Formula
LiOH
NaOH
KOH
RbOH
CsOH
Ca(OH)2
Sr(OH)2
Ba(OH)2		 Name
lithium hydroxide
sodium hydroxide
potassium hydroxide
rubidium hydroxide
cesium hydroxide
calcium hydroxide
strontium hydroxide
barium hydroxide


Degree of Acidity

pH

Defined

Since concentrated acids have  [ H+] equal to about 10M, very dilute acids have  [ H+] of about .00001M, and strong bases, about .000000000000001 M, we have a very wide range of numbers. Such numbers are difficult to work with and almost impossible to graph. Any graph that would show the point .000000000000001 would not be able to represent the number 10 on the same planet. If each 1E-15 represented 1 mm on a number line, then the number 10 would be .1E16 mm away. This is a distance of 621,000,000 miles.

number log(number)
10 1
1 0
0.1 -1
0.01 -2
.001 -3
E-10 -10


In order to simplify dealing with such a large range of numbers, we use an exponential scale based upon logarithms. In general, the log 10x   = x. From this equation, the table to the left may be generated. Notice, however, that most of the [ H+] will be negative. Only when [ H+] > 1M will the log be positive. Since we would prefer to work with positive numbers, pH has been defined as

                   pH = - log [ H+]

The effect of introducing the negative sign is that a high pH indicates a low degree of acidity.

  Calculating pH

1. What is the pH of a solution with [ H+] = .010M?
   (answer)  pH = - log[ H+] = - log 1.0E-2 = - (-2.0) = 2.0
2. What is the pH of a solution with [ H+] = .050M?
   (answer) pH = - log [ H+] = - log 5.0 E-2 = ......
        Since .050 is between .010 and .10, the pH must be between 2 and 1.
        Enter .050 into your calculator, then hit the log key, and retrieve an answer of -1.301
        So then, pH = - log 5.0E-2 = - (-1.301) = 1.301
3. The pH of a solution is 3.301. What is the [ H+] ?
   (answer) pH = - log [ H+]           3.301 = - log [ H+]           log [ H+]  = -3.301
        Enter -3.301 into your calculator. (Sometimes you must enter 3.301, then +/-)
        Then hit the 10x key. The answer is .00050 M

pOH

 pOH                       pOH = - log [OH-]

We noted earlier that   [H+]  [OH-] = 1.0 E -14. We can then calculate the  [OH-] from the [H+].

[H+] [OH-] pH pOH  
1.0E-7 1.0E-7 7 7 in pure water
1.0E-1 1.0E-13 1 13 in .10M HCl
1.0E-3 1.0E-11 3 11 in .0010M HCl
1.0E-13 1.0E-1 13 1 in .10M NaOH

One consequence of H+]  [OH-] = 1.0 E -14 is that pH + pOH = 14
This can be derived formally:

  [H+]  [OH-] = 1.0 E -14
take the log of each side log [H+] + log [OH-] = -14
note that we used the concept, log(ab) = log a + log b
multiply each side by -1 pH + pOH = 14
  Some problems  Determine the [H+],  [OH-], pH and pOH for each of the following solutions
       (a) .0010M HBr, (b) .010M NaOH,  (c) 1.0M HClO4,   (d) 10M HCl, (e) 1.0M KOH
    [H+]  [OH-] pH pOH
a .0010M HBr .0010 1.0E-11 3 11
b .010M NaOH 1.0E-12 .010 12 2
c 1.0M HClO4 1.0 1.0E-14 0 14
d 10M HCl 10 1.0E-15 -1 15
e 1.0M KOH 1.0E-14 1.0 14 0

Note that the term in bold is the first term entered in each row.

Some more problems: Determine the [H+],   [OH-], pH and pOH for each of the following solutions
       (a) .0040 M HCl,    (b) 1.2M NaOH     (c) pure water,    (d) .00034M KOH

    [H+]  [OH-] pH pOH
a .0040 M HCl .0040 2.51E-12 2.40 11.60
b 1.2M NaOH 8.34E-15 1.2 14.079 -.079
c pure water 1.00E-7 1.00E-7 7.00 7.00
d .00034M KOH 2.95E-11 3.4E-4 10.53 3.47

For example, (a) [H+] = .0040, -log (.0040) = 2.40,
.0040 x = 1E-14, x = 2.51E-12, -log (2.51E-12) = 11.6

N

Normality

The normality of an acid is simply its [H+] . The normality of a base is its [OH-]. A 1 M NaOH solution is 1 N. A 1 M H2SO4 solution is more than 1 N but less than 2 N. This is because 1 mole of H2SO4 ionizes 100% to 1 M HSO4- and 1M H+ , but the resulting HSO4- is a weak acid which only partially ionizes to SO4-2 + H+. The normality would then be between 1 and 2 and depends upon the degree of ionization of HSO4-.

Some questions which you should be able to figure out and explain:
   1. Is the normality of a strong monoprotic acid equal to, less than, or greater than its molarity?
   2. Is the normality of a strong polyprotic acid equal to, less than, or greater than its molarity?
   3. Is the normality of a weak, monoprotic acid equal to, less than, or greater than its molarity?
Answers: (1) equal to  (2) greater than  (3) less than


 

Neutralization

When an acid combines with a base, neutralization occurs. H+ from the acid combine with OH- of the base to produce water. If the number of H+ and OH- are equal, then complete neutralization occurs and the resulting solution has a pH = 7.

A neutralization reaction has the form:
     acid    +  base   =>       salt     + water
   H2SO4 + NaOH =>  Na2SO4 + H2O

 

Titration

In an acid-base titration, acid and base are combined in such a way that the end solution has a pH = 7. Solutions are added dropwise from burets until an indicator which changes color near pH=7 just undergoes a color change. If the normality of either the acid or base solution is known, then the normality of the other solution may be calculated, since the experimental results give the volumes of acid and base used.

At pH = 7,                                           (1) n(H+) = n(OH-)
Since NA = n(H+) / LA, then               (2) n(H+) = NA LA
Similarly, NB = n(OH-) / LB and          (3) n(OH-) = NBLB
Then it follows from (1) that                (4) NALA = NBLB

Since, in the titration experiment, LA and LB can be determined from the buret readings, if either NA or NB are known, the other may be calculated. In this way, we can use a titration to determine the strength of an unknown acid. We can multiply both sides of equation (4) by 1000mL/L to yield the more familiar form of the equation:

    (5) NA mLA = NB mLB

A sample problem: If 10 mL of a 2.5 M HCl solution neutralizes 35 mL of an unknown NaOH solution, the N of the NaOH solution is calculated as follows:

         NA      mLA   = NB   mLB
       (2.5N)(10mL) = NB (35mL)
              .71N         =    NB

 

 

 

 

Acid-Base 3

3. Bronsted Theory

 An acid is a proton donor. A base is a proton acceptor. This is more general than, but does not contradict the Arrhenius definition. When HCl is placed in solution, it donates H+ and is therefore a proton donor. When NaOH is placed in solution, is supplies OH- . The effect of adding OH- is to increase the pOH by decreasing the pH. In other words it removes H+ from solution. NaOH is a proton acceptor.

    HCl   +   NaOH    =>   NaCl + H2O
    acid         base
   proton     proton
   donor     acceptor

 

Conjugate Acids and Bases

The acid, HCl, which has given up the proton becomes NaCl, a conjugate base (for the reverse reaction it would be accepting the proton). Thus NaCl is a conjugate base. The base, NaOH, which accepts the proton, becomes H2O, a conjugate acid (it is now capable of donating a proton).

    HCl   +   NaOH   =>    NaCl    +   H2O
    acid         base              conjugate conjugate
   proton     proton               base          acid 
   donor     acceptor

However the Bronsted Theory greatly expands our definition of acids and bases. In the reaction below, NH3 is now defined as a base.

HCl     +     NH3      =>     NH4+      +     Cl-
acid            base              conjugate     conjugate
                                          acid             base

Another excample:

HC2H3O2  +     H2O      =>   H3O+      +     C2H3O2-
     acid               base          conjugate          conjugate
                                             acid                  base

 

 

 

Acid-Base 4

4. Lewis Theory

 An acid is an electron pair acceptor. A base donates an electron pair.

    HCl   +   NaOH   =>   NaCl + H2O
    acid          base
electron     electron
    pair           pair
acceptor     donor

Notice that when H+ combines with :OH- , the H+ accepts a pair of electrons from the OH-. Thus HCl is a Lewis acid and NaOH is a Lewis base. Similarly, when the H+ from HCl combines with :NH3 to form NH4+ , the H+ accepts N's electron pair, making HCl a Lewis acid and NH3 a Lewis base.