Acids Bases and Salts

Acids Bases and Salts

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

  1. 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

  1. 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

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

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