HCl =>

H⁺¹

+ Cl^-1

moles/liter before ionization

.020

moles/liter after ionization

none

.020

.020

Ba(OH)₂ =>

Ba⁺²

+ 2OH^-1

moles/liter before ionization

.015

moles/liter after ionization

none

.015

.030

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

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

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^–

K_w

The equilibrium constant for the reaction is given by the expression

[H⁺]   [OH^–]

K_eq = —————-

[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 K_eq is also constant, the equation may be rewritten

K_w = K_eq [H₂O] =  [H⁺]   [OH^–]  or more simply K_w =  [H⁺]   [OH^–]

Experimentally, K_w is found to be 1.0 E -14

[H⁺], [OH^–]

If K_w =  [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.

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 K_w = 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

K_w = 1 E-14 = (.050-x)(1E-7-x)
x must be less than 1E-7
so .050-x = .050
K_w = 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 K_w. K_w = 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.

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.

In order to simplify dealing with such a large range of numbers, we use an exponential scale based upon logarithms. In general, the log 10^x   = 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 10^x 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⁺].

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

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

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

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 N_A = n(H⁺) / L_A, then               (2) n(H⁺) = N_A L_A
Similarly, N_B = n(OH^–) / L_B and          (3) n(OH^–) = N_BL_B
Then it follows from (1) that                (4) N_AL_A = N_BL_B

Since, in the titration experiment, L_A and L_B can be determined from the buret readings, if either N_A or N_B 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) N_A mL_A = N_B mL_B

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:

N_A      mL_A   = N_B   mL_B
(2.5N)(10mL) = N_B (35mL)
.71N         =    N_B