Acids and Bases

 

Here 3 acid-base concepts are presented.

• The concept of Arrhenius
• The concept of Lewis
• The concept of Bronsted

The Brönsted concept is the most important one, it will be gone through more deeply.

Generally

What is a base. A base is an element or a molecule possessing a lone pair of electrons. Typical example: Ammonia, NH3, other examples, the ethanoate anion CH3CH2O- , the chloride anion, Cl-, the fluoride anion, F-. Important to know : A base is never positively charged, it is never a cation!

 

The Arrhenius concept

The Arrhenius concept is quite similar to the Brönsted concept, which we’ll go through further down. It describes hydroxide-anions, OH-, as base and protons, H+, as acid in a watery solution, but only just.

spell: H+ (aq) and OH- (aq)

Base = OH- (aq) ,

Acid = H+ (aq)

The strength of the acid or base depends on how completely a compound dissociates in water to protons, H+, or hydroxide anions, OH-, just in water.

More information, see Brönsted concept.

 

The Lewis concept

With the concept of Gilbert N. Lewis owns the base, as just mentioned above, at least one lone pair of electrons. The fluoride anion is a Lewis base. The fluoride anion can provide a lone pair of electrons. It is an electron pair donor.

The Lewis acid on the other hand often lack such a pair of electrons because the octet (see octet rule) is not fulfilld. With boron trifluoride, BF3, the element of Bor is missing 2 electrons. The BF3 can now accept an electron pair by reacting with a fluoride anion. BF3 is a Lewis acid.

In some compounds, although the octet rule is satisfied, but for structural reasons in certain places the electron density can be reduced. Because of an apparently incurred electron deficiency the compound can react as a Lewis acid.

Sulfur dioxide, SO2, reacts with hydroxide anion, OH-, to hydrogen sulfate anion, HSO3-, SO2 is the Lewis acid, hydroxide anion, the Lewis base.

In complex compound as mentioned above (see chemical compounds) the central atom often works as a Lewis acid, but its ligands may function as Lewis bases.

Fe3+ + 6CN- ===> Fe (CN)6(3),  hexacyanoferrate

Fe3+ = Lewis acid

CN = Lewis base

Generally

Lewis bases are electron pair donors

Lewis acids are electron pair acceptors

More examples of reactions with the Lewis concept see following table:

 

The concept of Bronsted Lowry

 

The base, in turn, necessarily has a lone electron pair and may have a negative charge or may be an anion . However, a base can never be a cation with this concept.

In contrast to the Lewis concept it is not the electron pair going to be moved, but the proton.

 

The Bronsted acid is a proton donor.

A Bronsted acid must therefore necessarily have a H-atom and may also be a cation. Boron trifluoride , BF3 cannot be an acid after Bronsted, because a H-atom missing.

 

The base acts in return as proton acceptor. 

The base absorbs the proton of the acid.

Acids and bases are mostly in water, H2O, dissolved. Substances that are dissolved in water, H2O, are marked with the symbol (aq).

In the reaction

XH (aq)  +  H2O ===> H3O +  +  X- (aq) (generally, here X is any element)

Here XH is the acid an plays the role as the proton donor and H2O the base as the proton acceptor.

 

Important! Acid-base reactions are equilibrium reactions. Repeat “chemical Equilibrium”. Because the balance can also be moved to any position to the right or to the left, the acid-base reaction is considered reversible, because of the reverse reaction

 

H3O +  +  X- (aq) ===> H2O + XH (aq)

H3O+ acts as the acid and X- (aq) as the base. As you can see each 2 acids XH and H3O+ (hydrogenium cation) and 2 bases H2O and X- are involved in the reaction process.

In the forward reaction XH gives away a proton and becomes the base X-. Similarilly happens to the H2O molecule. This produces the acid H3O+. Both XH and X- and H2O and H3O+ vary by absence of a proton.

XH and X- and H3O + and H2O are called conjugated acid-base pairs.

 

Useful examples

Acetic acid + water  ===>   Hydrogenium cation + acetate ion

CH3COOH (aq) + H2O ===> H3O+ (aq) + CH3COO- (aq)

Acid a base b   ===>    acid b base a

 

Attention! here water acts as an acid!

Ammonia + water ===> ammonium ion + hydroxide ion

NH3(aq) + H2O  ===>  NH4+(aq)  +  OH-

Base a acid b → acid a base b

 

There are countless molecules containing hydrogen atoms, however, unable neither to accept nor to give away protons under normal conditions. We repeat:  protons are H+ cations, the proton contains no electron, which is why it is also very fragile.

Methane, CH4, ethane, C2H6, propane, C3H8. These and other alkanes cannot be deprotonated. To remove a hydrogen atom it needs special industrial methods such as pyrrolysis with Cl2 and UV or in laboratory scale with halogensuccinimides. The reaction proceeds via a free radical mechanism. More in higher lessons.

 

However, there are molecules which function either as acids and as bases. Water and ammonia:

Water / hydroxide ion and hydrogenium ion / water

Ammonia : acid–base : NH3 / NH2- and NH4+ / NH3

Ammonia / ammine ion and ammonium ion / ammonia

Substances with having both functions, acids and bases are called amphoteric substances.

 

Strength of acid

The strength of a Bronsted acid depends on how big the tendency is to donate a proton. Hydrochloric acid, HCl is a strong acid. In one liter of water 1 mol HCl is practically completely dissociated to H3O+ and Cl-. By contrast, the same amount of a weak acid, acetic acid, CH3COOH just 00:42% is dissociated in one liter of water

What depends the acid strength on? There are 2 factors:

• The electronegativity
• The atomic size

In the periodic table the strength of Bronsted acid increases within a period from left to right, as the electronegativity does.

NH3 <H2O <HF.

Methane, CH4, ethane, C2H6 mentioned above do not dissociate. Attention! sodium hydride, NaH, sodium borohydride, NaBH4. These are not as acids but hydrides! The hydride anion, H-, (H atom with an electron pair!) is not a proton.

Also in the subsequent or 3rd period the acidity is increasing due to strength of the electronegativity:

P < S < Cl

PH3 <H2S <HCl

While hydrochloric acid, which is content of swimming pools, is a strong acid, PH3 reacts hardly with water. H2S is a weak acid.

Within the group, the acid strength increases mainly due to stronger increasing atomic size. It is the outermost electrons, which are further away from the nucleus and therefore involve less the proton.

H2O < H2S < H2Se < H2Te

HF < HCl < HBr < HI

Hydrofluoric acid, HF is a weak acid. HF content of rinsing water for teeth.

HI, however, is a strong acid, as well HCl and HBr

 

strength of oxoacids

As stated, the acid strength depends on the electronegativity. In many cases, hydrogen is bound to an oxygen atom, especially in the organic substances. The formula of oxoacids is generally

H-O-X

X is any atom. It may be a metal element.

When X is an alkali metal (Li, Na, K) with low electronegativity, of the bond between O and X or the pair of electrons belongs to oxygen. It is predominantly an ionic compound. Na+OH-, Li+OH-, K+OH-. Sodium, lithium and potassium hydroxide.

If X is a nonmetallic element and it has a relatively high electronegativity, the bond O-X is covalently. The X atom attracts electrons from the oxygen itself, which also weakens the bond between O and H. Result: dissociation of proton, H+, is facilitated. You have to imagine that the X-atom can deduct electrons of the oxygen atom depending on electronegativity of X. The proton is so easily removed.

Examples: HOI, HOBr HOCl: because of the electronegativity of X-atoms like iodine, I, bromine, Br and chlorine, Cl, decreases in this order as is known, however increases the acid strength.

When the X-Atom is fixed with more oxygen atoms, it affects the acid strength accordingly:

H-O2-X  < H-O3-X < H-O4-X

The acid strength increases in this order.

Example: hypophosphorous acid  < phosphorous acid  < phosphoric acid

(When X = P)

Attention! These phosphorus acids are multiproton acids . Up to 3 protons can be issued by phosphorous and phosphoric acid . As soon as the first proton is off, all 3 acids are virtually the same in strength but more medium-strong, for the delivery of the second proton. The reason lies in the opposite formal charges . P+ and O-. Pay attention to the structures . Other samples of multiproton acids:

Sulfuric acid, H2SO4, arsenic acid, H3AsO4, carbonic acid, H2CO3, hydrogen Sulfide, H2S

 

Generally: The extra oxygen atoms that are not fixed to the hydrogen atom, occupy electrons of the X-atom itself. X is thus positive, which in turn attracts electrons of the oxygen fixed to H. This effect also increases the acid strength.

Small list of the strongest acids:

Hydrogen Chloride, (hydrochloric acid), HCl <  nitric acid, HNO3 < Perchloric acid, HClO4

Perchloric acid is the strongest acid in this list.

What about the strength of bases? There you are: the conjugated bases of strongest acid are the weakest

ClO4- < NO3- < Cl-

ClO4- is the weakest base.

In organic chemistry, ethanol, CH3CH2OH has an electron-withdrawing group, the OH bond. A proton is hardly cleaved here, it is a very weak acid. By contrast, acetic acid , CH3COOH is quite sour, C is fixed with 2 oxygens

However the ethanoat-anion, CH3CH2O- is a strong base and often used in organic Synthesis.

 

 

The calculation of the acid strength

Here some mathematics is required . Deal with decade logarithmic ! logx = 2, x = 100,

And repeat chemical equilibrium

 

Intermezzo: What does the unit, mol/l, mean?

If x moles of an element or compound is dissolved in one liter of water. 10 g of sodium chloride, NaCl is dissolved in 1 liter of water. 1 mol NaCl weighs = 58.44g.

The concentration is therefore 10g / 58.44g = 00:17 mol / l

 

A liter of water consists at 25 ° C  of 55.55 mol H2O particles. 1 mol H2O weighs 18g. 1000/18 = 55.55 mol. 25C ° or 298.15 K (Kelvin) is the standard temperature. More about that see thermodynamic lesson in physics. (1dm³ = 1 lt, water density is 1, weight = volume x density, 1dm³ x 1 = 1 kg = 1000 g)

 

Self-dissociation of water:

2H2O ===> H3O+  +  OH‾

This is an equilibrium reaction, but which is located on the left side. Do you remember the game between Mario and Bruno. However there are equal quantities  H3O + and OH- on the rigth side.

Applied on the law of mass action, MWG, it means:

c (H3O+) * c (OH)/c²(H2O) = K

Instead of c (H3O+) you can write c (H+)

As c(H+) and c(OH) occur in equal parts, the MWG can be simplified as follows:

c²(H +) / c²(H2O)

The exact calculation of K: 0.0000001² / 55.55² = 3:24 * 10exp-18 mol²/l²

However, this result is meaningless, because the concentrations of H3O + and OH- are very much smaller than the 55.55 mol H20 in one liter of water. The result is defined 1-10exp-14 mol²/ l².

 

In one liter of water 0.0000001 mol Hydrgenium cations, H3O+, and 0.0000001 mol hydroxide anions, OH-, are dissociated.

each 10exp-7 mol / l.

This is a definition and is the basis for calculating the acid and base strength.

Therefore, the result for K is = 10exp-14

 

To symplify these very small figures logarithms can be used,

decadic logarithm like. Log 0.0000001 = -7

Log 10exp-14 = -14

To get rid of the minus sign you write

c (H+) = -logc (H +) = pH

if c (H +) = 0.0000001 mol / l = 1 + 10exp-7 mol / l, pH = 7

c (OH) = -logc (OH) = p OH,

analogous p OH = 7

And

pOH + pH = pK = 14

We now have an important basis on how to calculate the acid and base strength.

All we drink most of it water, the rain, the snow, the rivers, the lakes and the sea are neutral and have each pH = pOH = 7

 

Calculation of the pH value:

We can now determine the concentration c(H+) and the pH of an acid which dissociates to 100%,.

 

For 0.01 mol of hydrochloric acid, HCl in 1 liter of water the pH is = -log 0,01 = 2

The solution is acidic, because pH <7

How many hydroxide ions, OH-, are there now water in 1 liter?

The pOH value is 14-2 = 12 since pOH + pH = 14

We therefore have 10exp¯¹² mol hydroxide ions in one liter of water

 

Also sodium hydroxide, NaOH, dissociates to 100% in 1 liter of water.

We solve 0.02 mol of this substance in 1 liter of water.

How many grams of NaOH must be weighed?

The weight of 1 M NaOH (rounded, see the periodic table) Na = 23 g, O = 16 g, H = 1 g. 23 + 16 + 1 = 40g

40g = 1 mol

.02 Mole NaOH = .02 * 40 = 0.8g

What is the worth of pOH and pH?

pOH = -Log0.02 = 1.7

pH = 14 – 1.7 = 12.3

The solution is basic because pH > 7

And consists of 10exp-12.7 = 5* 10exp-13 Hydrogenium cations, c (H+).

 

What is Ka and pKa?

In the examples above it can be assumed that the dissociated acid or base is up to 100%. In a solution of HCl in water only a negligible proportion of undissociated HCl -molecules can be found.

Other acids dissociate, however, only to a small percentage.

 

Ka, pKa and Kb, pKb values are only acid and base concentration of not more than 0.1 mol/l and are not exact.

We first come back to the equation:

Generally:

HX + H2O ===> H3O + + X-

To simplify, you can omit H2O, because water is present in excess, and instead H3O+, H+ can be written, as already mentioned above.

HX ===> H+  +  X-

 

As soon as balance has been set, according to the law of mass action, Ka can be determined as follows

c(H+)*c(X-)/c (HX) = Ka

Since c(H+) and c (X-) are formed in equal amounts, one can write:

c²(X-)/c(HX) = Ka

Ks is the acid dissociation constant.

This value is represented as a logarithm because of its small size :

Analog pH:

Ks = -logKs = pKa

but pKa is not the same as pH!

 

Only when we know Ka we can calculate the acid strength c(H+) and the pH.

We denote the initial concentration of HX with co.

We look for c(H +)

The law of mass action is now

c²(X-)/c(HX)

or

c²(H+)/c(HX)

 

The initial concentration is

c(HX) = co(HX)

After equilibrium is reached

c(HX) = co(HX) – c H+)

We know Ka and are looking for c (H+) The following equation must be solved:

c²(H +) /co (HX) – c (H +) = Ka

we replace c(H) with y, and co (HX) with co

y² / (co – y) = Ka,

co and Ka are known. This yields the following quadratic equation:

y² + Ka*y – Ka*co = 0 (see course Mathematics)

as a solution only a positive value can be the right result because negative concentrations do not make sense.

y = -½Ka  +  √(¼Ka² +  Ka*co)  = y = c(H+)

pH = -logy =  -logc(H+)

 

Practical example:

Benzoic acid is a flaky, pale yellow substance at 25 ° C. and is mainly used as a preservative. From this aromatic 0.1 mol are dissolved in 1 lt water. 0.1 mol of benzoic acid, (1 mole = 122 g, 0.1mol = 12.2g) dissociates to 2:42%.

This can be determined experimentally by decanting, after the balance of the solution has been set. Before weighing remaining benzoic acid non dissolved on the bottom it should be dried, otherwise the result won’t be exact enough.

Ka, pKa and pH ? See table below!

Weak electrolytes have a small. The calculation of c (H+) can be simplified as follows :

c (H+) = √ (Ka * co)

for the above example this would thus

c (H+) = √ (6 * 10exp-5 * 0.1) = 0.00244 mol / l

and

pH = -log.00244 = 2.61

 

Kb, pKb, pOH

The same derivation as the acid, we also receive, if a base is dissolved in water:

We formulate again the first equation

X + H2O ===> OH + HX

Water we omit because abound

X ===> OH + HX

As above, we get on the law of mass action at equilibrium

Kb = c(OH-)*c(HX) / c(X)

and

pKb = – logKb

c(OH) = is obtained by solving the equation

Kb = c(OH)² / co(X) – c(OH)

c(OH) = -½Kb +  √(¼Kb² + Kb*co(X)

and

p(OH) = -logc(OH)

Here too, the approximation is valid for weak electrolytes:

p OH = ½ (pKb – logco (X))

 

Important relationship

If we multiply out the law of mass action of the acid and base reaction, of an acid and its corresponding (conjugate) base

Acid HX ===> H+ +  X-

Base X ===> OH + HX

So we get the following relationship:

Ks*Kb = [c (H +) * c (X -) / c (HX)] * [c (OH -) * c (HX) / c (X)] = 10exp-14

and

pKa + pKb = 14 analogous to p OH + pH = 14 (see above)

Table of some Ka and pKa

• HF, hydrogen fluoride, Ka =  6.7*10exp-4, pKa =  3.2
• HCOOH, methane acid, “formic acid”, Ka = 1.8*10exp-4, pKa = 3.7
• HCN, hydrogen cyanide, “cyanide”, Ka = 4.0*10exp-10, pKa = 9.4

Table of some Kb and pKb

• NH3, ammonia, Kb = 1.8*10exp-5, pKb = 4.7
• N2H4, hydrazine, Kb = 9.8*10exp-7, pKb = 6.0
• C6H5NH2, aminobenzene, “aniline”, Kb = 4.3*10exp-10, pKb = 9.3

 

indicators

There are some quite complicated organic substances which change the color at a specific pH value:

A solution of thymol has at low pH, the color red and changes at pH 1.2 – 2.8 to yellow and from pH 8 – 9.6 on blue.

Phenolphtalein is up to pH 8.3 – 10 colorless and then changes to red.

Table of some indicators with color and pH transition range

• Methyl orange changes from red to yellow from pH 3.1 – 4.5
• bromocresol changes from yellow to blue from pH 3.8 – 5.5
• Methylred changes from red to yellow from pH 4.2 – 6.3
• litmus changes from red to blue from pH 5.0 – 8.0

At Unishops yellow acid detection strips are available, with this, the pH of any liquids can be measured. If the solution is neutral (pH7), the color of the strip does not change. In acidic solutions pH changes color from orange to dark red, according to acid strength, basic solution changes color from light green to dark blue. These strips contain several indicators.

 

What is a buffer?

Water and solutions of a certain value are subject to changes. pH solutions are simple to prepare, but they hardly remain constant. Air or the smallest particles of the vessel wall can affect the pH. Swimmingpools would have to be emptied and refilled at least daily without the appropriate additions.

With the buffer the pH can be maintained by adding a relative large amount of acid and base. The solution consists of an acid and a base in high concentrations e.g. 1 mol. A swimming pool has a pH in the neutral range pH 6.8 to 7.2, which is maintained with hydrochloric acid, HCl, and sodium hydroxide, NaOH.

 

We create a solution pH = pK

We now give 1 mol of the acid, HA, in 1 liter of water.

HA ===> H+ + A-

Determining the acid constant, Ks

c(H+)*c(A-) / c(HA) = Ks

for c(H+) we get

c(H+) = Ks*c(HA)/c(A-)

Thus  c(H+) = Ks

must apply

c(HA)/c(A) = 1

What means that the acid and their (mostly conjugated) base are of the same concentraton 1 : 1.

This can be expressed logarithmically

pH = pK – logc(HA)/c(A-)

if c(HA) = c(A-)

logc(HA) /c(A-) = 0 and pH = pK

This is the Henderson-Hasselbalch equation.

Using this equation, you can now determine the molar ratio for the desired pH .

Coming back to the benzoic acid . The pH should be 5. What is the concentration ratio of benzoic acid / benzoate?

logc(benzoic acid ) / c(benzoate) = pK – pH

pK = 4.2 (see example above)

4.2 – 5 = -.08

the molar ratio then

10exp -0.8 = 0.158 rounded 0.16

You can now, for example, put 1 mol of benzoic acid and 6.25 mol of benzoate in 1 liter of water . 1 / 6.25 = 0.16

Now giving 0.1 moles of HCl in this solution. What is happening?

The concentration of benzoic acid increases by 0.1 and now amounts to 1.1 mol / l

The concentration of benzoate reduces by 0.1 and is now 6.25 – 0.1 = 6.15 mol / l

We calculate the pH using the Henderson – Hasselbalch equation :

pH = 4.2 – log 1.1 / 6.15 = 4.94

The pH varies up 0.06 units

If you add 0.1 mol H+ in pure water (pH = 7) it would lower the pH by 6 units to:

pH = -log 0.1 = 1

 

Buffer play not only in the industry from tanning, electroplating and the manufacture of dyes an important role, but also in biology. The blood can barely stand the pH of 7.4 without buffers: The human blood buffer is consisting of bicarbonate, phosphate and proteins.

 

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