Acid And Bases

Provided By http://www.ionsclub.com

Acid and Bases Fluids in our bodies are solution and colloidal dispersions. The concentrations of solutes in these solutions are balanced in a healthy person. Particularly important is the balance between acids of hydrogen and hydroxide ions in aqueous solution.

The pH scale:

The pH is a way to state hydrogen ion concentration [H + ] . pH = log

[ ]

1 = − log H + + H

[ ]

[H ] = 1×10 +

− pH

Ex: The [H + ] of pure water at 25˚C is 1× 10 −7 M. its pH can be calculated as:

[H ] = 1×10 +

∴ pH = 7

−7

, M = 1×10 − pH

The values of pH range from zero for [H + ] =1M( 1×10 0 ) To 14 for [H + ] = 1×10 −14 M. The pH and [H + ] are inverse of each other. An acidic solution has a pH values less than 7, and a basic solution has a pH value more than 7. A change of one unit in pH from 1 to 2 represents a change of factor of 10 in [H + ] , a change of 2 units from 1 to 3 represents a 100-fold change in [H + ] . [H + ] and [OH − ] are related by the ion product constant of water (Kw). If we know the molar concentration of either [H + ] or [OH − ], we can easily calculate the other and the same for pH and pOH.

[OH ] = 1×10 −

− pOH

(1× 10 − pH )(1× 10 − pOH ) = K w = 1× 10 −14 pH + pOH = 14 Ex: if the pH=5 , its pOH=? Answer: pOH= 14− pH= 14− 5 = 9

The molar concentration of all substance, and [H + ] are written as in the following: m +

[H ] = n × 10

[ ]

Ex: calculate the pH of a solution whose H + is 1.5× 10−3 . Answer: 1 1 = log(6.66× 102 ) pH= log + = log −3 H 1.5× 10

[ ]

2 pH= log6.66+ log10 pH= 0.82+ 2 = 2.82 Ex: Calculate the H + of a solute whose pH = 5.25.

[ ]

pH = - log H

[ ] =-

log H

+

[H ] +

=

[ ]

+

pH = - 5.25 = - 6+0.75 -6 5.62X10

Body fluids differ in their acidity stomach acid is the most acidic ( pH=1-3).

Blood plasma is slightly basic (pH = 7.35 – 7.45), if pH changes outside the range the ability of blood to transport oxygen is reduced. So maintaining the pH of blood plasma within a narrow range is important to human life. Urine pH (7- 8.4) is a wide range, it can be acidic, basic or neutral. This is due to the fact that many acidic and basic substances are removed from the body through the urine to help maintain the pH of blood plasma. Measuring pH: 1.the most accurate method of measuring the pH in the laboratory is by a pH meter. 2. using the indicator. The indicator is an organic compound that changes color with a pH range, like the litmus. The color of litmus is red at pH < 5 (basic) and blue at pH > 7.8 (alkaline).

Acid- Base titrations: Titration is a method of determining the amount reaction between an acid and a base. Ex: it takes 35.2 ml of a 0.1 M HCl to neutralize 25ml of NaOH. What is the concentration of NaOH solution. 35.2X 0.1 = 25 X M 35.2× 0.1 = 0.141 of NaO M = 25

Normality : HCl + NaOH → NaCl + H 2 O

1 mole HCl neutralizes 1 mole NaOH. H 2 SO 4 + 2 NaOH → Na 2 SO 4 + H 2 O

1 mole H 2 SO4 neutralizes 2 moles so 1mole H 2 SO4 equivalent to 2 moles HCl. The gram-equivalent weight of an acid is : The weight of an acid, in grams, that supplies 1 mole of hydrogen ions is defined as the gram-equivalent weight of that acid. Also: The weight of a base, in grams, that supplies 1 mole of OH- or reacts with one mole of H + is defined as the gram-equivalent weight of that base. Gram-equivalent weight of an acid: Gram-equivalent weight of an acid:

=

gram.M .wt.of an acid no.of acidic hydrogen per molecule

=

gram.M .wt.of a base no.of basic hydroxide per molecule

Ex: Determine the gram-equivalent weight of H2CO3. M.wt. of H2CO3=62g. 62 = 31g. Gram-equivalent wt.= = 2 Ex: Determine the gram-molecular wt. of Ca(OH)2. M.wt. = 74.1 g 74.1 = 37g Gram-equivalent wt.= = 2

The ratio of the gram-molecular wt. of an acid or base to its gram-equivalent wt. is the number of equivalents per mole. no . of equivalent per mole of acid or base =

gram molecular weight gram - equivalent weight

Ex: How many equivalents are there in 24 gram H2SO4? Answer: 98.1 = 49g/Eq. Gram equivalent wt. of H2SO4 = 2 24 = 0.49Eq.H No. of equivalents = 2 SO 4 49 Ex: How many equivalents are there in 52 g of Al(OH)3 ? 52 = 2 Eq.Al(OH) Gram-equivalents = 3 26

Equivalents can be extended to solution by expressing the amount of acid or base present in a litter of solution. Normality (N) : is the no. of equivalents of acid or base per liter of solution. N=

no. of equivalents of acid or base volume of solution in liters

Ex: what is the normality of a solution made by adding enough water to 20 g NaOH to make 500 ml of solution?. Answer: Gram-molecular weight of NaOH = 40g. 40 Gram-equivalent weight of NaOH = = 40g/Eq. 1 20 = 0.5 Eq Number of equivalents = 40 0.5Eq N= = 1N. 0.5

Ex: It takes 52 ml of 0.2 N NaOH solution to neutralize 25 ml of H2SO4? N1V1 = N2 V2

0.2× 52.5= N× 25

Bronsted Acids and Bases: Bornoted define an acid as any compound or ion that donates a proton. And the base as any compound or ion that accepts a proton. All acids and bases according to Arrhenius definition are acids and bases according to Bronsted definition HCl, H 2 SO 4 , H 2 CO 3 are acids. NaOH, KOH, are bases. Bronsted allows us to classify acids and bases many compounds and ions that don't fit the Arrhenius definition. Ex: HCl in water. HCl donates a proton to H 2 O to form H 3 O and Cl-. HCl is an acid according to Arrhenius and bronsted definition. Cl- that remains after the proton is donated, is a base according to Bronsted (but not Arrhenius ) definition because it accept a proton to form HCl. The two parts of the same molecule are called ( a conjugate acid-base pair) Ex: Conjugate-acid-base pair HCl

+

H 2O

H 3 O + Cl -

Conjugate acid-base pair

Ex:

NH 3 +

H 3O +

H 2 O + NH4 +

Certain ions and molecules can be both a Bronsted and a base like in the ionization of water. One water molecule donates a proton to another molecule of water to form hydronium ion and a hydroxide ion. One water molecule acrs as an acid and the other acts as a base:

Conjugate-acid-base pair

NH 3 +

H 3O +

H 2 O + NH4 +

Conjugate acid-base pair

H2O OHH 3 O+ H2O

– Acid 1 – Base 2 – Acid2 – Base 2

Conjugate-acid-base pair

NH 3 +

H 3O +

H 2 O + NH4 +

Conjugate acid-base pair

Base1 acid2 base2 acid 1 Many compounds and ions show such amphoteric behavior. The bicarbonate ion reacts as a base as in the above equation. It also reacts as an acid, according to the following reaction: HCO -3 +

OH −

H 2 O + CO3 -2

Acid 1 base2 acid2 base1 The compound or ions involved in acid-base reactions can be divided into two conjugate acid-base pairs. This means that there are two acids and two bases in solution. Ex: H 2O +

HCl

H 3 O + + Cl -

The two bases in this reaction are Cl- + H 2 O. There is a competition between two bases for the proton thus Cl- + H 2 O compete to protons in the aqueous HCl solution. The HCl in water is completely ionized to H+ + Cl- which means that H 2 O capture most of the protons. So water is stronger base than the chloride ion. A strong acid is : a compound that is readily donates a proton. A strong base is : is a compound that holds its proton tightly. A weak base is : is reluctantly accepts a proton. The stronger the acid, the weaker is its base. The weaker the acid, the stronger is its conjugate base.

NH 3 +

OH - + NH4 +

H 2O

The reaction is in equilibrium because the weaker acid and base, ammonia and water form, so the equilibrium lies to the left.

Ionization constants of Acid and bases : Strong acids are more completely ionized in a solution than are weak acids. The degree of ionization of any acid is given by Ka which is the ionization constant. The equilibrium constant is defined as its ionization constant (Ka). CH 3 COOH +

H 3 O + + CH3COO

H 2O

Acetic acid

acetic ion

[H O ][CH COO ] +

The equilibrium constant K=

[H 2 O] = is constant

[H O ]= [H ] +

+

3

−

3

[H 2 O][CH 3 COOH]

+

3

K [H 2 O] = Ka =

[H ][CH COO ] (ionization constant for acetic acid) +

−

3

[CH 3 COOH]

Ex: Diprotic acids ionize in 2 steps :

[H ][HCO ] Ka=

H 2 CO 3 +

H 2O

H 3 O + + HCO 3

+

H 2O

H 3 O + + CO 3

-

−

+

3

[H 2 CO 3 ]

[H ][CO ]

-

-2

−2

+

Ka=

HCO 3

[HCO ] 3 −

3

Weak acids have small Ka values and strong acids have large Ka values. Usually the 2nd Ka value for diprotic acid is smaller than the 1st. and the third for phosphoric acid is even smaller indicting that the concentration of PO4-3 is very low in an aqueous solution of phosphoric acid (H 3 PO 4 ). Ka= 7.52 × 10 −3 H 3 PO 4 H 3 O + + H 2 PO 4 H 2 PO 4 HPO 4

H 3 O + + HPO 4

-

H 3O + + PO 4

-2

-3

-2

Ka= 6.23 ×10 −8 Ka= 2.2 ×10 −13

Ionization constant of strong acids are : H + + Cl -

HCl

H + + HSO 4

H 2 SO 4 HSO 4

H + + SO 4

-

-2

-

Ka= 1× 10 3 Ka= 1× 10 2 Ka= 1.2 × 10 −2

Ka can be expressed as pKa : pKa= - log Ka Ex: Which is the stronger acid, HCN or HF ? Ka for HF = 3.53× 10−4 Ka for HCN= 4.93× 10−10 4.93× 10−10is smaller than 3.53× 10−4 so HCN is the weaker

The strength of bases are expresses as base ionization constant Kb. NH 3 +

[NH ][OH ] K= +

−

4

[H2O][NH 3 ]

[H2O] =Kb = [NH ][OH ] +

4

[NH 3 ]

−

H 2O

OH - + NH4 +

Weak bases have small Kb, strong bases have large Kb. Phosphate ion is a strong base while bicarbonate ion is a weak one. PO 4

-3

+ H 2O

HPO 4

CO 3

-2

+ H 2O

HCO 3

HCO3 + H 2 O

−2 −

+ OH -

K b = 4.52 ×10 −2

+ OH -

K b = 1.84 ×10 −4

H2CO 3 + OH -

-

K b = 2.3 ×10 −8

Ka can be expressed as pK b : pK b = - log K b

Buffer solutions: A buffer solution is a mixture of either a weak acid + salt of this acid, or a weak base + a salt of that base. Such a mixture reacts with both acids and bases so small addition of either strong acids or bases cause little change in its pH. Ex: Mixture of equimolar quantities of acetic acid and solution acetate dissolved in water is a buffer solution. If small amounts of a strong acid added to the buffer it will react with the conjugate base (acetate): H 3O + + C 2 H 3O 2

-

C 2 H 3O 2 H + H 2O

Weaker acid weaker base The equilibrium lies to the right because water is a weaker base and acetic acid is a weaker acid. So most of the added hydrogen ion is removed from the solution and the pH hardly changes. Hydroxide ions added to the buffer solution react with molecules of acetic acid to form acetic acid ions and water: OH - + C 2 H 3 O 2 H

C 2 H 3O 2

-

+ H 2O

The equilibrium lies to the right so most of the OH- are removed form solution. The pH of the buffer solution is determined according to ( HendersonHasselbalch) equation: pH=pKa + log

[conjugate base] [acid]

a buffer solution has a limited ability to react with acids and bases without changing its pH. A solution acts as a buffer because its contains both

members of a conjugate acid-base pair. Removal of one of them destroys the buffer action. So if enough strong acid is added to a buffer to react with all the acetate in acetic acid-acetate buffer, the solution loses its ability to act as a buffer. Buffer solutions are important in body, because they maintain the acid-base balance in the blood.

Acid-Base balance in Blood: The pH of various body fluids is maintained by buffers. There are several different buffer systems in the body. Dihydrogen phosphate (H 2 PO 4 -) and monohydrogen phosphate (HPO 4 -2) are one weak acid-base conjugate pair that act as buffer in blood. HPO4 -2 + H 3 O +

H 2 PO4 - + H 2 O

Dihydrogen phosphate is a weak acid that reacts with any base as follows: HPO4 -2 + OH -

HPO4 -2 + H 2 O

Another conjugate acid-base pair that acts as a buffer is carbonic acidbicarbonate ion. Carbonic acid (H 2 CO 3 ) is formed by dissolving CO 2 in body fluids. CO 2 + H 2 O

H 2 CO 3

HCO 3

-

+ H 2O

Normally in body fluids as blood there is 24 mEq./L of bicarbonate ion : 1.2 mEq./L of carbonic acid. The pH of blood 7.35- 7.45 when this ration is maintained (20:1) .

[HCO ] 〈 20 . -

The pH of blood becomes acidic when

3

[H 2 CO 3 ]

1

The acidic condition of blood signified by a pH. Less than 7.35 is called acidemia. The pH of blood becomes more basic when the ratio.

[HCO ]/[H CO ] 〈 20 / 1. -

3

2

3

The alkaline condition of blood signified by a pH > 7.45 called alkalemia. Death occurs if the pH of blood is > 6.8 or < 7.8. Buffers in the body differ from those in the laboratory. In that body can replenish components of the buffer as they are used up or remove from the body any excess component.

Ex: Consider a patient who has an illness that causes an increase in the concentration of acidic products in blood. The physiologist processes causing a cidemia are called acidosis. The acidic products react with bicarbonate ions to produce carbonic acid. This cause a down in the ration [HCO3 - ]/[H 2 CO3 ]. , so acidosis will occur. One of the functions of the lungs and the kidneys is to maintain the pH of blood by replenishing or removing excess components from the body. The circulation of air into and out of the lungs called (ventilation) produces the quickest response. An increase in [H 2 CO3 ] in blood causes increase in (CO 2 ) formed from the decomposition of H 2 CO 3 . To lose this excess CO 2 , deeper and faster breathing called (hyperventilation) occurs. This causes decrease in the acidity of blood as CO 2 will be lost through the lungs. If this doesn't return the pH to the normal, the kidneys can help by releasing more bicarbonate ion into the blood and removing hydrogen ions so the body tries to return the [HCO3 - ]/[H 2 CO3 ]. ratio to normal value. Ex: A patient who has an illness that causes increasing in the concentration of basic products in blood. The physiologic process causing alkalemia are called alkalosis. These basic products react with carbonic acid to form bicarbonate ions. The ratio [HCO3 - ]/[H 2 CO3 ]. increasing. To prevent the ratio form increasing is to use CO 2 to produce more carbonic acid. To do this the loss of CO 2 from lungs is minimized by slower breathing called (hypoventilation). The kidneys can help by removing the bicarbonate ions and adding hydrogen ions to blood.