1. Analytical Chemistry Acid-Base

2. Arrhenius Theory: H+ and OH- This theory states that an acid is any substance that ionizes (partially or completely) in water to give hydrogen ions (which associate the solvent to give hydronium ions, H3O+): HA + H2O H3O+ + A- A base ionizes in water to give –hydroxyl ions. Weak (partially ionized) bases generally ionize as follows: B + H2O BH+ + OH-

3. Bronsted-Lowry Theory: Taking and Giving Protons This theory states that an acid is any substance that can donate a proton, and a base is any substance that can accept a proton. Thus, we can write a “half-reaction”: 1. acid= H+ + base The acid and base of a half-reaction are called conjugate pairs. Free protons do not exist in solution. 2. There must be a proton acceptor (base) before a proton donor (acid) will release its proton. HA H+ + A- (acid) (proton)(conjugate base) B + H+ BH+ (base) (proton)(conjugate acid) 3. There must be a combination of two half-reactions. HA + BBH+ + A- (acid 1) (base 1) (acid 2) (base 2) Note that (acid 2), (base 2) are weaker than (acid 1), (base 1).

4. Lewis Theory: Taking and Giving Electrons In the Lewis theory, an acid is a substance that can accept an electron pair, and a base is a substance that can donate an electron pair.

5. Acid- Base Equilibria in Water 1. Acid-Base Ionization in Aqueous Solution: H 2 O + HCl H 3 + O + Cl - H 2 O + NH 3 OH - + NH 4 + Pure water ionizes slightly to give hydronium and hydroxide ions 2H 2 OH 3 + O + OH - For simplifying purposes the equation above is written in this way: H 2 OH + + OH - Constant equilibrium is written: k eq = [H + ] [OH - ] = 1.82 x 10 -16 [H 2 O]

6. 2. Since the amount of dissociated water is very little, it is assumable that its concentration is stable and equal to the concentration of water (1g/ ml) (1000g/1) /18 = 55.56 H 2 O k eq [H 2 O] = 1.82 x 10 -16 x 55.56 = k w = [H + ] [OH - ] k w = [H + ] [OH - ] = 10 -14 where kw is the thermodynamic autoprotolysis of water or self ionization of water. Accordingly, we find that the hydrogen ion concentration is equal to the hydroxide ion concentration in water: [H + ] = [OH - ] = k w = 10 -14 = 10 -7 M Acid- Base Equilibria in Water

7. 3. Water also has another characteristic, which is the ability to equalize different strengths of strong acids such as, HCl and HNO 3. That is due to its reaction with these acids that produces H 3 + O (acid 2), which is weaker than the original acid. HCl + H 2 O H 3 O + + Cl - (acid 1) (base 1)(acid 2) Because they all transform to hydronium ion acid, they are all equal in strength. This transformation is known as the leveling effect. The same situation applies to strong bases, where they all transform to the base [OH-]. By that, all strong bases appear equally strong in water. Acid- Base Equilibria in Water

8. When [H + ] = [OH - ], then a solution is said to be neutral. If [H + ] [OH - ] then the solution is acidic. And if [H + ] [OH - ], the solution is alkaline. The hydrogen ion hydroxyl ion concentrations in pure water at 25 C are each 10 -7 M, and the pH of water is 7. A pH of 7 is therefore neutral. The scientist Sorensen invented what is known as pH, which is equal to negative logarithm for [H] for the specific solution. pH = -log [H + ] Because [H + ] and [OH - ] are related to each other by the k w equation, knowing one of them can determine the other. So, it has been customary that pH is used to determine the acid-base surroundings. k w = [H + ] [OH - ] = 1 x 10 -14 by taking the logarithm for both ends: log k w = log [H + ] + log [OH - ] when multiplying by (-): -log k w = log [H + ] + (-log [OH - ]) = -log 1 x 10 -14 since log is known as p, we find that:pk w pH + pOH = 14 Acid- Base Equilibria in Water

9. This equation strongly bonds together pH, pOH. That is why the neutral aqueous solution’s pH number is: pH = log 10 -7 = pOH if the surrounding is an acid pH 7 If the surrounding is a base pH 7 Acid- Base Equilibria in Water

10. pH scale

11. 1. pH Calculation for Aqueous Solution A. Strong Acid and Base Solutions: Strong acids and bases are ionized completely in water, for e.g., HCl H + + Cl - C a O O before ionization O C a C a after ionization pH = -log [H + ] = log C a C a = HCl concentration Dissolving of NaOH: NaOH Na + + OH - C b O O before ionization C b C b C b after ionization pOH = -log [OH - ] = log C b C b = NaOH concentration

12. Strong Acid and Base Solutions:

13. B. Weak Acids and Bases: Weak acids and bases are ionized incompletely in water: Example 1: CH 3 COOH CH 3 COO - + H + C a O O before ionization C a - after ionization 1. pH Calculation for Aqueous Solution

14. Example 2: Acid: HA H + + A - K a = [H+] [A-] = [H+] [A-]...........  [HA] C a - [H + ] K a = weak acid contuent dissociation [ ] = Molar concentration C a = acid initial concentration before dissociation [HA]= remaining acid concentration at equilibrium = [A - ] = [H + ] K a = [H + ] C a – [H + ] If we assume that the acid is very weak and K a very small, then we are neglecting [H + ] compared to C a, we obtain: K a = [H + ] 2 C a [H + ] = K a C a pH = -log K a C a 1. pH Calculation for Aqueous Solution

15. Example 3: Base: The weak base solution is treated the same way. B + H 2 O [BH + ] [OH - ] k b = [BH + ] [OH - ] = [BH + ] [OH - ] [B] C b - [OH - ] C b = weak base concentration before dissociation C b - [OH] = weak base concentration after dissociation at equilibrium If we assume that the base is very weak and k b very small, then we neglect [OH - ] concentration compare to C b k b = [OH - ] 2 C b = [OH - ] = [BH + ] at equilibrium [OH - ] = K b C b pOH = - log K b C b k b = weak base contuent C b = initial base concentration [B]= remaining weak base after dissociation 1. pH Calculation for Aqueous Solution

16. K a and K b equal the equilibrium constituent K eq for the dissociation reaction multiplied by the water concentration (55.56 molar) considering water concentration is constant as long as acid and base solutions in water are diluted. Now if we look at the conjugate base A - and the conjugate acid BH + we will find that both of them have good basic and acidic qualities based on Bronsted’ s defined of acid and base. For instance, the base A - reacts with water: A - + H 2 O HA + OH - k b - = [HA] [OH - ]...........  [A - ] and the conjugate acid BH + reacts with water: BH + B + H + k a - = [B] [H + ] [BH + ] K b - = conjugate base constant dissociation K a - = conjugate acid constant dissociation 1. pH Calculation for Aqueous Solution

17. And they are concluded from values of k a and k b of the conjugate acid and base, by multiplying  by  : k a x k b - = [H+] [A-] x [HA] [OH - ] [A - ] [HA] k a x k b - = [H+] [OH - ] = k w k b - = k w k a by using the same method we can conclude that: k a - = k w k b 1. pH Calculation for Aqueous Solution

18. The pH of blood at body temperature (37 o C) is 7.35 to 7.45. This value represents a slightly more alkaline solution relative to neutral water than the same value would be at room temperature. Since a neutral blood solution at 37 o C would have pH 6.8, a blood pH of 7.4 is more alkaline at 37 o C by 0.2 ph units than it would be at 25 o C. 2. Blood pH

19. A buffer solution consists of a maximum of a week acid and its conjugate base of a weak base and its conjugate acid at predetermined concentrations or ratios. Which resists change in pH when a small amount of an acid or base is added or when the solution is diluted. A. Buffer Capacity: The buffer capacity describe how much the buffer solution is able to resist the change in the pH. And it is determined quantitatively by the amount of moles of the strong base or acid needed to be added to one liter of buffer solution in order to increase or decrease the pH –respectively- of the buffer solution by one unit. And if it is at its maximum when the ratio of the weak acid concentration and its conjugate base is unity. B. The Buffering Mechanism: If we assume that buffer solution consists of a maximum of acetic acid and sodium acetate, the following reactions will occure: Buffer Solutions

20. Bufferd solution

21. - Weak acid ionization: CH 3 COOH CH 3 COO - + H + - Complete salt ionization: CH 3 COONa Na + + CH 3 COO - - Very small water dissociation: H 2 O H + + CH 3 COO -  Here, we can observe that if an amount of hydrochloride acid was added to this mixture, then the mixture will suffer a complete ionization to chloride and hydrogen ions. Which would have caused a sharp decrease in the pH if the mixture above did not exist. Since the mixture does exist, the hydrogen on will react with the acetate (which we mentioned as a weak base), and acetate acid will be produced: H + + CH 3 COO - CH 3 COOH this means we transformed the strong acid (hydrochloric) into weak acid. Buffer Solutions

22. On the other hand, if we added strong base such as hydroxide sodium, the produced hydroxide ions will increase the pH highly if the buffer solution did not exist. But, here it reacts with the hydrogen ions to produce neutral water: H + + OH - H 2 O The same mechanism is observable when using an alkaline buffer solution, such as ammonia and ammonium chloride: NH 3 + H 2 O NH 4 + + OH - NH 4 Cl NH 4 + + Cl - By adding HCl to this solution, the H + from this acid will react with [OH - ] and will produce water H 2 O. Meaning that, the acid HCl has been transformed to water by the buffer solution. While, when adding NaOH to this buffer solution the following reaction will occur: NaOH + NH 4 + NH 3 + Na + + H 2 O meaning, that strong base transformed to a weak base. Buffer Solutions

23. 1. The pH for a weak acid and its salt: For example, acetic acid with sodium acetates: CH 3 COOH H + + CH 3 COO - k a = [H + ] [CH 3 COO - ] [CH 3 COOH] If we assume that the molar salt concentration is C s and that molar acid concentration is C a, we will find that: K a = [H + ] C s C a k a  [H + ] C s C a [H + ] = k a C a C s pH = pk a + log C s C a 3. Calculation for the pH the Buffer Solution 2. The pH for a weak base and its salt: It is calculated the same way [OH - ] = k b C b C s pOH = pk b + log C s C b

24. The pH of the blood in a healthy individual remains remarkably constant at 7.35 to 7.45. This is because the blood contains a number of buffers that protect against pH change due to the presence of acidic or basic metabolites. The buffering capacity of blood for handling CO 2 is estimated to be distributed among various buffer systems as follows: hemoglobin and ox hemoglobin, 62%; H 2 PO 4 -, 22%; plasma protein, 11%; bicarbonate, 5%; proteins contain carboxylic and amino groups; which are weak acids and bases respectively. Certain diseases cause disturbances in the acid balance of the body. For e.g.; diabetes may give rise to “acidosis” which can be fatal. An important diagnostic analysis in the CO 2 /HCO 3 - balance in blood. This ratio is related to the pH of the blood by the Henderson_Hasselbalch equation: pH = 6.10 + log [HCO 3 - ] [H 2 CO 3 ] Where H 2 CO 3 can be considered to the concentration of dissolved CO 2 in the blood; 6.10 is pk a of carbonic acid in blood at body at body temperature (37 o C). Normally, the bicarbonate dioxide (CO 2 ) is 1.3 mmal/ L.Accoringly for the blood: pH = 6.10 + log 26 mmal/ L = 7.40 1.3 mmal/ L Physiological Buffers

25. The HCO 3 - \H 2 CO 3 buffer system is the most important one in buffering blood in the lung (alveolar blood). As oxygen from inhaled air combines with hemoglobin. The oxygenated hemoglobin ionizes, releasing a proton. This excess acid is removed by reacting with HCO 3 - : H + + HCO 3 - H 2 CO 3 But note that the [HCO 3 - \H 2 CO 3 ] ratio at pH 6.4 is 26/ 1.3 = 20: 1 this is not a very effective buffering ratio; and significant amounts of HCO 3 - are converted to H 2 CO 3,, The pH would have to decrease to maintain the new ratio. But fortunately, the H 2 CO 3,, Produced is rapidly decomposed to CO 2 and H 2 O by the enzyme decarboxylase, and the CO 2 exhaled by the lungs. Hence, the ratio of HCO 3 - \H 2 CO 3 remains constant at 20: 1