Acids and Bases 1 PDF

MED-102 General Chemistry Acids and Bases 1  LOBs covered Describe the two definitions of acids and bases Determine acid/conjugate base pairs Identify whether an acid is strong or weak Discuss the relative strengths of acids and their conjugate bases Describe the dissoc...

MED-102 General Chemistry Acids and Bases 1  LOBs covered Describe the two definitions of acids and bases Determine acid/conjugate base pairs Identify whether an acid is strong or weak Discuss the relative strengths of acids and their conjugate bases Describe the dissociation of distilled water Predict the approximate pH of some common substances Discuss the various ways of making pH measurements and their accuracy Estimate the pH of a solution using acid-base indicators Calculate the pH of strong acid and strong base solutions  Definitions of Acids and Bases We will examine two definitions of acids and bases: 1. Brönsted-Lowry 2. Lewis  Brønsted-Lowry theory Acid – substance that donates a proton (H+ ion) – a proton donor Base – substance that accepts a proton – a proton acceptor  Brønsted-Lowry theory – Revision Slide In the above example, we see that H2O donates a proton (H+ ion) to the NH3 molecule. Therefore, H2O is a proton donor, or a Brönsted-Lowry acid. The NH3 accepts a proton and is therefore considered a Brönsted-Lowry base. Now, examine the reverse reaction in this equilibrium situation. The NH4+ ion donates a proton to the OH- ion, and it is therefore a Brönsted-Lowry acid. But it is not the main acid. We call it the conjugate acid. Similarly, the OH- accepts a proton and is therefore acting as a Brönsted-Lowry base. It is not, however, the main base, so we call it the conjugate base. This theory gives rise to conjugate acid-base pairs. H2O/OH- and NH3/NH4+ are the two conjugate acid pairs in this equilibrium.  Brønsted-Lowry theory Another example:  Brønsted-Lowry theory What is the main disadvantage of the Brönsted-Lowry theory? A substance like FeCl3(s) is known to dissolve in water to produce a highly acidic solution, even though it does not directly donate a proton Lewis Acids and Bases Lewis acid – a lone pair acceptor Lewis base – a lone pair donor Lewis Acids and Bases – Revision Slide In the top example, we see that the NH3 donates a lone pair to the BF3 to form a covalent bond. Thus, the NH3 is a lone pair donor, or Lewis base. The BF3 is acting as a lone pair acceptor, or a Lewis acid. In the second example, the oxygen atom of OH- donates a proton to the H atom of HCl, and forms a bond with it to yield H2O. Therefore, we see that OH- is acting as a Lewis base. The HCl, which is the recipient of the lone pair of electrons, is acting as a Lewis acid. Therefore, we see that the Lewis definitions are more comprehensive than the previous two definitions of acids and bases. The Lewis definitions find wide use in Organic Chemistry.  Lewis Acids and Bases An aqueous solution of FeCl3 turns out to be very acidic. Why? The hydrogen atoms in water become highly + as electron density flows towards the metal cation. The O – H bonds become weaker and can be abstracted very easily by free water molecules, forming many H3O+ ions in the solution, making it more acidic Lewis acids – Metal cations  Lewis Acids and Bases Electron-rich molecules or anions are good bases Nucleophiles Electron-poor molecules or cations are good acids Electrophiles  Acid and Base Strength View following reaction as an equilibrium Since there is an equilibrium situation here, the main acid H3PO4 and the conjugate acid H3O+ are both competing. H3PO4 is a weak acid, dissociating only partially. The equilibrium lies heavily to the left. This means that H3O+ must be a stronger acid than H3PO4. Since the equilibrium lies heavily to the left, this means also that the conjugate base, H2PO4-, must be a fairly strong base. Conclusion: the weaker the acid, the stronger its conjugate base, and vice versa.  Strong and weak acids Strong acid – Dissociates essentially completely in water Equilibrium lies on the far right Very large equilibrium constant – For example, the equilibrium constant for HCl is 1.3  106. Conjugate base (Cl-) is very weak  Weak acids A weak acid dissociates only partially in water Equilibrium constant is very small – For example, the equilibrium constant for acetic acid is 1.8  10-5. Equilibrium lies on the left hand side Conjugate base (acetate anion) is strong Strengths of acids and their conjugate bases  Dissociation of water Water is an amphoteric substance – sometimes acts as an acid, sometimes as a base Water as a base – proton acceptor HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq) Water as an acid – proton donor NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) So it is easy to think of a reaction between two water molecules, where one is the acid and the other is the base  Dissociation of water Imagine two water molecules reacting together – one acting as an acid, the other acting as a base 2 H2O(l) H3O+(aq) + OH-(aq) Water dissociation as an equilibrium 2 H2O(l) H3O+(aq) + OH-(aq) + − Kw = [H3O ][OH ]  Water dissociation Distilled Water is a very weak electrolyte [H3O+] = 1.0  10-7 M at 25 oC [OH-] = 1.0  10-7 M at 25 oC [H 2O]dissociated 1.0 10−7 M = = 1.8 10−9 [H 2O]undissociated 55.4 M 2 in a billion molecules dissociate  Water dissociation – Revision Slide Distilled Water is a very weak electrolyte [H3O+] = 1.0  10-7 M at 25 oC [OH-] = 1.0  10-7 M at 25 oC These are experimental measurements [H 2O]dissociated 1.0 10−7 M = = 1.8 10−9 [H 2O]undissociated 55.4 M Where does the 55.4 M come from? This is the molarity of pure water. Consider 1.00 L of water. It weighs 1.00 kg = 1000 g Moles = 1000 g / 18.02 g/mol = 55.4 moles, in 1.00 L Molarity = 55.4 mol / 1.00 L = 55.4 M  Ion-product constant Kw = [H3O+ ][OH− ] = (1.0 10−7 )2 = 1.0 10−14 + − Acidic [H3O ]  [OH ] Neutral [H3O+ ] = [OH − ] + − Basic [H3O ]  [OH ] 5-Minute Break  The pH Scale pH = − log[H3O+ ] + − pH [H3O ] = 10 Acidic pH < 7 Neutral pH = 7 Basic pH > 7 It is important to note that the pH can be negative, and greater than 14 These numbers are valid for 25 oC Recall that equilibrium constants vary with temperature pH of some common substances  Measuring pH pH electronic meter Offers a very accurate way of making pH measurements ( 0.01 units) Needs to be calibrated versus a solution with known pH (e.g. distilled water, or buffer solution)  Measuring pH Acid-base indicators Can approximate the pH to  0.2 units  Measuring pH – Revision Slide Let us look at an example of how these acid-base indicators work. We have a solution of unknown pH. It gives a yellow colour with methyl orange and a yellow colour with bromocresol purple. What is the approximate pH of this solution? First, since it is yellow with methyl orange, its pH must be higher than about 4.2, as this is the colour that methyl orange gives above this pH. Second, it is yellow with bromocresol purple. This indicator is yellow for pH less than 5.6, approximately. Therefore, we conclude that the pH of the solution must be in the range 4.2 – 5.6. We could do additional tests to narrow down even further the pH range. For example, if we use bromocresol green and the solution turns green, then we know that the pH must be between 4.4 and 4.6.  Measuring pH Universal indicator (paper strips or liquid solution). ( 1 unit)  pH for Strong Acids Strong monoprotic acid (100% dissociation) HA(aq) + H2O(l ) → H3O+ (aq) + A- (aq) Initial concentration gives [H3O+] (and pH) Strong diprotic acid K1 is very large K2 is small  pH for Strong Acids A solution of hydrochloric acid is 0.050 M. What is the pH of this solution? SOLUTION: HCl is a strong acid and dissociates completely. Thus [H3O+] = 0.050 M. pH = -log10[H3O+] = -log10(0.050) = 1.30  pH for Strong Bases Alkali metal hydroxide, MOH → MOH(aq) ⎯⎯⎯→ M+ (aq) + OH− (aq) H O 100% MOH(s) ⎯⎯⎯ 2 Once we find [OH-] we can calculate [H3O+] Kw = [H3O+ ][OH− ] = 1.0 10−14  pH for Strong Bases We make a solution of NaOH (MM = 40.00 g/mol) by dissolving 2.50 g of solid NaOH in enough water to make a 100.0 mL solution. What is the pH of this solution? SOLUTION: Calculate the moles of NaOH as 2.50 g / 40.00 g/mol = 0.0625 moles. Molarity = moles / Liters solution = 0.0625 mol / 0.1000 L = 0.625 M Kw = [H3O+][OH-], therefore 1.0 × 10-14 = [H3O+](0.625 M) [H3O+] = 1.60 × 10-14 and pH = -log10(1.60 × 10-14) = 13.80  pH for Strong Bases We make a solution of Ca(OH)2 (MM = 74.09 g/mol) by dissolving 5.00 g of the solid in enough water to make 1250.0 mL of solution. What is the pH of this solution? SOLUTION: Moles of Ca(OH)2 = 5.00 g / 74.09 g/mol = 0.0675 mol. Moles of OH- = 2 × Moles of Ca(OH)2 = 2(0.0675) = 0.135 mol Molarity = moles/Liter solution = 0.135 mol / 1.2500 L = 0.108 M Kw = [H3O+][OH-], therefore 1.0 × 10-14 = [H3O+](0.108 M) [H3O+] = 9.26 × 10-14 and pH = -log10(9.26 × 10-14) = 13.03  Strong Bases Alkaline-earth metal hydroxides, M(OH)2 These are strong bases, but they are not fully soluble like MOH. We can use Coulomb’s Law to predict solubility Q1Q2 E=k d Since the metal cation has a charge of +2, the Coulombic energy of attraction between the ions is relatively large. This makes the metal hydroxide difficult to break apart and dissociate. Alkaline-earth hydroxides may be strong bases in principle, but due to incomplete dissociation, we get fewer OH- ions, and a lower pH.  Strong Bases Metal oxides are alkaline (e.g. CaO) 2− (aq) + H2O(l ) ⎯⎯⎯→ OH − (aq) + OH − (aq) 100% O 2+ − CaO(s) + H2O(l ) → Ca (aq) + 2 OH (aq)  Summary for Revision There are three definitions of acids and bases: Arrhenius, Brönsted-Lowry, and Lewis. In the Arrhenius definitions, an acid is a substance that increases the H+ concentration when it dissolves in water, and a base is a substance that increases the OH- concentration when it dissolves in water. A Brönsted-Lowry acid is a proton (H+) donor, and a Brönsted-Lowry base is a proton acceptor. The Brönsted-Lowry definitions allow for the concept of acid-base conjugate pairs. A Lewis base is a lone pair (of electrons) donor, and a Lewis acid is a lone pair acceptor. Strong acids dissociate essentially 100%, whereas weak acids dissociate only partially. Strong acids have weak conjugate bases and vice versa. Water is amphoteric, that is, it can act as either a base or an acid, depending on what it reacts with. Water has a very small but finite self-dissociation, producing very small amounts of H+ and OH- ions. Kw = 1.0  10-14 at 25 oC. This is the equilibrium constant for the self-dissociation of water. We can use the pH scale to describe the degree of acidity of solutions. The pH can be measured in several ways: electronic pH meter, universal indicator, or acid-base indicators. Strong acids and strong bases dissociate 100%. This fact is useful in calculating the pH of such solutions. Metal oxides are basic and form alkaline solutions.