Acids and Bases Acids release H+(aq) when dissolved in water and bases release HO-(aq). Strong acids and bases dissociate completely in water. Weak acids and bases are in equilibrium with H+(aq) and HO(aq), respectively. Acids react with bases to form water in neutralization reactions. Outline • Strong Acids and Bases • Weak Acids • Homework Litmus paper for pH determination Strong Acids and Bases Strong Acids in Water Strong acids are those that completely dissociate in water. You've seen them all in a previous section: HCl, HBr, HI, H2SO4, HNO3, and HClO4. Chemistry 102 Prof. Shapley page 1 The concentration of solvated protons is equal to the moles of the acid added to each liter of water. When 0.05 mol of HNO3 is added to 50 mL water: [H+] = (0.05 mol)/(50 mL)(1 L/1000 mL) = 1.0 M pH = -log(1.0) = -log(100) = 0 When 0.05 mol of H2SO4 is added to 10 L water: [H+] = (0.05 mol)/(10 L) = 0.005 M = 5 x 10-3 = 100.70 x 10-3 pH = -log(10-2.3) = 2.3 Strong Bases in Water The strongest base that can exist in water is HO-. There are many hydroxide salts that provide HO- when dissolved in water. These include NaOH, KOH, Mg(OH)2, Ca(OH)2, and Al(OH)3. There are other bases that are even stronger than hydroxide but these react with water to make hydroxide. [Li+][NH2-] + H2O [Na+][H-] + H2O NH3 + Li+ + HOH2(g) + Na+ + HO- Neutralization Reactions The reaction between a strong acid and an equal number of moles of a strong base produces water. This is a neutralization reaction because the resulting solution is neutral, neither acidic nor basic, with a pH of 7. When the molar quantity of acid does not equal the molar quantity of base, the pH of the solution depends on the quantity of acid or base that remains. 1. What is the pH of a solution resulting from the addition of 0.10 mol HCl and 0.07 mol NaOH to 1.0 L of water? Chemistry 102 Prof. Shapley page 2 2. What is the pH of a solution resulting from 50 mL of 0.02 M HCl and 50 mL of 0.10 M NaOH? Weak Acids Why are acids weak or strong? Every acid reacts with water. When it loses a proton, it transfers that proton to water. Water acts as a base when it solvates the proton. The conjugate acid is protonated water and it is represented as (H3O)+, (H5O2)+, (H2nOn)+, or just as H+(aq). The conjugate base is the acid minus its proton. A strong acid is completely dissociated in water while a weak acid is in equilibrium with its conjugate base in water. Let's look at the reaction of a strong acid, nitric acid or HNO3, and a weaker acid, nitrous acid or HNO2, in water. Chemistry 102 Prof. Shapley page 3 Nitric acid is a stronger acid than nitrous acid because its conjugate base is more stable. The NO3- ion has excess negative charge spread out over 3 oxygen atoms while the excess negative charge in NO2- is spread out over on 2 oxygen atoms. A conjugate base is more stable when the negative charge is on an electronegative element and when the charge is delocalized over multiple atoms. The more stable the conjugate base, the stronger the acid. A stable conjugate base is not very basic. A very strong acid has a very weak conjugate base and a very weak acid has a very strong conjugate base. Weak Acid Equilibrium Nitrous acid is in equilibrium with a proton and its conjugate base, NO2-. Because the concentration of water is constant, we can define a constant equal to the product of the concentration of water and the equilibrium constant. This constant, the Ka is 10-3.29 for HNO2. We can use the Ka to determine the pH of a solution of nitrous acid. For example, let's determine the pH of a solution made from 0.01 mole of HNO2 in 1.0 L of water. Some of the acid will dissociate into H+ and NO2- but we don't know w=how much. We'll call the molar concentration of solvated protons x so [H+] = x. This means that the concentration of NO2- must also be x because they form in equal amounts by the dissociation of the acid. The concentration Chemistry 102 Prof. Shapley page 4 It is always possible to solve for x by using the quadratic equation but most of the time we can solve it using an approximation. If we assume that the extent of dissociation is small, then the equilibrium concentration of acid doesn't change very much from its original concentration. We can then use the original acid concentration in the Ka expression. The approximation is a good one in this case because we get the same pH value using it as we did from the quadratic equation. Reaction between Weak Acids and Strong Bases Chemistry 102 Prof. Shapley page 5 Strong bases react completely with weak bases to form the conjugate base of the weak acid. Then there will be an equilibrium between the weak acid and its conjugate base in water. For example, let's look at the solution formed by mixing 0.010 mol HNO2 and 0.004 mol NaOH in 1.0 L water. Again, we can use an approximation to solve this. The approximation is valid because the extent of dissociation will be small when there is already some product present. Weak Acids and Strong Acids The acid dissociation constants of a selection of acids is in the tables. In the table are listed pKa values. pKa = -log(Ka) The very strong acids, those that are completely dissociated in water, are distinguished by examining their acid dissociation equilibrium in non-aqueous solvents and the pKa for water is estimated. The equilibrium constants for the weaker acids (pKa >0) are determined from pH measurements of solutions. 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