Presentation on theme: "ACID BASE EQUILIBRIA Dr. Harris Ch 20 Suggested HW: Ch 20: 5, 9, 11*, 18*, 19*, 21, 29**, 35, 56**, 59, 66 * Use rule of logs on slide 10 ** Use K a and."— Presentation transcript:
ACID BASE EQUILIBRIA Dr. Harris Ch 20 Suggested HW: Ch 20: 5, 9, 11*, 18*, 19*, 21, 29**, 35, 56**, 59, 66 * Use rule of logs on slide 10 ** Use K a and K b tables in ch 20 text.
Introduction In chapter 10, we defined an acid as a substance capable of producing H + (aq) ions in solution. For strong acids, this equilibrium lies so far to the right that you can make the mathematical assumption that [HX] ≈ 0. These “simple” acids are known as Arrhenius acids. The Arrhenius definition of an acid is restricted to aqueous acid solutions. HX aq H + aq + X - aq An Arrhenius base is a substance that produces OH - (aq) ions in aqueous solution. MOH aq M + aq + OH - aq
Bronsted-Lowry Acids and Bases Until now, H + (aq) has been used to represent the species formed when an acid is dissolved in water. In the Bronsted-Lowry definition of acids and bases, an acid is a H + (proton) donor. A base is a proton acceptor. Here, we define the hydronium ion, H 3 O + (aq). This species is formed when a water molecule accepts a proton. In a chemical equation taking place in aqueous solution, H + and H 3 O + are interchangeable. For example: In the example above, HA is a Bronsted acid, and water acts as a Bronsted base.
Autoionization of Water Pure water contains very low concentrations of hydronium and hydroxide ions that arise from the equilibrium In the reaction above, one water molecule transfers a proton to another. Thus, water can act as BOTH a Bronsted acid and Bronsted base The equilibrium constant expression for the autoionization of water is: At 25 o C, the value of K w is 1 x 10 -14 M 2. In pure water, the concentrations of hydronium and hydroxide are equal. H 2 O(l) + H 2 O(l)H 3 O + (aq) + OH − (aq)
Strong Acids and Bases (RECAP) By now, you should know the strong acids and bases. If not, make sure that you learn them. They are listed in the table to the right.
Conjugate Acid-Base Pairs In any acid-base equilibrium, the forward and reverse reactions involve proton transfers. When an acid, HA, loses its H + (aq), the A - (aq) left behind can behave as a base. Likewise, when a base accepts the H + (aq), it can now act as an acid. An acid and a base such as HA and A - that differ only in the presence of a proton are called a conjugate acid-base pair. Every acid has a conjugate base, and visa-versa.
Relative Strengths of Acids and Bases The stronger an acid, the weaker the conjugate base. Same for bases and conjugate acids. A strong acid completely transfers its protons into water. Its conjugate base has a negligible tendency to be protonated. A weak acid partially dissociates, and so it exists as a mixture of reactants and products. The conjugate of a weak acid is a weak base. HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl − (aq) x
pOH and the PH scale In terms of basicity, we can introduce a quantity pOH. For a basic solution, you can not directly calculate pH. You must calculate pOH first. pH and pOH are related by the following: * at 25 o C
pH scale and Important Notes For this class, we will assume the temperature of all aqueous solutions to be exactly 25 o C so that the previous expression holds true. It is important to know the following: An acidic solution is defined as one in which [H 3 O + ] > [OH - ] (pH < 7) A neutral solution is defined as one in which [H 3 O + ] = [OH - ] (pH = 7) A basic solution is defined as one in which [OH - ] > [H 3 O + ] (pH > 7) this will come in handy
Example What is the pH of a 0.040M solution of HClO 4 at 25 o C? Perchloric acid is a strong acid, so we can assume complete dissociation. The dissociation of the acid may also be written as: It is a matter of preference.
Example What is the pH of a 0.028M solution of NaOH at 25 o C? NaOH is a strong base. We can therefore assume that it completely dissociates, so [OH - ] = 0.028 M. We must calculate pOH first.
Acid Dissociation Constant, K a Let’s consider acetic acid, CH 3 COOH, a weak acid. The equilibrium expression of this dissociation is defined by K a : CH 3 COOH (aq)+ H 2 O(L) H 3 O + (aq) + CH 3 COO - (aq) The size of K a describes mathematically how readily an acid dissociates. The stronger the acid, the higher the value of K a.
Example The pH of a 0.40 M formic acid solution (HCOOH) is 2.08. What is the concentration of H 3 O + ? What is K a ? What % of the acid dissociates? Formic acid is a weak acid. Weak acids do not fully dissociate HCOOH (aq)+ H 2 O(L) H 3 O + (aq) + HCOO - (aq) I 0.40 M 0 0 C -x +x +x E 0.40-x x x
Group Example 1. We set up the ICE table to determine the concentration of H 3 O + ConcentrationCH 3 COOH H3O+H3O+ CH 3 COO - Initial0.050 M00 Change- x+ x Equilibrium0.050 – xxx CH 3 COOH (aq) + H 2 O(L) H 3 O + (aq) + CH 3 COO - (aq) 2. Solve for x. We see that K a is very small because CH 3 COOH is a weak acid. 3. Calculate pH Calculate the pH of a 0.050 M acetic acid solution given that K a = 1.8 x 10 -5 M
Method of Successive Approximations Weak acids tend to dissociate quite poorly, so the percentage of dissociation is very small. Hence, K a is very small as well. Let’s revisit the previous example. Since x is so small: Therefore: Use this method when K < 10 -4
pK a K a values range over many orders of magnitude. Therefore, it is more convenient to use the quantity pK a Lower pK a means stronger acid! Physically, pK a represents the pH at which 50% of a weak acid is deprotonated.
Interpreting pK a By knowing the pK a of an acid, we can predict its form at given pH values. At pH > pK a +1, the acid is primarily in a deprotonated form. At pH = pK a, the acid and conjugate base are present in equal ratio At pH < pK a -1, the acid is primarily protonated Ex. Predict the form of Acetic acid CH 3 COOH (pK a = 4.74) at the following pH values: pH = 3 pH = 4.74 pH = 5.3 pH = 9
Amines Similar to acids, we can describe the protonation of bases. For example, lets look at the equilibrium involving ammonia in water. The lone electron pair of electrons on the Nitrogen allows it to accept an H + ion (proton). Ammonia falls under a class of bases known as amines, which are composed of a central nitrogen atom with 3 bonding domains. Amines react with acids to form ammonium salts (positively charged central N atom with 4 bonding domains).
K b and pK b We can use the base protonation constant K b, or pK b to describe the degree of the dissociation of a base. Higher K b means stronger base Example: Determine the pH of a 0.75 M hydroxylamine, HONH 2 (aq) solution given that K b = 8.7 x 10 -9 M We can write out the equilibrium expression showing the base protonation. HONH 2 (aq) + H 2 O(L) HONH 3 + (aq) + OH - (aq)
Example, contd. Set up ICE table ConcentrationHONH 2 HONH 3 + OH - Initial0.75 M00 Change- x+ x Equilibrium0.75 – xxx Solve x THIS IS A BASE. That means that you must calculate pOH FIRST. Then, determine pH.
Acidic and Basic Salts Suppose we dissolve NaF(s) in water. The result would be an aqueous solution of Na + (aq) and F - (aq). You would expect this solution to be neutral. However, F - (aq) is the conjugate base of a weak acid, HF acid conjugate base base conjugate acid Therefore, a solution of F - (aq) ions will be weakly basic. weak base weak acid strong base negligible acid Na + (aq) is a cation of a strong base, NaOH. Thus, it has negligible acidity and does not affect the pH. So it is considered a neutral cation.
Acidic and Basic Salts 1.Neutral cations are the cations of strong bases. 2.Neutral anions are the anions of strong acids. 3.Basic anions are the conjugate bases of weak acids. 4.Acidic cations are the conjugate acids of weak bases, and hydrated metal ions, as shown on pg 760.
Polyprotic Acids A polyprotic acid is an acid than can donate more than one H +. For example, sulfuric acid, H 2 SO 4, is a strong, diprotic acid. Thus, it can undergo two deprotonations: We can assume 100% dissociation of the H 2 SO 4 in step 1). However, the remaining HSO 4 - is a much weaker acid. EACH PROTON IS SUBSTANTIALLY HARDER TO REMOVE THAN THE PREVIOUS. Therefore, the degree of dissociation of the weaker acid is determined from its K a
Polyprotic Acids K a values of polyprotic acids are labeled K a1, K a2, and so forth to denote each deprotonation. The table to the left shows pK a values of common polyprotic acids. Remember, pK a = -log K a
Example Calculate the pH of a 0.10 M H 2 SO 4 (aq) solution. 1.) We first have the initial deprotonation. Because sulfuric acid is a strong acid, you can assume that all of it is dissociated at equilibrium. 2.) Now, we have the second deprotonation. Use the concentrations of H 3 O + and HSO 4 - from above as your initial values. Use pK a values from the table to determine the final concentration of [H 3 O + ] ConcentrationHSO 4 - H3O+H3O+ SO 4 2- Initial(0.10 M) 0 Change- x+ x Equilibrium0.10 – x0.10+ xx