Acid/Base Equilibria

Acid/Base Equilibria Acid-Base Definitions: Bronsted-Lowry Acid: proton donor (H+) Bronsted-Lowry Base: proton acceptor (H+) Conjugate Acid/Base Pair: going backwards, the original acid is now the base (such as: H2O + HCl → H3O+ + Cl-) Conjugate one is the one on the side that is going in the reverse direction Lewis Acid: electron pair acceptor Lewis Base: electron pair donor Lewis Acid/Base Example → Chemistry of Buffers & Buffers in our Blood: Buffer: an aqueous solution that resists changes in pH upon the addition of an acid or a base; even adding or removing water from the buffer won’t change pH drastically either Typically consist of a large volume of weak acid and its conjugate base or a large volume of weak base and its conjugate acid This therefore means that when a small amount of base is added, the acid present in the buffer will neutralize the hydroxide ions (OH-) and when a small amount of acid is added, the base present in the buffer will neutralize the hydronium (H3O+) ions Whenever hydronium or hydroxide ions are neutralized, they are still converted into a constituent of what is already in the buffer In order to make an effective buffer: The # of moles of weak acid and conjugate base must be significantly larger than the number of moles of acid/base added to the buffer Best buffering occurs when the ratio of [HA] and [A-] is 1:1, where pH = pKa , however buffers are typically referred to as “effective” when the ratio of [HA] and [A-] is anywhere between 10:1 and 1:10. Buffering of our Blood: involved the equilibrium of H2CO3 + H2O → HCO3- + H3O+ weak acid conjugate base This equilibrium maintains the pH of the blood to be around 7.4 (physiological pH). Ka and Acid Strength: Ka: the acid ionization (or even dissociation) constant (simply the equilibrium of an acid/base reaction) and is still products over the reactants We do NOT include water in the equilibrium expression (because we don’t include pure liquids or solids in the expression) - For example: H2O + HA → H3O+ + A- | Ka = ["3 #! ][%" ] ["%] The larger the Ka the more acidic the reaction is (because you have more products than reactants, meaning that it favors the forwards direction) Ka >> 1 is recognized as a strong acid Ka << 1 is recognized as a weak acid The stronger the acid, the weaker the conjugate base Autoionization of Water: Amphoteric: can act as an acid or a base (i.e. water) Autoionization of water: water automatically forms acid/base equilibria on its own | H2O + H2O → H3O+ + OHInstead of Ka we used Kw (autoionization constant) such that: Kw = [H3O+][OH-] (remember, there is no water in the denominator because we don’t include pure solids/liquids) [H3O+] at 25°C is 1.0x10-7 M [OH-] at 25°C is 1.0x10-7 M Therefore, Kw = [H3O+][OH-] = 1.0x10-14 M (since Kw is so low, you can think of the equilibrium as lying far to the left, not wanting to really proceed in the forwards direction Also, since [H3O+] = [OH-], then this is a situation where the solution is neutral - Definition of pH: pH = -log[H3O+] For the pH of water, pH = -log(1.0x10-7) = 7.00 Just grab the opposite sign of the exponent (-7 → 7) If you want to find the concentration of [H3O+], then [H3O+] = 10-pH Also, pOH + pH = 14 Strong Acids & Bases: Strong Acid/Base: ionize basically 100% Strong Acids: perchloric acid (HClO4), HX (HCl, HBr, HI, etc.), sulfuric acid (H2SO4), and nitric acid (HNO3) Remember, if you have the concentration of a strong acid, then you know the concentration of [H3O+], as shown in the picture to the right → Strong Bases: sodium hydroxide (NaOH), potassium hydroxide (KOH), calcium hydroxide (CaOH)2, and magnesium hydroxide (MgOH)2 Weak Acid Equilibrium: 3 main weak acids: hydrofluoric acid (HF), acetic acid (CH3COOH), and methanol (CH3OH) pKa = -log(Ka) → the lower the pKa the stronger the acid For a weak acid equilibrium, what you want to do is use the equilibrium constant Ka and any change in concentration to (')(') be a variable (x). From here, you can make the Ka = and when x << [HA], you can get rid of x on the denominator to get ["%])' '2 Ka = ["%] Weak Base Equilibrium: Kb is known as the base ionization (dissociation) constant B + H2O → BH+ + OH- | Kb = [*" ! ][#" " ] [*] and therefore the higher the Kb the stronger the base (because overall [product] > [reactant]) 2 weak bases: ammonia (NH3) and aniline (C6H5NH2) pKb = -log(Kb) → the lower the pKb the stronger the base For a weak base equilibrium, what you want to do is use the equilibrium constant Kb and any change in concentration to (')(') be a variable (x). From here, you can make the Kb = and when x << [B], you can get rid of x on the denominator to get Ka [*])' '2 = [*] Relationship Between Ka and Kb: By simply adding the conjugate acid/base pairs together, you end up getting the autoionization of water, which has a Kw = 1.0x10-14 Therefore, (Ka)(Kb) = Kw Also, pKa + pKb = 14 Acid-Base Properties of Salts: Acid-Base Neutralization Reaction: where you start with an strong acid/base and end up with water and a salt (HCl + NaOH → H2O + NaCl) These salts therefore form neutral solutions which means they simply have a pH = 7 Now, if a weak acid/strong base react and end up with water/salt, then the conjugate base will be strong and will increase the pH of the water (pH>7) Now, if a strong acid/weak base react and end up with water/salt, then the conjugate acid will be strong and will lower the pH of the water (pH<7) pH of Salt Solutions: See weak acid/base equilibrium sections in order to understand how to come up with pH of different salt solutions (pretty standard tbh) Common Ion Effect & Buffers: Le Chatelier's Principle: if you increase the concentration of products, the equilibrium will shift to the left and vice versa Common Ion Effect: the shift in equilibrium due to the addition of an ion already involved in the equilibrium reaction Buffer Solutions: By having weak acid and weak base present, this allows for the pH to remain relatively constant even when H+ or OH- is added to the buffer (the acid or base will react with the H+ or OH- in order to keep concentration of H+ or OH- constant) Remember though that you need at least as many moles of weak acid/weak base as whatever acid/base you are adding in order to keep this pH relatively constant [%" ] Henderson-Hasselbalch Equation: pH = pKa + 𝑙𝑜𝑔(["%] Buffer Solution pH Calculations: Be comfortable with calculating stuff like this. If 0.01M of NaOH is added, make sure to know that 0.01M of the OH- ions present will become neutralized by the acid (NH4+) and therefore 0.01M of NH4+ will be lost and converted into NH3 NOTE: whether a chemical is an acid or a base is dependent on whether Ka or Kb is greater!!!