ACID-BASE EQUILIBRIUM WHAT IS AN ACID AND A BASE? Arrhenius Theory Acids – are solutes that produce hydrogen ions H + in aqueous solutions ex. HCl (aq) H + (aq) + Cl - (aq) Bases – are solutes that produce hydroxide ions, OH - in aqueous solutions ex. NaOH (s) Na + (aq) + OH (aq) Bronsted-Lowry Theory A Bronsted-Lowry acid is a proton donor A Bronsted-Lowry base is a proton acceptor Arrhenius Acids Arrhenius Bases HBr (aq) H + (aq) + Br - (aq) LiOH (aq) Li + (aq) + OH - (aq) H 2 SO 4(aq) H + (aq) + HSO 4 - KOH (aq) K + (aq) + OH - (aq) (aq) HClO 4(aq) (aq) H+ (aq) + ClO 4 - Ba(OH) 2(aq) Ba 2+ (aq) + 2OH - (aq) PROBLEMS WITH ARRHENIUS THEORY The Salt problem: The theory failed to account for the basic properties of compounds that do not contain the hydroxide ion, such as ammonia (NH 3(aq) ) NH 3(g) + H 2 O (l) NH 4 + (aq) + OH - (aq) The Solvent Problem The theory failed to account for a solvent ’s key role in acid-base properties - for example, the aqueous solution of hydrogen chloride (hydrochloric acid) conducts electricity, while hydrogen chloride in an organic solvent does not [therefore, latter solution does not contain ions]. - The solvent plays a key role in acid -base properties. BRONSTED-LOWRY THEORY \ Acid (H+ donor) HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) B a s e ( H + a c c e p to r ) H+ donor (acid) CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq) H+ acceptor (base) NH3(g) + H2O(l) ↔ NH4+ (aq) + OH-(aq) H+ acceptor (base) H+ donor (acid) AMPHOTERIC ** A substance can be classified as a Bronsted-Lowry acid or base only for a specific reaction Protons may be gained in a reaction with one substance, but lost in a reaction with another H 2O (l) + H 2O (l) H 3O +(aq) + OH −(aq) H+ acceptor (base) H+ donor (acid) Water is amphoteric, meaning it can act as both an acid and a base. one molecule of water acts as an acid, donating a H+ ion and forming the conjugate base, OH - , and a second molecule of water acts as a base, accepting the H + ion and forming the conjugate acid, H3O + . CONJUGATE ACID-BASE PAIRS According to the Bronsted-Lowry concept, acid -base reactions involve the transfer of a proton Usually reversible and result in an acid-base equilibrium Every base has a corresponding acid (conjugate acid) and every acid has a corresponding base (conjugate base) Two molecules or ions that are related by the transfer of a proton are called a conjugate acid-base pair. Conjugate acid-base pairs differ in formula by one hydrogen ion H2SO4 – HSO4 H2O – OH- COMMON CONJUGATE ACID-BASE PAIRS COMMON CONJUGATE ACID-BASE PAIRS HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) CH 3 COOH (aq) + H 2 O (l) ↔ CH 3 COO - (aq) + H 3 O + (aq) NH 3(g) + H 2 O (l) ↔ NH 4 + (aq) + OH - (aq) H 2 SO 4(aq) + H 2 O (l) HSO 4 - (aq) + H 3 O + (aq) HCl (aq) + NH 3(g) Cl - (aq) + NH 4 + (aq) A COMPETITION FOR PROTONS The stronger the acid, the weaker its conjugate base, and conversely, the weaker an acid, the stronger its conjugate base. Acid-base reactions – competition for protons between two bases Acetic acid equilibrium – between acetate and water CH 3 COOH (aq) + H 2 O (l) ↔ CH 3 COO - (aq) + H 3 O + (aq) Ability of acetate ion to hold on to its proton (H + ion) is much higher than the ability of H 2 O to pull the proton away Percent ionization Is low Equilibrium lies far left – acetic acid is a weak acid (acetate ion is a strong base!) AUTOIONIZATION OF WATER Autoionization is when water molecules ionize one another. The transfer of a proton (H + ) from one molecule of water to the other, producing a hydronimum ion (H 3 O + ) H 2O (l) + H 2O (l) H 3O +(aq) + OH −(aq) Simplified into… H 2O (l) H +(aq) + OH −(aq) H2O(l) H+(aq) + OH−(aq) The equillibrium constant for the ionization of water: Kw = [H+(aq)][OH-(aq)] @ SATP; Kw= 1.0 x 10-14 *an endothermic process ∴ ↑ 𝐾𝑤 ↑ 𝑇 RECALL FROM GR. 11… [H + (aq) ] = [OH - (aq) ] = pH 7 NEUTRAL [H + ACIDIC (aq) ] > [OH - (aq) ] = low pH [H + (aq) ] < [OH - (aq) ] = high pH BASIC STRONG ACIDS HClO 4(aq), HI (aq), HBr (aq), HCl (aq) , HNO 3(aq) , H 2 SO 4(aq), etc.. % ionization for strong acids > 99% [We assume 100% dissociation in calculations] 1. A 0.15 mol/L solution of hydrochloric acid at SATP. Calculate the concentration of the hydroxide ions using K w = [H + (aq) ][OH - (aq) ]. HCl (aq) H + (aq) + Cl - (aq) Hydrochloric acid is a strong acid – assume 100% ionization [*tiny contribution made by autoionization of water can be ignored] 1: 1: 1 mole ratio If [HCl (aq) ] = 0.15 mol/L then.. [H + ] = 0.15 mol/L @ SATP; K w = [H + (aq) ][OH - (aq) ] K w = 1.0 x 10 -14 [OH - (aq) ] = K w / [H + (aq) ] = 1.0 x 10 -14 / 0.15 mol/L = 6.7 x 10 -14 mol/L The concentration of hydroxide ions is 6.7 x 10 -14 mol/L. 2. Calculate the hydroxide ion concentration in a 0.25 mol/L HBr (aq) solution. HBr (aq) H + (aq) + Br - (aq) Hydrobromic acid is a strong acid – assume 100% ionization [*tiny contribution made by autoionization of water can be ignored] 1: 1: 1 mole ratio If [HBr (aq) ] = 0.25 mol/L then.. [H + ] = 0.25 mol/L @ SATP; Kw = [H + (aq) ][OH - (aq) ] Kw = 1.0 x 10 -14 [OH - (aq) ] = K w / [H + (aq) ] = 1.0 x 10 -14 / 0.25 mol/L = 4.0 x 10 -14 mol/L The concentration of hydroxide ions is 4.0 x 10 -14 mol/L. STRONG BASES LiOH (aq), NaOH (aq), KOH (aq_, RbOH (aq) , CsOH (aq) , etc.. Dissociate completely in water to release hydroxide ions All group 1 hydroxides are strong bases When dissolved in water, these bases produce 1 mole of hydroxide ion for every mole of metal hydroxide NaOH (aq) Na + (aq) + OH - (aq) 3. Determine the hydrogen ion and hydroxide ion concentration in 500.0mL of an aqueous solution, containing 2.6 g of dissolved sodium hydroxide. NaOH (aq) Na + (aq) + OH - (aq) m of NaOH (a) solute = 2.6g v of NaOH (aq) solution = 500.0 mL n = m/M = 2.6g / (40 g / mol) = 0.065 mol c = n/v = 0.065 mol / 0.5L = 0.13 mol/L 1 mol NaOH (aq) : 1 mol Na + (aq) : 1 mol OH - (aq) @ SATP; K w = [H + (aq) ][OH - (aq) ] K w = 1.0 x 10 -14 [H + (aq) ] = K w / [OH - (aq) ] = 1.0 x 10 -14 / 0.13 mol/L = 7.7 x 10 -14 mol/L The concentration of hydrogen ions is 7.7 x 10 -14 mol/L and the concentration of hydroxide ions is 0.13mol/L. [H + ] & PH pH: A way to express the ACIDIT Y/BASICIT Y of an aqueous solution the amount of hydronium ion H 3 O + in solution • pH > 7 basic • pH = 7 neutral • pH < 7 acidic [ H + ( aq) ] < [ OH - (aq) ] [H + ( aq) ] = [ OH - ( aq) ] [H + ( aq) ] > [ OH - (aq) ] pH = -log[H 3 O + ] [H 3 O + ] = 10 -pH 1 . ) If [H 3 O + ] = 1.0 x 10 -7 mol/L, then pH is… pH = -log[H 3 O + ] = -log(1.0 x 10 -7 mol/L) = 7.0 2. Given the pH = 10.33, calculate the [H + (aq) ]. [H 3 O + ] = 10 -pH = 10 -10.33 = 4.7 x 10 -11 mol/L [OH - ] & POH pOH = -log[OH - ] [OH - ] = 10 -pOH pH + pOH = 14 Can simplify pH calculation – if either pH/pOH is known, the other can be found! Calculate the pH, pOH and [OH - (aq) ] of a 0.042mol/L HNO 3(aq) solution. HNO 3(aq) NO 3 - (aq) + H + (aq) 1:1:1 mole ratio [HNO 3(aq) ]= [H + (aq) ] = 0.042 mol/L pH = -log [H + ] = -log(0.042 mol/L) = 1 .40 pH + pOH =14 pOH = 14 – pH = 14 – 1 .40 = 1 2.60 [OH - ] = 10 -pOH = 10 -12.60 = 2.5 x 10 -13 mol/L The pH of the solution is 1 .40; the pOH is 1 2.60; and the [OH - ] = 2.5 x 10 -13 mol/L. • A solution was made by dissolving 0.627 g Ba(OH) 2 in 100.0 mL final volume. If Ba(OH) 2 is fully broken up into its ions, what is the pOH and the pH of this solution? • G: 0.627 g of Ba(OH) 2 dissolved in 100 .0 mL (final volume) • M of Ba(OH) 2 = 171.63 g / mol • R: pOH = ? pH = ? • A: n = m / M , c = n/ v • pOH = -log[OH - ], pH + pOH = 14 m = 6 . 27 g/L x 0 .1 L = 0 . 6 27 g n = m / M = 0 . 6 27 g / ( 171 . 63 g / m o l ) = 0 . 0 0 3 65 m o l B a ( OH ) 2 B a 2 + + 2 OH M o l e ra t i o 1 B a ( O H ) 2 : 2 OH n o f O H - = 2 x 0 . 0 0 3 6 5 = 0 . 0 07 31 mol [ O H - ] = 0 . 0 07 31 mol/0.1L = 0 . 07 31 M p O H = - l o g [ OH - ] = - l o g (0. 07 31) = 1 .14 pH + pO H = 14 pH = 1 2 . 9 P : Th e s o l ut ion h a s a fi n a l pOH o f 1 .14 a n d a fi n a l pH o f 1 2 . 9 . HOMEWORK Pg. 532 #1 , 2 Pg. 537 #4-6 Pg. 549 #17-19