ACID-BASE EQUILIBRIUM

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 WHAT IS AN ACID AND A BASE?

Arrhenius Acids HBr (aq)  H + (aq) + Br - (aq) H 2 SO 4(aq)  H + (aq) + HSO 4 - (aq) HClO 4(aq)  H + (aq) + ClO 4 - (aq) Arrhenius Bases LiOH (aq)  Li + (aq) + OH - (aq) KOH (aq)  K + (aq) + OH - (aq) Ba(OH) 2(aq)  Ba 2+ (aq) + 2OH - (aq)

 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. PROBLEMS WITH ARRHENIUS THEORY

\ Acid (H + donor) HCl (g) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) Base (H + acceptor) BRONSTED-LOWRY THEORY

H + donor (acid) H + acceptor (base)

H + donor (acid) H + acceptor (base)

 ** 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 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH − (aq) 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 +. AMPHOTERIC H + acceptor (base) H + donor (acid)

 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  H 2 SO 4 – HSO 4 -  H 2 O – OH - CONJUGATE ACID-BASE PAIRS

COMMON CONJUGATE ACID-BASE PAIRS

 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!) A COMPETITION FOR PROTONS

 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 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH − (aq)  Simplified into…  H 2 O (l)  H + (aq) + OH − (aq) AUTOIONIZATION OF WATER

 [H + (aq) ] = [OH - (aq) ] = pH 7  [H + (aq) ] > [OH - (aq) ] = low pH  [H + (aq) ] < [OH - (aq) ] = high pH RECALL FROM GR. 11… NEUTRAL ACIDIC BASIC

 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] STRONG ACIDS

 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  [OH - (aq) ] = K w / [H + (aq) ] = 1.0 x / 0.15 mol/L  = 6.7 x mol/L  The concentration of hydroxide ions is 6.7 x 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; K w = [H + (aq) ][OH - (aq) ]  K w = 1.0 x  [OH - (aq) ] = K w / [H + (aq) ] = 1.0 x / 0.25 mol/L = 4.0 x mol/L  The concentration of hydroxide ions is 4.0 x mol/L.

 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) STRONG BASES

 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 = mL  n = m/M  = 2.6g / (40 g/mol) = mol  c = n/v = 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  [H + (aq) ] = K w / [OH - (aq) ] = 1.0 x / 0.13 mol/L  = 7.7 x mol/L  The concentration of hydrogen ions is 7.7 x mol/L and the concentration of hydroxide ions is 0.13mol/L.

 pH: A way to express the ACIDITY/BASICITY of an aqueous solution  the amount of hydronium ion H 3 O + in solution pH > 7 basic [H + (aq) ] < [OH - (aq) ] pH = 7 neutral [H + (aq) ] = [OH - (aq) ] pH [OH - (aq) ]  pH = -log[H 3 O + ]  [H 3 O + ] = 10 -pH  1. ) If [H 3 O + ] = 1.0 x mol/L, then pH is…  pH = -log[H 3 O + ]  = -log(1.0 x mol/L)  = 7.0 [H + ] & PH

 2. Given the pH = 10.33, calculate the [H + (aq) ].  [H 3 O + ] = 10 -pH  =  = 4.7 x mol/L

 pOH = -log[OH - ]  [OH - ] = 10 -pOH  pH + pOH = 14  Can simplify pH calculation – if either pH/pOH is known, the other can be found! [OH - ] & POH

 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) ] = mol/L  pH = -log [H + ] = -log(0.042 mol/L)  = 1.40  pH + pOH =14  pOH = 14 – pH = 14 – 1.40 =  [OH - ] = 10 -pOH  =  = 2.5 x mol/L  The pH of the solution is 1.40; the pOH is 12.60; and the [OH - ] = 2.5 x mol/L.

A solution was made by dissolving g Ba(OH) 2 in 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: g of Ba(OH) 2 dissolved in mL (final volume) M of Ba(OH) 2 = g/mol R: pOH = ? pH = ? A: n = m / M, c = n/ v pOH = -log[OH - ], pH + pOH = 14  m = 6.27g/L x 0.1L = g  n = m / M = 0.627g / ( g/ mol) = mol  Ba(OH) 2  Ba OH -  Mole ratio 1 Ba(OH) 2 : 2OH -  n of OH - = 2 x = mol  [OH - ] = mol/0.1L = M  pOH =-log [OH - ] = - log(0.0731)  = 1.14  pH+pOH = 14  pH = 12.9  P: The solution has a final pOH of 1.14 and a final pH of 12.9.

 Pg. 532 #1, 2  Pg. 537 #4-6  Pg. 549 #17-19 HOMEWORK