K sp, K a and K b

 Much like with a system of equations, a solution is also an equilibrium  NaCl(aq)  Na + (aq) + Cl - (aq)  The ions in this solution are constantly dissociating and re-associating

 What is a saturated solution? ◦ A solution which has reached its capacity of a solute  What is a super-saturated solution? ◦ A solution which holds more than its full capacity of a solute ◦ Video demonstration…  Now onto the real stuff…

 K a – The acidity constant  K b – The Alkalinity (base) constant  K sp – Solubility product constant  K water – Water ionization constant

 An acid is a substance that dissociates in water to produce hydrogen ions (H + ) ◦ HCl (aq) -> H + (aq) + Cl - (aq)  A base is a substance that dissociates in water to produce hydroxide ions (OH - ) ◦ NaOH (aq) -> Na + (aq) + OH - (aq)

 Acids are neutralized by a base and vice versa ◦ NaOH + HCl -> NaCl + H 2 O  Acids and bases can be stronger or weaker  You need more of a weak base to neutralize a strong acid

 NaOH ◦ Base!  HCl ◦ Acid!  H 2 SO 4 ◦ Acid!  NH 3 ◦ Base!

 An acid is a substance in which a proton (Hydrogen atom, H + ) can be removed. An acid is seen as a proton donor. Seeing how a single H + cannot exist on its own, it can also be shown as a hydronium ion (H 3 O + )  A base is a substance that can remove a proton from an acid. A base is seen as a proton acceptor.

 In each acid/base reaction, there are 2 conjugate acid-base pairs  Ex: HCl (aq) + H 2 O (aq) -> H 3 O + (aq)+ Cl - (aq)  HCl and Cl- are one pair (A-B) ◦ HCl is the acid and Cl- is the base  H 2 O and H 3 O + are the other pair ◦ H 2 O is the base and H 3 O + is the acid

 NH 3 (aq) + H 2 O (aq) -> NH 4 + (aq) + OH - (aq)  NH 3 and NH 4 + ◦ NH 3 is the base and NH 4 + is the acid  H 2 O and OH - ◦ H 2 O is the acid and OH - is the base

 Looking back at the two conjugate Acid-Base pairs, is H 2 O an acid or a base?  In the Brønsted-Lowry theory of acids and bases, water can be considered an acid or a base, depending on its role in the reaction.

 The dissociation of ions in water is an equilibrium  The pH of the solution, which measures its acidity, is determined by where the equilibrium settles  This equilibrium can be quantified using the ionization of water constant K water

 This constant K water makes it possible to understand the interdependence of Hydronium ions (H 3 O + ) and Hydroxide ions (OH - )  Before we explore this wonderful relationship, let us go over what exactly the pH and pOH are...

 pH is a quantitative value attributed to the acidity of a solution  The lower the pH value, the higher the concentration of the hydronium ions (H 3 O+), and therefore the stronger the acid.  pOH is a quantitative value attributed to the alkalinity or basicity of a solution  The lower the pOH, the higher the concentration of the hydroxide ions (OH-) and therefore the stronger the base.

 pH and pOH can be expressed as the following mathematical expressions  pH = -log [H 3 O + ] ◦ [H 3 O + ] = 10 -pH  pOH = -log [OH - ] ◦ [OH - ] = 10 -pOH

 Express in the form of pH, the hydronium (H 3 O + ) concentration of 4.7 x mol/L in an aqueous solution. Is this solution acidic, neutral or basic?  Data ◦ [H 3 O+] = 4.7 x ◦ pH = ?

 pH = - log [H3O+]  pH = - log (4.7 x )  pH =  The solution is basic due to its pH being higher than 7

 Express the pOH of 3.60 in the form of the hydroxide (OH - ) concentration  Data ◦ pOH = 3.60 ◦ [OH - ] = ?

 [OH - ] = 10 -pOH  [OH - ] =  [OH - ] = 2.5 x  The concentration of hydroxide (OH - ) is 2.5 x mol/L

 What is [H 3 O + ] at pH 7? ◦ 1.00 x  What is [OH - ] at pOH 7? ◦ 1.00 x 10 -7

 The ionization of water follows this simple formula  2 H 2 O (l)  H 3 O + (aq) + OH - (aq)  Once this equation has reached equilibrium, we obtain the ionization of water constant K water

 K w = [H 3 O + ] x [OH - ]  If water is neutral, pH 7, then we know the concentrations of the hydronium and hydroxide ions.  The concentration of both ions is 1.00 x  Therefore...

 K w = [H3O+] x [OH-]  K w = 1.00 x x 1.00 x  K w = 1.00 x  This is always at 25°C

 By carrying out the logarithmic inverse of each side of the equation, the following equivalence can be obtained  -log [H 3 O + ] + -log [OH - ] = -log (1.00 x )  -log [H 3 O + ] + -log [OH - ] =14  pH + pOH = 14

 Knowing that K w is constant in all aqueous solutions, we can use this to determine the concentration H 3 O + and OH - ions in any acidic or basic solutions  Example!  At 25°C, a hydrochloric acid solution has a pH of 3.2. What is the concentration of each of the ions in this solution?

 [H3O + ] = 10 –pH  [H3O + ] = = 6.3 x  Kw = [H3O + ] x [OH - ] = 1.00 x  [OH - ] = 1.00 x / [H3O + ]  [OH - ] = 1.00 x / 6.3 x  [OH - ] = 1.58 x

Acidity and Basicity Constants

 Here we will be quantifying the strength of acids and bases  The stronger the acid or base depends on how it dissociates  The more dissociation, the stronger the acid

 When an acid comes into contact with water, a certain amount of dissociation takes place  In a strong acid, as much as 100% will dissociate ◦ Ex: HCl  In a weak acid, very little will dissociate. As little as 1% ◦ Ex: Acetic acid (Vinegar)

 Ionization percentage can be calculated by dividing the concentration of the H 3 O + ions by the concentration of the original acid and multiplied by 100  % = [H 3 O + ] eq / [HA] i * 100  Ensure that all of the concentrations are in the same units

 This can only be done using a weak acid, why? ◦ If there is none of the original acid left, you can’t calculate an equilibrium constant  Using the following general equilibrium, we can calculate the K a ◦ HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq)

 Ka = ([H 3 O + ] * [A - ]) / [HA]  Since water is a liquid, there is no concentration  We cannot do this with a strong acid because there is none of the original HA acid left and you cannot divide by 0  Would a weak acid have a higher or lower acidity constant? ◦ Lower!

 To find the acidity constant, all of the concentrations must be known  Also, if the Ka is known, then we can use that to predict either the final concentration of the [H 3 O + ], or the initial concentration of the [HA]  It can also be used to calculate the pH

 The basicity constant can also be calculated along the same lines  Using the following general equilibrium formula  B (aq) + H 2 O (l)  HB + (aq) + OH - (aq)  Kb = ([HB + ] * [OH - ]) / [B]  Again, the weaker the base, the smaller the constant

Solubility Product Constant

 A saturated solution that contains non- dissolved solute deposited at the bottom of a container is an example of a system at equilibrium.  The solubility of a substance corresponds to the maximum quantity of a substance that dissolves in a given volume of water  Usually given as g/100ml

 BaSO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq)  K sp = [Ba 2+ ] * [SO 4 2- ]  General formula  X n Y m(s)  nX + (aq) + mY - (aq)  K sp = [X + ] n * [Y - ] m

 The solubility of silver carbonate (Ag 2 CO 3 ) is 3.6 x g/100ml of solvent at 25°C. Calculate the value of the solubility product constant of silver carbonate.  Steps  1- Find the concentration of the Ag 2 CO 3 using M = m / n then the solubility ◦ Where M is molar mass, m is mass and n is the amount in moles of Ag 2 CO 3  2- Calculate the K sp

 1- M = m/n  n = m/M  n = 3.6 x g / g/mol  n = 1.3 x mol for 100 ml (0.1 L)  Solubility = 1.3 x / 0.1 L  Solubility = 1.3 x mol/L

 Ag 2 CO 3  2 Ag + (aq) + CO 3 2- (aq)  K sp = [X + ] n * [Y - ] m  K sp = [Ag + ] 2 * [CO 3 2- ]  [Ag + ] = 2 * [Ag 2 CO 3 ] = 2 * 1.3 x mol/L  [Ag + ] = 2.6 x mol/L  [CO 3 2- ] = [Ag 2 CO 3 ] = 1.3 x mol/L  K sp = [Ag + ] * [CO 3 2- ]  K sp = (2.6 x ) 2 mol/L * 1.3 x10 -4 mol/L  K sp = 8.8 x