Measuring Ka, Buffers and the pH of Ionic Solutions Stuff we need to know for the Weak Acid Lab in Chem V01BL

Anions as Weak Bases every anion can be thought of as the conjugate base of an acid therefore, every anion can potentially be a base A−(aq) + H2O(l) ⇋ HA(aq) + OH−(aq) the stronger the acid HA is, the weaker the conjugate base A- is an anion that is the conjugate base of a strong acid is pH neutral Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq) an anion that is the conjugate base of a weak acid is basic F−(aq) + H2O(l) ⇋ HF(aq) + OH−(aq) since HF is a weak acid, the position of this equilibrium favors the right

Use the Table to Determine if the Given Anion Is Basic or Neutral NO3− the conjugate base of a strong acid, therefore neutral HCO3− the conjugate base of a weak acid, therefore basic PO43−

Polyatomic Cations as Weak Acids some polyatomic cations with a H in them can be thought of as the conjugate acid of a base (:B) BH+(aq) + H2O(l) ⇋ :B(aq) + H3O+(aq) the stronger the base :B(aq) is, the weaker the conjugate acid BH+ is a cation that is the counterion of a strong base (Na+, K+ etc) is pH neutral a cation that is the conjugate acid of a weak base is acidic NH4+(aq) + H2O(l) ⇋ :NH3(aq) + H3O+(aq) since :NH3 is a weak base, the position of this equilibrium favors the right

Metal Cations as Weak Acids While alkali metal cations and alkali earth metal cations pH neutral Cations of small, highly charged metals (especially transition metals) are weakly acidic Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+ (aq) + H3O+(aq)

Classifying Salt Solutions as Acidic, Basic, or Neutral if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base NH4F Ka of NH4+ is larger than Kb of the F−; therefore the solution will be acidic

Determine whether a solution of the following salts is acidic, basic, or neutral SrCl2 Sr2+ is the counterion of a strong base, pH neutral Cl− is the conjugate base of a strong acid, pH neutral solution will be pH neutral AlBr3 Al3+ is a small, highly charged metal ion, weak acid Br− is the conjugate base of a strong acid, pH neutral solution will be acidic CH3NH3NO3 CH3NH3+ is the conjugate acid of a weak base, acidic NO3− is the conjugate base of a strong acid, pH neutral

NH4+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq) Determine whether a solution of the following salts is acidic, basic, or neutral and write out any hydrolysis reaction that happens if it is basic or acidic NH4Cl NH4+ is the conjugate acid of the weak base (NH3), so is acidic Cl− is the conjugate base of a strong acid (HCl), pH neutral Overall the solution will be acidic. Since the NH4+ is acidic it reacts with water (undergoes hydrolysis) NH4+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq) Have a go completing the first page on today’s handout

Measuring pKa of a weak acid by titration CH3COOH(aq) + NaOH(aq)  NaCH3COO(aq) + H2O(l)

Measuring pKb of a weak base by titration 𝐾 𝑏 = 𝑁𝐻 4 + 𝑂𝐻 − 𝑁𝐻 3 NH3(aq) + HCl(aq)  NH4Cl(aq) + H2O(l) −𝑙𝑜𝑔 𝐾 𝑏 =−𝑙𝑜𝑔 𝑁𝐻 4 + 𝑂𝐻 − 𝑁𝐻 3 Half equivalence point [NH3] = [NH4+] 𝑝𝐾 𝑏 =−𝑙𝑜𝑔 𝑂𝐻 − −𝑙𝑜𝑔 𝑁𝐻 4 + 𝑁𝐻 3 𝑝𝐾 𝑏 =𝑝𝑂𝐻 −𝑙𝑜𝑔 𝑁𝐻 4 + 𝑁𝐻 3 But we can only measure pH 14−𝑝𝐾 𝑏 =14−𝑝𝑂𝐻 −𝑙𝑜𝑔 𝑁𝐻 4 + 𝑁𝐻 3 𝑝𝐾 𝑎 =𝑝𝐻 −𝑙𝑜𝑔 𝑁𝐻 4 + 𝑁𝐻 3 At the ½ equivalence point 𝑝𝐾 𝑎 =𝑝𝐻 𝑝𝐾 𝑏 =14− 𝑝𝐾 𝑎

Solutions close to the 1/2 equivalence point: Buffers Aqueous Solution pH Δ pH Bottled Water (30 mL) 5.6 0.7 Bottled Water + 1 drop of 0.1M HNO3 4.9 Deionized Water (30 mL) 6.0 2.2 Deionized Water + 1 drop of 0.1M HNO3 3.8 pH Buffer 7.8 0.0 pH Buffer + 1 drop of 0.1M HNO3 What should the pH be when we add 1 drop of 0.1M HNO3 to 30mL of deionized water ? Assume 1 drop = 0.05mL When this 1 drop is dissolved in 30 mL of H2O

How Acid Buffers Work HA(aq) + H2O(l) ⇋ A−(aq) + H3O+(aq) buffers work by applying Le Châtelier’s Principle to weak acid equilibrium – buffers are solutions that resist change in pH Adding OH- to the buffer: the OH- reacts with the H3O+ to make 2H2O(l) so it removes H3O+ so drives the equilibrium to the right effectively doing the following Effectively 𝐻𝐴 (𝑎𝑞) + 𝑂𝐻 (𝑎𝑞) − → 𝐴 𝑎𝑞 − + 𝐻 2 𝑂 (𝑙) By Le Châtelier it drives the equilibrium to the right causing more HA to make A- and maintaining [H3O+] Adding H3O+ to the buffer: By Le Châtelier it drives the equilibrium to the left (because we are adding more product) causing A- to react with the H3O+ to make more HA maintaining [H3O+] Effectively 𝐴 𝑎𝑞 − + 𝐻 3 𝑂 (𝑎𝑞) + ⇢ 𝐻𝐴 (𝑎𝑞) + 𝐻 2 𝑂 (𝑙)

Calculating the pH of Buffers What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5? First let us try this using an ICE table

Henderson–Hasselbalch Equation Earlier we showed that 𝑝𝐾 𝑎 =𝑝𝐻−𝑙𝑜𝑔 𝐴 − 𝐻𝐴 Or 𝑝𝐻 = 𝑝𝐾 𝑎 +𝑙𝑜𝑔 𝐴 − 𝐻𝐴 This is really useful for buffer solutions it is called the Henderson – Hasselbalch equation when we replace the equlibrium values with initial concentrations 𝑝𝐻 = 𝑝𝐾 𝑎 +𝑙𝑜𝑔 𝐴 − 0 𝐻𝐴 0 This approximation works whenever x is small works with an ICE table