Acids and Bases. CHEM 102 T. Hughbanks

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1 Acids and Bases CHEM 102 T. Hughbanks According to the Brønsted Lowry theory, all acid base reactions can be written as equilibria involving the acid and base and their conjugates. All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.

2 When a weak acid is in solution, the products are a stronger conjugate acid and base. Therefore equilibrium lies to the left. All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base. In following equilibrium, will reactants or products be favored? H 3 PO 4 (aq) + CH 3 CO 2 (aq) H 2 PO 3 (aq) + CH 3 CO 2 H(aq)

3 In following equilibrium, will reactants or products be favored? K b = K b = H 3 PO 4 (aq) + CH 3 CO 2 (aq) H 2 PO 3 (aq) + CH 3 CO 2 H(aq) K a = K a = Strong acid (HCl) + Strong base (NaOH) HCl (aq) + NaOH (aq) H 3 O + (aq) + NaCl (aq) Net ionic equation H 3 O + (aq) + OH (aq) 2H 2 O(l) Mixing equal molar quantities of a strong acid and strong base produces a neutral solution.

4 Weak acid (HCN) + Strong base (NaOH) HCN(aq) + OH (aq) CN (aq) + H 2 O(l ) Mixing equal amounts (moles) of a strong base and a weak acid produces a salt whose anion is the conjugate base of the weak acid. The solution is basic, with the ph depending on K b for the anion. CN (aq) + H 2 O(l ) HCN(aq) + OH (aq) Strong acid (HCl) + Weak base (NH 3 ) H 3 O + (aq) + NH 3 (aq) H 2 O(l ) + NH 4 + (aq) Mixing equal amounts (moles) of a weak base and a strong acid produces a conjugate acid of the weak base. The solution is basic, with the ph depending on K a for the acid. NH 4 + (aq) + H 2 O(l ) H 3 O + (aq) + NH 3 (aq)

5 CH CO H(aq) + NH (aq) CH CO (aq) + NH (aq) Weak acid + Weak base Product cation = conjugate acid of weak base. Product anion = conjugate base of weak acid. ph of solution depends on relative strengths of cation and anion.

6 Determining K from Initial Concentrations and ph HNO 2 (aq) + H 2 O(l ) H 3 O + (aq) + NO 2 (aq) 0.10 M HNO 2 (aq) ph = 2.17 [HNO 2 ] [H 3 O + ] [NO 2 ] Initial 0.10 Change Equilibrium Determining K from Initial Concentrations and ph H 2 S(aq) + H 2 O(l ) H 3 O + (aq) + HS (aq) 1.00 M H 2 S(aq) K a = [H 2 S] [H 3 O + ] [HS ] Initial 1.00 Change Equilibrium

7 Calculations with Equilibrium Constants Determining K from Initial Concentrations and ph In general, the approximation that [HA] equilibrium = [HA] initial x [HA] initial is valid whenever [HA] initial is greater than or equal to 100 K a. If this is not the case, the quadratic equation must by used. Calculations with Equilibrium Constants Determining ph after an acid/base reaction: Calculate the hydronium ion concentration and ph of the solution that results when 22.0 ml of 0.15 M acetic acid, CH 3 CO 2 H, is mixed with 22.0 ml of 0.15 M NaOH.

8 Calculations with Equilibrium Constants Determining ph after an acid/base reaction: Polyprotic Acids & Bases Because polyprotic acids are capable of donating more than one proton they present us with additional challenges when predicting the ph of their solutions. For many inorganic polyprotic acids, the ionization constant for each successive loss of a proton is roughly 10 5 times smaller than the previous step. This implies that the ph of many inorganic polyprotic acids depends primarily on the hydronium ion generated in the first ionization step. The hydronium ion produced in the second step can be neglected.

9 Polyprotic Acids & Bases Sulfurous acid, H 2 SO 3, is a weak acid capable of providing two H + ions (pk a1 = 1.9, pk a2 = 7.0). (a) What is the ph of a 0.45 M solution of H 2 SO 3? (b) What is the equilibrium concentration of the sulfite ion, SO 3 2- in the 0.45 M solution of H 2 SO 3? Polyprotic Acids, cont.! As we have seen, several common acids can potentially donate more than one proton. Sulfuric (H 2 SO 4 ) and phosphoric (H 3 PO 4 ) acids are particularly prominent examples: H 2 SO 4 : pk a1 < 0; pk a2 = 1.92 H 3 PO 4 : pk a1 = 2.12; pk a2 = 7.21; pk a3 = What are the concentrations of all species in a 0.1 M phosphoric acid solution?

10 Features of Polyprotic Acids! H 3 PO 4 is a representative example: pk a1 = 2.12; pk a2 = 7.21; pk a3 = At what ph are the concentrations of H 3 PO 4 and H 2 PO 4 - equal? At that ph, what the concentrations of HPO 4 2- and PO 4 3-? Tough question to answer directly: What is the ph of a H 2 PO 4 - or HPO 4 2- salt solution? (e.g., K 2 HPO 4 or KH 2 PO 4 solution) Easier: what s the ph when [H 2 PO 4 - ] = [PO4 3- ]? K a2 = [H 3 O+ ][HPO 2-4 ] ; K [H 2 PO - a3 = [H 3 O+ ][PO 3-4 ] 4 ] [HPO 2-4 ] when [H 2 PO 4 - ] = [PO 4 3- ], 2- ] [H 3 O + ][HPO 2-4 ] [HPO = K 4 K a3 a2 [H 3 O + ] ph = 1 ( 2 pk + pk a2 a3) But at this ph, [HPO 4 2- ] >> [H 2 PO 4 - ] = [PO 4 3- ]

11 ph = (1/2)(pK a1 + pk a2 )" ph = (1/2)(pK a2 + pk a3 )" [H 3 PO 4 ] = [H 2 PO 4 - ]" [H 2 PO 4 - ] = [H 2 PO 4 2- ]" [H 2 PO 4 2- ] = [PO 4 3- ]" ph and composition in polyprotic systems! The Lewis Concept of Acids & Bases The concept of acid base behavior advanced by Brønsted and Lowry in the 1920 s works well for reactions involving proton transfer. However, a more general acid base concept, was developed by Gilbert N. Lewis in the 1930 s. A Lewis acid is a substance that can accept a pair of electrons from another atom to form a new bond. A Lewis base is a substance that can donate a pair of electrons to another atom to form a new bond.

12 The Lewis Concept of Acids & Bases A + B: B: A Acid Base Adduct The Lewis Concept of Acids & Bases

13 Reaction of a Lewis Acid & Lewis Base The Lewis Concept of Acids & Bases The formation of a hydronium ion is an example of a Lewis acid / base reaction H + A C I D O H H O H H H B A S E The H + is an electron pair acceptor. Water with it s lone pairs is a Lewis acid donor.

14 Lewis Acid Base Reactions Lewis Acids & Bases Metal cations often act as Lewis acids because of open d-orbitals. Co 2+ ACID O H Co 2+ H BASE O H H

15 Lewis Acids & Bases The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 leads to Coordinate Complex ions. Lewis Acids & Bases Aqueous solutions of Fe 3+, Al 3+, Cu 2+, Pb 2+, etc. are acidic through hydrolysis. [Al(H 2 O) 6 ] 3+ (aq) + H 2 O(l) [Al(H 2 O) 5 (OH)] 2+ (aq) + H 3 O + (aq)

16 Hydrolysis of Metal Complexes can give acidic solutions" pk a s for [M(H 2 O) 6 ] n+ [Fe(H 2 O) 6 ] [Cr(H 2 O) 6 ] [Al(H 2 O) 6 ] [Fe(H 2 O) 6 ] [Cu(H 2 O) 6 ] [Ni(H 2 O) 6 ]