Acid & Bases Arrhenius: Acids are proton (H + ) sources and bases are hydroxide ion (OH - ) sources. E.g. HCl is an acid and NaOH a base Brønsted-Lowry: Acids are proton sources and bases are proton acceptors. E.g. HCl is an acid and NH 3 a base, these form the conjugate base Cl - & NH 4 +. Lewis: Acids are electron pair acceptors and bases are electron pair donors. E.g. AlCl 3 & :NH 3. to form Cl 3 Al:NH 3

Acid & Bases Acid Strength Acid strength describes the position of the equilibrium or the size of K eq HX  H + + X -, K A = [H + ][X - ]/[HX] for example: H 2 SO 4 H + + HSO 4 - K A ~ 1.0*10 3 HSO 4 - H + + SO 4 - K A ~ 1.2*10 -2 In the first case the reaction essentially goes to completion, in the second HSO 4 - hardly dissociates.

Acid-Base Equilibria For now, an acid is a proton (H + ) donor and a base is either a proton acceptor or a hydroxide (OH - ) donor. Water will be the solvent throughout this discussion. Water dissociates to give both a proton and a hydroxide ion. This may be written several ways. I choose to write it in the simplest, least correct, way -- H 2 O  H + + OH -

Acid-Base Equilibria this equilibrium may be (poorly) modeled by the equation: K eq = [H + ][OH - ]/[H 2 O] where K eq = 1.0* at 25 o C & the activity of water is 1. One may rewrite the equation as: K eq *[H 2 O] = K W = [H + ][OH - ] = 1.0* Also, in neutral solution, [H + ]=[OH - ]= (1.0* ) 1/2 = 1.0*10 -7

Acid-Base Equilibria solving for [H + ], we get:[H + ]=[OH - ]= (K W ) 1/2 Or, taking the Log 10 of both sides we get: Log 10 K W = Log 10 [H + ] + Log 10 [OH - ] = -14 multiplying both sides by -1 gives -Log 10 K W = -Log 10 [H + ] + -Log 10 [OH - ] = 14

Acid-Base Equilibria defining “p” to mean "1/Log 10 " or -Log 10, we can rewrite our equation to become: pK W = pH + pOH = 14 Thus, for water or a "neutral" solution at 25 o C, the pH = pOH = 14/2 = 7 or both the [H + ] and [OH - ] = M

Acid-Base Equilibria For any other type solution, the hydrogen or hydroxide ion concentrations will depend on BOTH the dissociation of water and ions contributed by other components of a solution. For example, if we make a 0.20M solution of nitric acid the hydrogen ion concentration would depend on the hydrogen ion from the nitric acid and from the dissociation of water. Let’s model this on the blackboard.

Acid-Base Equilibria A MAJOR PITFALL for the lazy or unwary student is to ignore ion sources. This person would say, "we need consider only the strong acid or strong base concentration and may always ignore the contribution from water." NOT! If you remember that the ion concentrations in equilibrium expressions are ALWAYS TOTALS, Acid-base equilibria, indeed all solution equilibria, are much easier to understand.

Acid-Base Equilibria Let’s look at what happens if we have a weak acid instead of a strong one. HA  H + + A - K A = [H + ] T [A - ] [HA] where the “T” means total hydrogen

Acid-Base Equilibria Solving for [H + ] T, we have: [H+] T = [Ac - ] + [OH - ] [H + ] T = [Ac - ] + K w /[H+] [Ac - ] = [H + ] T - K w /[H+] & K A = [H + ][Ac - ]/[HAc] = [H + ] ([H + ] - K w /[H+] ) C A - ([H + ] - K w /[H+] ) = ([H + ] 2 + K w C A – {([H + ] 2 – K w )/[H+]}

Acid-Base Equilibria If [H + ] T > ~ 10 -5, this simplifies to: [H + ] T = K A *[HAc]/[Ac - ] Or, in this case, the water contribution is negligible compared to the proton from the acetic acid.

Acid-Base Equilibria Let's look at what's left: [H + ] T = K A *[HA]/[A - ] C A M 0 0 HA  H + + A - & K A = [H + ][A - ]/[HA] (C A -x)M xM xM or K A = X*X/(C A -X), expanding we find X 2 + C A X - K A C A = 0, a quadratic or, if X is much smaller than C A, K A = X 2 /C A, & X=[H+]= (C A K A ) 0.5

Acid-Base Equilibria If we have a very dilute SA solution: HA  H + + A - & [H + ] = [A - ] + [OH - ] [H + ] = [A - ] + K W / [H + ] ; [A - ] = C A solving for C A ; C A = ([H+] 2 – Kw)/[H + ] or [H+] 2 – C A [H + ] - Kw = 0 For example, if C A = 1.0*10 -7, pH ~ 6.80

Acid-Base Equilibria Alternatively: If one adds enough nitric acid to water to prepare a 1*10 -7 M solution, the only equilibrium is: H 2 O  H + + OH - K W = [H + ][OH - ] [H + ][OH - ] CiCi 2* *10 -7 ΔCΔC-x CfCf 2* x1* x

Digression – base 10 Logs Numbe r Log =2*2, log 4= = = 25*10 -1 = 5*5*10 -1 log 2.5= (-1) =.40 Antilog of – 0.60 = antilog(-1)+.40 = 2.5*10 -1 or 0.25 if you multiply #’s, logs are added divide #’s logs are subtracted take a # to a power, multiply log times power