# PH = pK a2 pH = pK a1 pK a2 = 2.33 pK a1 = 9.45 Acids and Bases in aqueous solution.

## Presentation on theme: "PH = pK a2 pH = pK a1 pK a2 = 2.33 pK a1 = 9.45 Acids and Bases in aqueous solution."— Presentation transcript:

1 pH = pK a2 pH = pK a1 pK a2 = 2.33 pK a1 = 9.45 Acids and Bases in aqueous solution.

2 The properties of water: H-bonds Water is held together by H-bonding, which gives it its surprisingly high B.Pt. and M.Pt. for such a small molecule. Note that the heavier H 2 S molecule is a gas at room temperature. Water is also weakly ionized into H + and OH - ions, as discussed next. ~ 2.75 Å

3 The self-ionization of water: Water ionizes weakly to produce OH - and H + ions, where K w is the ionization constant of water, according to : H 2 O( l )  H + ( aq ) + OH - ( aq ) K w = 10 -14 This means that the product of the concentration (molarity) of [H + ] and [OH - ] must always be 10 -14. K w =10 -14 =[H + ] [OH - ][1] Thus, for pairs of [H + ] and [OH - ] concentrations in solution we have: [H + ] = 1.0 M, then [OH - ] = 10 -14 /1.0 = 10 -14 M [H + ] = 0.01 M, then [OH - ] = 10 -14 /0.01 = 10 -12 M [H + ] = 10 -7 M, then [OH - ] = 10 -14 /10 -7 = 10 -7 M [H + ] = 10 -12 M, then [OH - ] = 10 -14 /10 -12 = 10 -2 M [H + ] = 10 -15 M, then [OH - ] = 10 -14 /10 -15 = 10 M

4 pH and pOH The concentration of [H + ] and [OH - ] can be expressed as the pH and pOH: (p in pH or pOH means ‘potency’, and means that any constant ‘K’ or concentration with a ‘p’ in front of it is represented as pK = -log K, or pH = -log [H + ]) pH = - log [H + ]pOH = - log [OH - ] We can show that: pK w = 14.0 =pH + pOH

5 Thus, for any solution where the pH is known, pH + pOH = 14.0, or pOH = 14 – pH. [H + ]pHpOH[OH - ] 10 M-1.015.010 -15 M 1 M 0.014.010 -14 M 0.01 M 2.012.010 -12 M 10 -5.7 M 5.7 8.310 -8.3 M 10 -7 M 7.0 7.010 -7.0 M 10 -9.3 M 9.3 4.710 -4.7 M 10 -12 M12.0 2.010 -2 M 10 -14 M14.0 0.01 M 10 -15.09 M15.09 -1.0912 M

6 Acid dissociation constants (K a ): A weak Brønsted acid such as acetic acid (CH 3 COOH) will dissociate to give off protons in aqueous solution according to: CH 3 COOH (aq)  CH 3 COO - (aq) + H + (aq) The extent of such dissociation is controlled by the acid dissociation constant, K a. K a =[ CH 3 COO - ] [ H + ] [ CH 3 COOH ] acetic acid acetate ion conjugate acid conjugate base proton

7 The acid dissociation constant of acetic acid: The value of K a for acetic acid is 10 -4.84, so the pK a is 4.84. This gives us the expression: 10 -4.84 =[ CH 3 COO - ] [ H + ][2] [ CH 3 COOH ] This expression can be used to solve problems relating to the ionization of acetic and other acids. For example, what is the pH of a 0.1 M solution of acetic acid? Equation 2 must be obeyed at all times, so we have: 10 -4.84 = [CH 3 COO - ][H + ]/[0.1] = [H + ] 2 /[0.1] so [H + ] = (10 -4.84 x 0.1) 0.5 = 10 -2.92 M, pH = 2.92.

8 Notice that in solving the above problem, ionization of acetic acid produces equal concentrations of H + and CH 3 COO - ions. One assumes that the amount of ionization is small, so that [CH 3 COOH] is not corrected for the small amount that it decreases, and we assume that [CH 3 COOH] is still ≈ 0.1 M. b) What is the pH of a 1 M solution of ammonia if the pK a of NH 4 + is 9.2? Note: pK a of an acid plus pK b of its conjugate base = pK w pK a (NH 4 + ) + pK b (NH 3 ) = pK w = 14.0. pK b NH 3 = 14.0 – 9.2 = 4.8, so K b = 10 -4.8. 10 -4.8 = [NH 4 + ] [OH - ] / [NH 3 ] so again we have: [OH - ] = (10 -4.8 x 1) 0.5, = 10 -2.4. If [OH - ] = 10 -2.4 M, then pOH = 2.4, and pH = 14-2.4 = 11.6

9 The species distribution diagrams of acids and bases: We can calculate the percentage of an acid (e.g. CH 3 COOH) or its conjugate base (CH 3 COO - ) that is present as the acid or the conjugate base at any given pH value if the pK a is known (4.84). Thus, for acetic acid at pH 3.7 we have: 10 -4.84 =[CH 3 COO - ] [H + ] / [CH 3 COOH] =[CH 3 COO - ] [10 -3.7 ] / [CH 3 COOH] [CH 3 COO - ] / [CH 3 COOH] = 10 -4.84 / 10 -3.7 = 10 -1.14 = 0.072 so percent of [CH 3 COO - ] at pH 3.7 = 100% x (0.072/(1+0.072)) = 7.2 %. % of [CH 3 COOH] = 100 – 7.2 = 92.8 %.

10 CH 3 COOHCH 3 COO - pH 50 = pK a

11 The relationship between pH 50 and the pK a The pH at which the concentrations of CH 3 COOH and CH 3 COO - in solution are equal, i.e. both are 50%, is known as the pH 50. On the previous slide it was indicated that the pK a of the acid equals pH 50. This arises simply as follows: K a =10 -4.84 =[ CH 3 COO - ] [ H + ] [ CH 3 COOH ] 10 -4.84 = [ H + ] or pK a = pH 50 Thus, we saw that for CH 3 COOH the crossover point for [CH 3 COOH] and [CH 3 COO - ] occurred at pH 4.84, which is the pK a. Similarly, on the next slide, we see that the crossover point for [NH 4 + ] and [NH 3 ] occurs at pH 9.22, which is the pK a of NH 4 +. [CH 3 COO - ] = [CH 3 COOH] at pH 50, so these cancel

12 NH 4 + NH 3 pH 50 = pK a The species distribution diagram of ammonia/ammonium ion:

13 Multiprotic acids and bases: Many acids have more than one ionizable proton, and many bases have more than one proton acceptor site. A familiar example of this is triprotic phosphoric acid: H 3 PO 4  H 2 PO 4 - +H + pK a3 = 2.15 H 2 PO 4 -  HPO 4 2- +H + pK a2 = 7.20 HPO 4 2-  PO 4 3- +H + pK a1 = 12.38 It can be shown that for a monoprotic acid, when the concentration of the acid and its conjugate base are equal (both 50%), then pK a = pH. For polyprotic acids the average extent of protonation at any pH is given by the function nbar. For the phosphate ion: nbar = average number of protons bound per phosphate

14 The relationship between nbar, pH, and the pK a values of multiprotic acids On a previous slide we showed that for a monoprotic acid such as CH 3 COOH, the pK a = pH 50. At pH 50 for a monoprotic acid, in fact, nbar = 0.5. This is the point at which half of the acetate ions have a proton on them, while half do not. For polyprotic acids, similar considerations apply. At nbar = 0.5 for phosphate, we have [HPO 4 2- ] = [PO 4 3- ], and the pH at nbar = 0.5 = pK a1. In fact the half-values of nbar give us the pK a values from the corresponding pH values: nbarpH = pK an 0.5 pH = pK a1 1.5pH = pK a2 2.5pH = pK a3 etc. Thus from a curve of nbar versus pH for phosphoric acid, we can estimate the pK a values from the pH values corresponding to nbar = 0.5, 1.5, and 2.5:

15 pH = pK a3 pH = pK a2 pH = pK a1 pK a3 = 2.15 pK a2 = 7.20 pK a1 = 12.38 Plot of nbar versus pH for phosphate: pH H 3 PO 4 H 2 PO 4 - HPO 4 2- PO 4 3-

16 H 3 PO 4 H 2 PO 4 - HPO 4 2- PO 4 3- Species distribution diagram for phosphoric acid pH

17 pH = pK a2 pH = pK a1 pK a2 = 2.33 pK a1 = 9.45 pH

18 pK a2 = 2.52 pK a1 = 9.46 pK a1 = 9.52 pK a2 = 6.13 pK a3 = 2.69 NTA (nitrilo- triacetate) EDTA (ethylene- diamine tetraacetate)

19 Lewis acids and bases: You will recall Brønsted acids and bases, where a Brønsted acid is a proton donor and a Brønsted base is a proton acceptor. A broader definition is that of Lewis Acids and Bases, where a Lewis Acid is an electron acceptor, and a Lewis Base is an electron donor. Gilbert Newton Lewis (1875-1946) BF 3, a Lewis Acid F -, a Lewis Base Lewis Acid accepts electrons from the Lewis Base to form a complex.

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