Chemical Equilibrium Acids & Bases in Aqueous Solution 1

Hey! There’s a test FRIDAY! Test material ends at LAST FRIDAY! Everything through General Equilibrium One 8-1/2x11 inch “cheat sheet” with whatever you want on it. 2

K is K is K is K No matter what type of reaction you are talking about – equilibrium properties remain the same. K c, K p, K a, K b, K w, K sp, K f The subscripts refer to certain specific TYPES of equilibria, but… 3

K is K is K is K 4

Acid Dissociation Reactions  This is just a specific type of reaction.  Referring to Bronsted-Lowry acids: proton donors  An acid is only an acid when in the presence of a base  Water is the universal base 5

General K a Reaction The general form of this reaction for any generic acid (HA) is: HA (aq) + H 2 O (l)  A - (aq) + H 3 O + (aq) Acid Base Conjugate Conjugate base acid 6

Shorthand Notation Sometimes the water is left out: HA (aq)  A - (aq) + H + (aq) This is simpler, but somewhat less precise. It looks like a dissociation reaction, but it doesn’t look like an acid/base reaction. 7

Familiar Ground Given this reaction: HA (aq) + H 2 O (l) ↔ A - (aq) + H 3 O + (aq) Can you write the equilibrium constant expression? 8

Equilibrium Constant Expression K a = [H 3 O + ][A - ] [HA] NOTE: This is just a K eq, there is nothing new here. It is just a specific type of reaction. So, ICE charts, quadratic formula, etc. all still apply! 9

A sample problem What is the pH of a M HOAc solution? The K a of HOAc = 1.8 x

Old Familiar solution 1 st we need a balanced equation: 11

Old Familiar solution 1 st we need a balanced equation: HOAc (aq) + H 2 O (l) ↔ H 3 O + (aq) + OAc - (aq) The we need to construct an ICE chart OAc - = C 2 H 3 O

ICE ICE Baby ICE ICE HOAc (aq) + H 2 O (l) ↔ H 3 O + ( aq ) + OAc - (aq) I C E What do we know, what do we need to know? ??? 13

A peek back at the problem. What is the pH of a M HOAc solution? The K a of HOAc = 1.8 x What do we know? What do we need to know? 14

A peak back at the problem. What is the pH of a M HOAc solution? The K a of HOAc = 1.8 x What do we know? The INITIAL CONCENTRATION of HOAc What do we need to know? The EQUILIBRIUM CONCENTRATION of H 3 O + (Recall, that’s what pH is: pH = - log [H 3 O + ] 15

ICE ICE Baby ICE ICE HOAc (aq) + H 2 O (l) ↔ H 3 O + ( aq ) + OAc - (aq) I C E How do we solve for x? M-00 -x-+x – x-xx 16

Use the Equilibrium Constant Expression K a = 1.8x10 -5 = [H 3 O + ][A - ] [HA] 1.8x10 -5 = [x][x] [0.100-x] How do we solve this? 17

2 Possibilities 1.8x10 -5 = [x][x] [0.100-x] 1. Assume x << Don’t assume x<<0.100 and use quadratic formula 18

The long way 1.8 x = (x)(x)/(0.1-x) = x 2 /0.1-x x 2 = 1.8 x (0.1-x) =1.8x10 -6 – 1.8x10 -5 x x x10 -5 x – 1.8 x = 0 Recall the quadratic formula: x = - b +/- SQRT(b 2 -4ac) 2a 19

The long way x x10 -5 x – 1.8 x = 0 x = - b +/- SQRT(b 2 -4ac) 2a x = - 1.8x /- SQRT((1.8x10 -5 ) 2 -4(1)(– 1.8 x )) 2(1) x = [-1.8x /- SQRT (7.200x10 -6 )]/2 x = [-1.8x / x ]/2 20

2 roots - only 1 makes sense 21

ICE ICE Baby ICE ICE HOAc (aq) + H 2 O (l) ↔ H 3 O + ( aq ) + OAc - (aq) I C E M-00 -x = x10 -3 M -+x=x = 1.33x10 -3 M M – 1.33x10 -3 = M -1.33x10 -3 M 22

pH = - log [H 3 O + ] = - log (1.33x10 -3 ) = 2.88 Was all of that work necessary? Let’s look at making the assumption! 23

Assume x<< x10 -5 = [x][x] [0.100-x] If x<<0.100, then x≈ x10 -5 = [x][x] [0.100] 1.8x10 -6 = [x][x] = x 2 x = 1.34x10 -3 M 24

Was the assumption good? We assumed that x<<0.100, is 1.34x10 -3 M << 0.100? The 1% rule applies and it is very close, but notice how little difference it makes in the final answer? And if I calculate the pH = - log (1.34x10 -3 ) pH = 2.87 This compares well with pH = 2.88 calculated the long way. And look at all the work we saved! 25

Base Dissociation Reactions  Acids and bases are matched sets.  If there is a K a, then it only makes sense that there is a K b  The base dissociation reaction is also within the Bronsted-Lowry definition  Water now serves as the acid rather than the base. 26

General K b Reaction The general form of this reaction for any generic base (B) is: B (aq) + HOH (l)  HB + (aq) + OH - (aq) Base Acid Conjugate Conjugate acid base 27

KbKb It is, after all, just another “K” K b = [HB][OH - ] [B] And this gets used just like any other equilibrium constant expression. 28

Water, water everywhere Both K a and K b reactions are made possible by the role of water. Water acts as either an acid or a base. Water is amphiprotic. If water is both an acid and a base, why doesn’t it react with itself? 29

Water does react with itself  Autoionization of water: H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) 30

Autoionization of water: H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq)  This is, in fact, the central equilibrium in all acid/base dissociations  This is also the connection between K a and K b reactions. 31

The Equilibrium Constant Expression K w H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ] = 1.0 x K IS K IS K IS K – this is just another equilibrium constant. Let’s ICE 32

ICE ICE Baby ICE ICE H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) I C E ??? 33

ICE ICE Baby ICE ICE H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) I C E Solve for x --1x x --1.0x x x 34

Evaluating K w K w = [H 3 O + ][OH - ] = 1.0 x [x] [x] = 1.0 x x 2 = 1.0 x x = 1.0x

ICE ICE Baby ICE ICE H 2 O (l) + H 2 O (l)  H 3 O + (aq) + OH - (aq) I C E What’s the pH? x =1.0x x

pH = - log [H 3 O + ] pH = - log (1.0x10 -7 ) pH = 7 This is why “7” is considered neutral pH. It is the natural pH of water. Neutral water doesn’t have NO acid, it has the EQUILIBRIUM (K w ) amount!!! 37

K b, K a, and K w It is the Kw of water (1.0 x 10-14­) which is responsible for the observation that: pOH + pH = 14 Since we’ve already established that pure water has 1x10 -7 M concentrations of both H + and OH - In an aqueous solution, this relationship always holds because K w must be satisfied even if there are other equilibria that also must be satisfied. 38

[OH-][H 3 O+]=1x log([OH-][H 3 O + ] = -log1x log[OH-] + (-log[H 3 O + ] = 14 pOH + pH = 14 39

What is the pH of a M solution of formic acid (HCHO 2 )? [K a =1.8x10 -4 ] 40

What is the pH of a M solution of NH 3 ? [K b = 1.8x10 -5 ] 41

K b, K a, and K w The general Ka reaction involves donating a proton to water. HA + H 2 O ↔ H 3 O + + A - where A- is the “conjugate base” to HA, and H3O+ is the conjugate acid to H2O. The general Kb reaction involves accepting a proton from water. A - + H 2 O ↔ HA + OH - 42

Writing the K for both reactions K a = [H 3 O+][A-] [HA] K b = [HA][OH-] [A-] If you multiply K a by K b : K a *K b = [H 3 O+][A-] [HA][OH-] [HA] [A-] = [H 3 O+][OH-] =K w So, if you know K b, you know K a and vice versa because: K a *K b =K w 43

Remember… K a and K b refer to specific reactions. For example, consider the acid dissociation of acetic acid: HOAc (aq) + H 2 O (l) ↔ H 3 O + (aq) + OAc - (aq) This reaction has a K a, it does not have a K b. BUT, its sister reaction is a base dissociation that has a K b : OAc - (aq) + H 2 O (l) ↔ OH- (aq) + HOAc (aq) It is this reaction that you are calculating the K b for if you use the relationship K w = K a *K b 44

What is the pH of M HCl? HCl is a strong acid! What does that mean? It means it almost completely dissociates. K a  [H 3 O + ] = [HCl] pH = - log [0.100 M] pH = 1 45

If you can’t find K a … …maybe it’s a strong acid. Or if you do find it, it is HUGE! Strong acids: HCl K a ~10 6 HBr HI HNO 3 HClO 4 H 2 SO 4 (diprotic – more on this next week) 46

If you can’t find a K b … …maybe it’s a strong base. (Or K b is huge.) List of strong bases: LiOH NaOH K b ~10 8 KOH Sr(OH) 2 Ca(OH) 2 Ba(OH) 2 47

So for a strong acid (or base) HCl (aq) + H 2 O (l) ↔ H 3 O + (aq) + Cl - (aq) This reaction has a K a that is huge (approximately 10 6 ).. Its sister reaction (reverse) is a base dissociation that has a K b : Cl - (aq) + H 2 O (l) ↔ OH- (aq) + HCl (aq) K w = K a *K b K b = K w /K a = 1x /10 6 = 1x It doesn’t happen!!! 48

So for a strong acid (or base) IT AIN’T AN EQUILIBRIUM! HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) M-00 -x-+x M ICEICE 49