Slide 1. Slide 2 Let s try another little problem: Slide 3 MORE ACIDS AND BASES. What is the ph of M formic acid (HCHO 2 )?

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1 Slide 1 MOR ADS AND BASS Slide Let s try another little problem: What is the ph of 0.13 M formic acid (HHO )? K a (HHO ) = 1.8x10-4 Slide 3 Why don t write it as H O? could, same molecule, but by writing it HHO m doing two things: 1. m emphasizing it s an acid by putting the H out front.. m indicating that only ON H can come off the molecule.

2 Slide 4 Not all H s are acidic H 4 methane t s got 4 hydrogens none of them are considered to be acidic because they don t easily come off. Generally, acids have the H bonded to something more electronegative like O or a halogen. Slide 5 H-O-H (acidic H bonded to O) H-l (acidic H bonded to halogen) H-S-H (acidic H bonded to S) H- (not acidic H bonded to ) Slide 6 Let s try another little problem: What is the ph of 0.13 M formic acid (HHO )? K a (HHO ) = 1.8x10-4

3 Slide 7 The 1 st thing we need is A BALAND QUATON! Slide 8 HHO What does the formic acid react with? H O How do you even know there s water? t s a solution! (M) What happens in the reaction? A proton moves from the acid (HHO ) to the base (H O): HHO (aq) + H O (l) HO - (aq) + H 3 O + (aq) Slide 9 Once have a balanced equation: HHO (aq) + H O (l) HO - (aq) + H 3 O + (aq) more parts:. K equation 3. ce chart!

4 Slide 10 K equation: HHO (aq) + H O (l) HO - (aq) + H 3 O + (aq) Slide 11 --BABY-- HHO (aq) + H O (l) HO - (aq) + H 3 O + (aq) What do know? of HHO is 0.13 M always know the line! Slide 1 Let s try another little problem: What is the ph of 0.13 M formic acid (HHO )? K a (HHO ) = 1.8x10-4

5 Slide 13 --BABY-- HHO (aq) + H O (l) HO - (aq) + H 3 O + (aq) 0.13 M x -x +x +x 0.13 M - x x Slide 14 Always worth trying the assumption X<<0.13 Slide 15 Always worth trying the assumption X<<0.13

6 Slide 16 Good assumption? 4.7x10-3 <6.15x10-3 so it s a good assumption! (although it s close) Slide 17 --BABY-- HHO (aq) + H O (l) HO - (aq) + H 3 O + (aq) 0.13 M x M ph=-log[h 3 O + ]=-log( M) =.33 Slide 18 Sample Problem alculate the ph of a 1x10-3 M solution of oxalic acid.

7 Slide 19 Solution As always, we 1 st need a balanced equation. Or, in this case, balanced equations! H O 4 (aq) + H O (l) H O 4 - (aq) + H 3 O + (aq) K a1 = 6.5x10 - H O 4 - (aq) + H O (l) O 4 - (aq) + H 3 O + (aq) K a = 6.1x10-5 quilbria = charts! Slide 0 Just take them 1 at a time H O 4 (aq) + H O (l) H O - 4 (aq) + H 3 O + (aq) 1x x - +x +x 1x x x -x Slide 1 K a1 = 6.5x10 - = Try x<<1x x10-5 = x x= 8.06x10-3 which is NOT much less than 1x10-3 We have to do it the Quadratic Way!

8 Slide 6.5x x10 - x = x 0 = x + 6.5x10 - x 6.5x10-5 x = - b +/- SQRT(b -4ac) a x = - 6.5x10 - +/- SQRT((6.5x10 - ) -4(1)( 6.5x10-5 )) (1) x = - 6.5x10 - +/- SQRT(4.485x10-3 ) x = - 6.5x10 - +/ x10 - x = 9.85x10-4 M Slide 3 Finish the first one H O 4 (aq) + H O (l) H O - 4 (aq) + H 3 O + (aq) 1x x x x x x x10-4 Slide 4 and start the second one. H O - 4 (aq) + H O (l) O - 4 (aq) + H 3 O + (aq) 9.85x x10-4 -x - +x +x 9.85x x 9.85x x + x

9 Slide 5 Let s try x<< 9.85x x10-5 = x 6.1x10-5 is NOT much less than 9.85x10-4 Dang it all! Slide x x10-5 x = 9.85x10-4 x + x 0 = x x10-3 x x10-8 x = - b +/- SQRT(b -4ac) a x = x10-3 +/- SQRT((1.046x10-3 ) -4(1)( x10-8 )) (1) x = x10-3 +/- SQRT(1.334x10-6 ) x = x10-3 +/ x10-3 x = 5.46x10-5 M Slide 7 Finishing up H O - 4 (aq) + H O (l) O - 4 (aq) + H 3 O + (aq) 9.85x x x x x x x x10-3 learly, the nd equilibrium makes a big difference here.

10 Slide 8 Do need to do this for all acids and bases? Most, but not all. There is a distinction between a strong acid and a weak acid. (Or, a strong base and a weak base. Slide 9 strong isn t STRONG, it s complete Would you rather drink a strong acid or a weak acid? Depends on the concentration. strong = complete dissociation weak = partial dissociation Slide 30 HA + H O = A - + H 3 O + Strong = Weak = omplete dissociation means it all reacts so there is ZRO HA left. n other words, K a is HUG Partial dissociation means there is some HA left. n other words, K a is a number.

11 Slide 31 Appendix (your BST friend) f you look at the Table of K a in Appendix you ll see numbers from 10-1 down to All are weak acids. f you look on page 665, you ll see a short list of strong acids. These actually have K a of 10 6 or higher. They are soooo big, they are usually considered infinite. Slide 3 Strong Acids H SO 4 HNO 3 Hl HlO 4 HBr H H with a big electronegative group. Slide 33 Strong Bases (p. 68) LiOH NaOH KOH Sr(OH) a(oh) Ba(OH) Alkali metals (hey! Where d the name come from! ) with hydroxide ions.

12 Slide 34 Question What is the ph of 1x10-8 M H SO 4? K a1 = infinite K a = 1.0x10 - Slide 35 Just take them 1 at a time H SO 4 (aq) + H O (l) HSO 4 - (aq) + H 3 O + (aq) t s strong! 1x x - +x +x 0-1x10-8 1x10-8 Slide 36 nd one starts where 1 st one ends! HSO - 4 (aq) + H O (l) SO - 4 (aq) + H 3 O + (aq) 1x x10-8 -x - +x +x 1x x - x 1x10-8 +x

13 Slide 37 K a = 1.0x10 - = [H 3 O + ][SO 4 - ] [HSO 4- ] 1.0x10 - = (1x10-8 +x)(x) (1x10-8 -x) an we assume x<<0.100?? Never hurts to try. 1.0x10 - = (1x10-8 )(x) (1x10-8 ) x=1.0x10 - which is NOT much less than 1x10-8 We have to do it the Quadratic Way! Slide 38 K a = 1.0x10 - = [H 3O + ][SO 4 - ] [HSO 4- ] 1.0x10 - = (1x10-8 +x)(x) (1x10-8 -x) 1.0x x10 - x = 1.0x10-8 x + x 0 = x x10 - x 1.0x10-10 x = - b +/- SQRT(b -4ac) a x = x10 - +/- SQRT(( x10 - ) -4(1)( 1.0x10-10 )) (1) x = x10 - +/- SQRT( x10-4 ) x = x10 - +/ x10 - x = x10-8 X= x10-9 = 1x10-8 Slide 39 Finish off the nd one! HSO - 4 (aq) + H O (l) SO - 4 (aq) + H 3 O + (aq) 1x x x x x10-8 1x x - 1x10-8 x10-8

14 Slide 40 AND START TH 3 RD ON!!!!!!! Slide 41 VRY dilute acid can t ignore K w H O (l) + H O (l) OH - (aq) + H 3 O + (aq) x x +x - - x x10-8 +x Slide 4 K w = 1.0x10-14 = [H 3 O + ][OH - ] = (.0x x)(x) 1.0x10-14 =.0x10-8 x +x 0 = x +.0x10-8 x 1.0x10-14 x = - b +/- SQRT(b -4ac) a x = -.0x10-8 +/- SQRT((.0x10-8 ) -4(1)( 1.0x10-14 )) (1) x = -.0x10-8 +/- SQRT(4.04x10-14 ) x = -.0x10-8 +/ x10-7 x = x10-7 X= x10-8 = 9.05x10-8

15 Slide 43 Finish off K w H O (l) + H O (l) OH - (aq) + H 3 O + (aq) x x x x x10-7 Slide 44 ph = - log[h 3 O + ] ph = - log (1.105x10-7 ) ph = 6.96 Slide 45 Suppose you have a really, really dilute acid say 1x10-7 MHl, what s the ph?

16 Slide 46 What do we know about Hl? t s a really strong acid! Suppose had M Hl, what s the ph? A. ph = 0.1 B. ph = 1.0. ph = -1.0 D. ph =.3. m still thinking about the test Slide 47 Strong acids, completely dissociate Hl + H O H 3 O + + l - Y x - +x +x 0 Y Y So M Hl yields M H 3 O +. ph = -log[h 3 O + ] = - log (0.100) = 1.0 ( don t even need the chart ) Slide 48 What is the ph of 1x10-7 M Hl? Hl is still a strong acid, so it completely dissociates. 1x10-7 M Hl gives you 1x10-7 M H 3 O + ph = - log (1x10-7 ) = 7 s that it, are we done? A really dilute acid is neutral. Seems reasonable.

17 Slide 49 There is another equilibrium! H O (l) + H O (l) H 3 O + (aq) + OH - (aq) K w = 1.0 x10-14 And H 3 O + is part of it! Slide 50 H O (l) + H O (l) H 3 O + (aq) + OH - (aq) - - 1X X -X +X +X x10-7 +x K w = 1.0 x10-14 = [H 3O+][OH-] 1.0 x = (1.0x x)(x) x Slide 51 H O (l) + H O (l) H 3 O + (aq) + OH - (aq) - - 1X X -X +6.18x x x x10-8 ph = - log (1.6x10-7 ) ph = 6.8 ompared to 1x10-7 and ph = 7 for the Hl alone

18 Slide 5 When do need to consider K w? 1. The acid is very dilute. The acid is very weak (K a less than 10-1 ) 3. Both 1 and Slide 53 A very weak acid problem What is the ph of a 1 x 10-7 M solution of HOAc? K a,hoac = 1.8 x 10-5 Slide 54 Baby HOAc (aq) + H O (l) H 3 O + (aq) + OAc - (aq) 1x x - +x +x 1x10-7 -x - x x

19 Slide 55 K a K a = 1.8x10-5 = [OAc-][H 3 O + ]/[HOAc] = (x) (x)/(1x10-7 -x) = x /(1x10-7 -x) will not assume x is small since 1x10-7 is pretty small itself 1.8x x10-5 x = x 0 = x + 1.8x10-5 x 1.8x10-1 Slide 56 Solving for x 0 = x + 1.8x10-5 x 1.8x10-1 x = - b +/- SQRT(b -4ac) a x = - 1.8x10-5 +/- SQRT((1.8x10-5 ) -4(1)( 1.8x10-1 )) (1) x = - 1.8x10-5 +/- SQRT(3.4x x10-1 ) x = - 1.8x10-5 +/- SQRT(3.31x10-10 ) x = - 1.8x10-5 +/ x10-5 x = 9.95 x10-8 M Slide 57 Suppose already have 1x10-7 M [H 3 O + ] from the K w? HOAc (aq) + H O (l) H 3 O + (aq) + OAc - (aq) 1x x x - +x +x 1x10-7 -x - 1x x x

20 Slide 58 K a K a 5 [ OAc ][ H3O ] [ HOAc] 7 5 [ x][1 10 x] K a [1 10 x] 1.8x x10-5 x = x + 1x10-7 x 0 = x x10-5 x 1.8x10-1 Slide 59 Solving for x 0 = x x10-5 x 1.8x10-1 x = - b +/- SQRT(b -4ac) a x = x10-5 +/- SQRT((1.81x10-5 ) -4(1)( 1.8x10-1 )) (1) x = x10-5 +/- SQRT(3.76x x10-1 ) x = x10-5 +/- SQRT(3.348x10-10 ) x = x10-5 +/ x10-5 x = 9.88 x10-8 M Slide 60 But already have 1x10-7 M [H 3 O + ] from the K w before even add the HOAc HOAc (aq) + H O (l) H 3 O + (aq) + OAc - (aq) 1x10-7 M - 1x10-7 M x10-5 M x10-8 M x10-8 M x10-8 M 9.88 x10-8 M

21 Slide 61 omparing the numbers Without considering K w, calculate from K a : [H 3 O + ] = 9.95 x10-8 M onsidering K w and K a, calculate: [H 3 O + ] = x10-7 M A significant difference!!