Concept Test: Identify the acid/base and conjugate acid and base in each of the following reactions: Indicate who the conjugates are connected to!
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1 Concept Tests: Chapter 17 and 18 Identify the acid/base and conjugate acid and base in each of the following reactions: Indicate who the conjugates are connected to! NH 3 + HCl NH Cl -1 H 2 S + NH 3 NH HS -1 OH -1 + H 2 PO 4-1 H 2 O + HPO 4-2 Given the following acids/bases and their strengths, write the conjugate acid/base and indicate its strength HCl Cl -1 OH -1 H 2 O NO 3-1 HNO 3 HS -1 S -2 1
2 In solution, strong acids dissociate almost completely into ions in water: HA (g, l) + H 2 O (l) H 3 O + (aq) + A - (aq) Writing a K expression for the above reaction: K = remember!! Water as a liquid does NOT appear in a Q/K expression! In solution, weak acids dissociate to a much lower extent: HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) K = What is the ph of a solution whose [H +1 ] = 1.12 x 10-2 M? What is the [H +1 ] if the 4.41? [H +1 ] = 2
3 Relationship of [H +1 ] to ph Calculate the ph of the following 4 solutions: Comment on [H +1 ] concentration and how it relates to ph and whether or not the solution is acidic or basic... If [H +1 ] = x 10-3 M If [H +1 ] = x 10-6 M If [H +1 ] = x 10-9 M If [H +1 ] = x M As the [H +1 ] goes down (becomes less) the ph is greater. Lower ph values correspond to higher [H +1 ] Lower ph values represent acidic systems Knowing that the pk w = Verify this using [H +1 ] = [OH -1 ] = 1.00 x log(1.00 x 10-7 ) = poh = -log(1.00 x 10-7 ) = ph + poh = K w 3
4 Use the above [H +1 ] to determine the poh (HINT: use the pk w relationship!) 2.82 poh = 5.82 poh = 8.82 poh = poh = Finding K a from the ph of a solution Given 0.12 M phenylacetic acid (C 6 H 5 CH 2 COOH) which dissociates to yield a ph of 2.60 what is the K a? 1.) set up ICE table 2.) fill in values 3.) what does ph mean? 4.) does the autodissociation of water matter? 5.) solve for K a I C E C 6 H 5 CH 2 COOH + H 2 O H 3 O +1 + C 6 H 5 CH 2 COO -1 K a = [H +1 ] = 4
5 comparing [H +1 ] to 1.0 x 10-7 does 1.0 x 10-7 matter???? Yes or No? If no, then ignore the autodissociation of water! I C E C 6 H 5 CH 2 COOH + H 2 O H 3 O +1 + C 6 H 5 CH 2 COO -1 x = Therefore, since H 3 O +1 and C 6 H 5 CH 2 COO -1 are in a 1:1 ratio the equilibrium concentrations of the two are the same [H +1 ] = = [C 6 H 5 CH 2 COO -1 ] [C 6 H 5 CH 2 COOH] = K a = K a = 5
6 determining concentrations when given the K a value 0.10 M propanoic acid has a K a value of 1.3 x 10-5, what is its [H 3 O +1 ]? prop means 3 oic acid means OOH group I C E CH 3 CH 2 COOH + H 2 O H 3 O +1 + CH 3 CH 2 COO -1 K a = 1.3 x 10-5 = assume x is small 1.3 x 10-5 = x = check assumption: = Therefore the equilibrium concentrations are [H +1 ] = [CH 3 CH 2 COO -1 ] = [CH 3 CH 2 COOH] % dissociation = [HA] dissociated x 100 [HA] initial 6
7 What is the ph of 0.25 M sodium acetate in water. The K a of acetic acid is 1.8 x ) Write and balance the chemical reaction 2.) consider how acetate will react in water 3.) fill in values 4.) use the K a appropriately Sodium acetate will dissolve completely in water and will behave as a base I C E CH 3 COO -1 + H 2 O CH 3 COOH + OH -1 Since the acetate ion behaves as a base, we need the K b K b = K b = b = K b = K w K a assume x is small and x = [OH -1 ] = poh = -log[oh - ] poh = ph + poh =
8 Addition of a small amount of HCl reacts with the acetic acid/sodium acetate solution how? What is ph defined as? -log[h +1 ] What happened to the free [H +1 ] that were added to the solution? Would you expect the ph of the system to change? yes no We use weak acids or weak bases and the salts thereof to make a buffer system: Why won t strong acids or strong bases and their salts work as a buffer system? (HINT: Think about the equilibrium direction of the dissociation!) 8
9 Calculate the ph of a solution that is both 1.00 M acetic acid and 1.00 M sodium acetate (K a = 1.8 x 10-5 ) CH 3 COOH HOH H + + CH 3 COO -1 I 1.00 M C - E - K a = assume x is small! x = Check assumption! If x = [H +1 ] then 9
10 If we add base to the solution, it will react with CH 3 COOH. It will react in such a way with the CH 3 COOH that the proton from CH 3 COOH will be lost and combine with the OH -1 to form water. Write the chemical equation for the addition of base to the buffer system (1.00 M acetic acid, 1.00 M sodium acetate) CH 3 COOH + OH -1 HOH + CH 3 COO -1 I 1.00 M Added - C - E - Addition of 0.02 M sodium hydroxide will react with 0.02 M CH 3 COOH This will use up the added hydroxide ion. In turn, it will form an additional 0.02 M of acetate ion. As long as the amount of strong base (in M) is less than the concentration of acid, the base will be completely neutralized. Now we have a new starting concentration of species! CH 3 COOH + H 2 O H 3 O +1 + CH 3 COO -1 I - 0 C - E - K a = K a = 1.8 x 10-5 = Assume x is small! 1.8 x 10-5 = x = 10
11 pk a + log [base] [acid] Calculate the ph of a solution that is 0.40 M in NH 3 and 0.30 M NH 4 Cl The K a for NH +1 4 = 5.65 x pk a + [NH 3 ] [NH +1 4 ] Concept Check: NH H 2 O H 3 O +1 + NH 3 I C - E - K a = [H 3 O +1 ][NH 3 ] [NH 4 +1 ] 5.65 x = assume x is small 5.65 x = x = 11
12 You want to create a buffer system to maintain the ph around 7.00 Given the following acids, choose which species would be an acceptable buffer system. THINK about BOTH species that need to be present in the buffer! HF K a = 3.53 x 10-4 pk a = H 3 PO 4 K a = 7.52 x 10-3 pk a = H 2 PO -1 4 K a = 6.23 x 10-8 pk a = 12
13 How many moles of solid Na 2 HPO 4 must be added to 250 ml of 0.25 M NaH 2 PO 4 to make the buffer system have a 7.00? pk a + log [base] [acid] is the base = but the question asked how many moles! L x = moles of 13
14 Calculate the ph at the equivalence point when ml of M CH 3 COOH is titrated with M NaOH The problem approach is very similar to what we did last term CH 3 COOH + NaOH CH 3 COONa + HOH how much NaOH was added? but the acetate ion is a strong conjugate base from a weak acid that means that the base can abstract a proton from water we need to know the concentration of acetate ion in solution to find out how it reacts with water to change the ph mole CH 3 COO -1 = M CH 3 COO -1 (25.00 ml NaOH ml CH 3 COOH) 14
15 The CH 3 COO -1 can and will act as a base to abstract protons from water. What is the ph? How is the acetate ion acting as a base therefore we must use the K b!! CH 3 COO -1 + H 2 O CH 3 COOH + OH -1 I C - x - + x + x E x - x x K b = 5.68 x = assume x is small x = poh = poh = 15
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