Mixtures of Acids and Bases CH202, lab 6 Goals : To calculate and measure the ph of pure acid and base solutions. To calculate and measure the ph of mixtures of acid and base solutions. Safety : Hydrochloric acid (HCl) and sodium hydroxide (NaOH) solutions are corrosive. Flush with lots of water if any solutions come into contact with they eyes or skin. Waste : Solutions can be flushed down the sink with plenty of water.
Calculating ph The formulas use to calculate ph are easy. The difficulty lies in deciding WHICH formulas to use. Students will need this skill for the next three labs. The key to choose a formula is determining which species are in solution, and which reaction, if any, occurs.
Calculating ph of Strong Acids in Water Strong Acids have a K a > 1 and completely react with water. HCl(aq) + H 2 O(l) Cl 1- (aq) + H 3 O 1+ (aq) initial 0.10 M 0 0-0.10M +0.10M +0.10M final 0.00M 0.10M 0.10M ~none all all Therefore, [strong acid] = [H 3 O 1+ ] ph = - log[h 3 O 1+ ] = - log(0.10) = 1.00 An acid has ph <7
Calculating ph of Weak Acids in Water. Weak Acids have a K a < 1 and only partially react with water. HC 2 H 3 O 2 (aq) + H 2 O(l) C 2 H 3 O 1-2 (aq) + H 3 O 1+ (aq) initial 0.20 M 0 0 -x +x +x equilibrium 0.20 - x x x most a bit a bit K a = [C 2H 3 O 2 1- ][H 3 O 1+ ] [HC 2 H 3 O 2 ] = (x) (x) (0.20 - x) = x 2 0.20 =1.8. 10-5... x = (0.20) (1.8 10-5 )= C wa K a =1.9 10-3 M concentration of weak acid 1.9. 10-3 M=[C 2 H 3 O 1-2 ]and[h 3 O 1+ ] ph = - log [H 3 O 1+ ]=-log(1.9. 10-3 )=2.72 assume x is small compared to 0.20 acid dissociation constant An acid has ph <7
Calculating ph of Strong Bases in Water Strong Bases are usually Group 1A ionic salts that completely dissolve in water. NaOH (s) Na 1+ (aq) + OH 1- (aq) initial 0.30 M 0 0-0.30M +0.30M +0.30M final 0.00 0.30 M 0.30 M ~none all all Therefore, [strong base] = [OH 1- ] poh = - log[oh 1- ] = - log(0.30) = 0.52 ph = 14 - poh = 14-0.52 = 13.48 A base has ph >7
Calculating ph of Weak Base in Water Weak Bases have a K b < 1 and only partially react with water. C 2 H 3 O 1-2 (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH 1- (aq) initial 0.20 M 0 0 -x +x +x equilibrium 0.20 - x x x most a bit a bit K a K b = K w (1.8 10-5 ) K b = 1.0 10-14 K b = 5.6 10-10
K b = [HC 2H 3 O 2 ][OH 1- ] [C 2 H 3 O 2 1- ] = (x) (x) (0.20 - x) = x 2 0.20 =5.6. 10-10 assume x is small compared to 0.20 x = (0.20) (5.6. 10-10 )= C wb. K b =1.1. 10-5 M concentration of weak base base dissociation constant 1.1. 10-5 M=[HC 2 H 3 O 2 ]and[oh 1- ] poh = - log [OH 1- ]=-log(1.1. 10-5 )=4.98 ph = 14 - poh = 14-4.98 = 9.02 A base has ph >7
Formula Summary Species in Solution Formulas Strong Acid (SA) [H 3 O 1+ ] = [SA] ph = -log [H 3 O 1+ ] Weak Acid (WA) [H 3 O 1+ ] = (K a x c wa ) 1/2 ph = -log [H 3 O 1+ ] Strong Base (SB) Weak Base (WB) Weak Acid and Weak Base (presented as mixture #6) [OH 1- ] = [SB] poh = -log [OH 1- ] ph = 14 - poh K b = K w /K a [OH 1- ] = (K b x c wb ) 1/2 poh = -log [OH 1- ] ph = 14 - poh ph = pk a + log (mmol base / mmol acid)
Mixtures of Acids and Bases in Water There are six combinations (Figure 6-2 from CH202 Lab Manual)
1 Strong Acid and Strong Base Consider : H 3 O 1+ (aq) + OH 1- (aq)! H 2 O(l) + H 2 O(l) Reacting acid Produced acid Recall from CH101 that: K reaction = K a reacting acid K a produced acid = 1.0 1.0 10-14 = 1.0 10 14 This K indicates the reaction will produce ~ 100% product. Treat this reaction as a limiting reactant problem.
1 Strong Acid and Strong Base Consider the reaction of 10.0 ml of 0.10 M HCl and 15 ml of 0.20 M NaOH H 3 O 1+ (aq) + OH 1- (aq) H 2 O + H 2 O given 10.0 ml x 0.10 M 15mL x 0.20 M initial 1.0 mmol 3.0 mmol -1.0 mmol -1.0 mmol final 0.0 mmol 2.0 mmol Species remaining in solution : Strong base in water (slide 5) [OH 1-2.0 mmol ] = = 0.080 M (10.0 + 15)mL poh = -log[oh 1- ] = -log(0.080) = 1.10 Notes : ph = 14 -poh = 14-1.10 = 12.90 If H 3 O 1+ were the only remaining species, treat as a strong acid in water. (slide 3) If neither H 3 O 1+ or OH 1- remains, the ph = 7.00
2 Strong Base and Weak Acid In most cases, K reaction will be > 1000, such that the reaction goes to completion. Consider : HC 2 H 3 O 2 (aq) + OH 1- (aq) C 2 H 3 O 2 1- (aq) + H 2 O(l) reacting acid produced acid K = K a reacting acid K a produced acid =. 1.8 10-5 = 1.8 10 1.0. 9 10-14. Treat this reaction as a limiting reactant problem.
Strong Base and Weak Acid Consider the reaction of 15.0 ml of 0.20 M NaOH and 10.0 ml of 0.30 M HC 2 H 3 O 2 HC 2 H 3 O 2 (aq) + OH 1- (aq) C 2 H 3 O 2 1- (aq) + H 2 O(l) given 10.0 ml x 0.30 M 15.0 ml x 0.20 M initial 3.0 mmol 3.0 mmol 0 2-3.0 mmol - 3.0 mmol + 3.0 mmol final 0.0 0.0 3.0 mmol Species remaining in solution: weak base in water (slide 6+7) [OH 1- ] = C wb. 3.0 mmol K b = (10 + 15) ml.. 1.0 10-14 = 8.16. 10 1.8. -6 M 10-5 poh = - log(8.16. 10-6 ) =5.09 ph = 14-5.09 = 8.91 Notes : If OH 1- were the only remaining species, treat as strong base in water (slide 5). If weak acid and weak base remain, use the Henderson- Hasselbalch (slide 16).
3 Strong Base and a Weak Base Two bases will not react. Instead the strong base both suppresses the reaction of weak base with water and has the major effect on the ph. Consider the addition of 20.0 ml of 0.20 M NaOH to 10.0 ml of 0.10 M NaC 2 H 3 O 2. The strong base, [OH 1- ], is diluted C 1 V 1 = C 2 V 2 (0.20 M) (20.0 ml) = C 2 (20.0 + 10.0 ml). 0.13 M = C 2 [OH 1- ] = 0.13 M poh = - log (0.13) = 0.88 ph = 14 - poh = 14-0.88 = 13.12
4 Strong Acid and Weak Base In most cases, K reaction will be > 1000, such that the reaction goes to completion. Consider : H 3 O 1+ (aq) + C 2 H 3 O 2 1- (aq) H 2 O(l) + HC 2 H 3 O 2 (aq) reacting acid produced acid K = K a reacting acid K a produced acid = 1.0 = 5.6 10 1.8. 4 10-5. Treat this reaction as a limiting reactant problem.
Strong Acid and Weak Base Consider the reaction of 15.0 ml of 0.20 M HCl and 20.0 ml of 0.20 M NaC 2 H 3 O 2 H3O 1+ (aq) + C2H3O2 1- (aq) H2O(l) + HC2H3O2(aq) given 15.0 ml x 0.20 M 20.0 ml x 0.20 M initial 3.0 mmol 4.0 mmol 0 4-3.0 mmol - 3.0 mmol + 3.0 mmol final 0 mmol 1.0 mmol 3.0 mmol Species remaining in solution : weak base and weak acid Henderson- Hasselbalch. mmols base ph = pk a + log mmols acid ph = - log(1.8. 10-5 1 mmol ) + log = 4.26 3 mmol Notes : If HC 2 H 3 O 2 were the only remaining species, treat as a weak acid in water (slide 4). If H 3 O 1+ and HC 2 H 3 O 2 were remaining, treat as a diluted strong acid in water (slide 3).
5 Strong Acid and Weak Acid Two acids will not react. Instead the strong acid both suppresses the reaction of weak acid with water and has the major effect on ph. Consider the addition of 30.0 ml of 0.20 M HCl to 20.0 ml of 0.15 M HC 2 H 3 O 2. The strong acid, [HCl] = [H 3 O 1+ ], is diluted C 1 V 1 = C 2 V 2 (0.20 M) (30.0 ml) = C 2 (30.0 + 20.0 ml) 0.12 M = C 2 ph = -log [H 3 O 1+ ] = - log (0.12) = 0.92
6 Weak Acid and Weak Base Weak acid and the conjugate weak base are on the opposite side of an equilibrium reaction. They will form a buffer. Consider the addition of 10.0 ml of 0.10 M NH 4 Cl to 15 ml of 0.10 M NH 3. NH 4 1+ (aq) + H 2 O(l) NH 3 (aq) + H 3 O 1+ (aq) given 10.0 ml x 0.10 M 15 ml x 0.10 M When weak acid and weak base are present, ph is calculated using the Henderson-Hasselbalch equation. ph = pk a + log mmols base mmols acid = - log (5.6. 10-10 1.5 mmol ) + log = 9.43 1.0 mmol
Experimental Overview Students will prepare various mixtures of acids and bases and measure the ph. 10 5 Na 1+ OH 1- H 2 O
Tips for Collecting Good Data Double-check the calibration of the ph electrode by measuring the ph of the ph = 4.00 and ph = 10.00 calibration solutions. Keep the electrodes in ph = 7.00 buffer solution when not in use. Electrodes that have dried out will initially give incorrect readings. Rinse electrodes with DI water and blot dry, prior to measuring test solution ph. This is to avoid contamination of the test solution with other solutions or the buffers.