Question

Key Ouestion #2: Determine the initial concentrations for the following solutions. Solution Preparation Acetic Acid Acetate Ion Concentration Concentration Solution #1 50.0 mL of 0.0500 M acetic acid and 0.205 grams of sodium acetate Solution #2 25.0 mL of 0.0500 M acetic acid, 25.0 mL of deionized water, and 0.205 grams of sodium acetate 50.0 mL of 0.0500 M acetic acid and 0.102 grams of sodium acetate Solution #3 Solution # 4 50.0 mL of 0.0500 M acetic acid Solution # 5 50.0 mL deionized water and 0.205 grams of sodium acetate

Answer 1

solution #1 . : : so.nl of 0.05oom acetic acid and 0:285 gm of sodium acefate. Total volume of solunasonil | Acetic acid Acetate Ion Iconen determination n . Iconen determination / 10.05TO MY 0.205gm = 0.25 mol = o'cro 25 mol : So, the comer of acetate ion 1:2 (0:082521000) .-10.05-M Solution #2 25 ml of 0.05m acetic acid ; .25 ml of deionized water and 0.205 gm of sodium acetate. Total volume of soluna 50 ml i acetic acid : Acetate Ion. l cone" determinatio W,S; ZV, S2 0:205gm = 0.0025 mol conen of acetate ion =(0.0025x1600) M = 10-0574 . so I conen determination Sa z vs

iS, 2 2530.05 S₂2 50 :: 10.025. MI solution #3 50 mil of 0.0500m acetic acid and 0.102 gm of sodium acetate. is Total volume of soluna 50 ml 1 Acetic acid konen determination 10.0500M) ... 82 . Acefaite ion : conen determination/ 0.1029m = 0.102 mol 20:001244 mol The cone" of acetate ion - (0-001244x1000) M : 20.02488 M solution #4 some of oosoom Acetic acid comer = 00500M acetic acid Acetate in conero

solution #5 so mi deionized water and 0.205.gm of sodium acetate .. Acetre acidi conch determinationſ conen of acetic acid = 0 1 Acetate on I Teonen determination 9.205gm = 0.0025ml The conch of acetate jon 3.0.0025x1000 :- 6.05M SO :