Question

Question 25 1 pts Determine the pH at the equivalence point of a titration between 50.0 mL of 0.133 M (CH3)2NH solution with 0.133 MHCIO4. Remember that only the decimals in a pH count as significant figures. Please enter your answer with three decimal places.

Answer 1

solution Contains As [ (CH3)₂NH] = 0.133 m; 1000 me (CH3)2NH 50.0 mL 01 33 M (CH3), PH 0.133 mole mole (CH3)₂NH 50.0 0.133 mele = 0.00665 1000 as HClou mole nato reacts with (C H3, NH (CH3)NH₂ CLOU NN HClO4 (CH3), hence, 0.133M (CH3)2 NIA ( that Will exactly 50 mL 0.133 M Helou that Contains 0.00 665 mole is required for t thating 50 mL Contains 0.00665 mole (CH3)NH). 0.00665 mole Produced at equivalence Point in total volume of (50 +50) mL mL equivalence point, Hence, (CH3)NH2 Clou of 0.00665 x 1000 (-1) = of (CH3) Nha ciou, Concentration 0.0665 (11) of Salt of PH due solution and base made up of weak to hydrolysis strong acid is expressed Kw = -log Kue ; kwa ionic Walen; pkb = dissociation Product of OS Pkb - 12 logelog ky; bE 1/2 pk w Constant the weak basel PH Sall made up of pH of aald Acion, since (CH3)₂NH (104 und base strong weak solution due to this using the abote ( H) N 4 salt Con hydrolysis of equation. Calculated

pku = 14 (CH3)₂NH, ko = = 5.90 x1024 pKb = -log ko = -log(5.90x10 x104) a 3.229 Concentration of (CH₂)₂NH, c = 0.0665 M pH = pkno - I DNG - 1 loge 3x3.229 blog (0.0665) fax4) 3.229) 5.974 be Hence, PM at equiralence Point 5.974.