Question

Part A Calculate the pH of a solution that is 0.055 M in potassium propionate (C H3COOK or KC Hs02) and 0.080 Mi propionic acid (CH3COOH or HC3H5O2). Express your answer to two decimal places. Yo AS Roe? pH = Submit Request Answer Part B Calculate the pH of a solution that is 0.075 Min trimethylamine. (CH), N. and 0.13 Min trimethylammonium chloride. (CH3)2NHCI). Express your answer to two decimal places. Activity Details

Answer 1

(A) [CH₃ CH₂ COOH = 0.000 M [CH₃ CH₃COO- ] = 0.055 M pka of CH₂ cool = – log Ka = -log (1.3x105) = 4.89 According to Henderson equation - pH = pka & log [CHACH2000-1 [CH₂ CH₂COOH pH = 4.89 + log (0,055) = 4.73 70.000 (B) pk of (CH3)₂N – – log kh = -log (6.4X105) = 4.20 pka of (CH3)2N = 1410 - pkb = 14.0 - 4.20 = 9,00 concentrations - [(CH3)3NH+] = 0.13 M [(CH3)3N] = 0.075 M and Henderson equation – pH = pka & log [(CH3)3NT [(CH3)₂NHt] pH = 9.00 + log (0.875) = 9.56 1 o.13 (C) 50m L x 0.16M - = 0.00M (50+50) mL [CH₃COO-] = 50 ML X 0.23M = 0.115M (50+50) mL pka of CH₃COOH = -logka = -log (1.8x10-51 = 4.74 pH = pka + log [CH3COO-] = 4.74 + log (oolls) = 4.898 orchscob ] ~ 4.90