Question

A 27.3 mL sample of 0.355 M triethylamine, (C2H5)3N, is titrated with 0.355 M hydrochloric acid. After adding 38.2 mL of hydrochloric acid, the pH is Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

A) mmo I d (2445)2N = concen bertz'on x volume (mi) 0.355 MX 27.3 mmol = 9.6915 mmol mmol d HCl = 0.355MX 38.2 m/ = 13.561 mmol so, mmol of rich left after mixing = 13.561-9.6915 = 3.86 15 mmol Total volume = (27'3+ 38.2 ml = 65.5ml Thelle -2.0615, 00sqm - 635 pxz. -109 14+) = -109 (0:051) = 1.23 0 C, H, O, N 18,0 = (a, MgNH + --- 1 0 .38 0 e x te te E 0.3-4 ^ Ks = a2 K6 = 82 0 0.3-2 5.3810-4 = x² 22 0.0123 M 0-3-4 alon')= 0 - 0123 pu- 14-pou - = 14+loqlony_ = 14 F 10910.0123) _ p4 3 12.09_ to this please check vana d kla for (GH5N - if its different then put it in egno - & get answer