Question

A chemist titrates 160.0mL of a 0.3289 M lidocaine (C14H21NONH) solution with 0.3708M HCl solution at 25 degrees celsius. Calculate the pH at equivalence. The pKb of lidocaine is 7.94.

Answer 1

Comment in case of any doubt.

= 0.160m2 of Bihitsallies For convenience lets write lidocaine as B. [B ] initial = 0.3289 M, volume = 160.00mL Using MiVi= Ma Ya we find the required. 0-32897x160mL = 0.3708MX VHU -> Vice = 141.92 mL Vtotal at equivalence = (160+141.92) mL = 301.92 m2 At equivalence almost all & converts to BHt. So, at equivalence [877] x 301.92 m2=0.3289 MX160.om M (as moles of 87= moles Hi => [BH+] = 0.1743M pka of BHT = 14-plze of 8 = 14.7.94= 6:06 - - Ka=16806 (as pka = -logika] ICE Table BH+149) + H2O(l) + Blaz) + Hyot (29), ka I': 0.1743 - P 0 0 c: -0 . +7 +7 +43-71) 10 = Ka= CBIC Hy ot]. (10 (20) = 2? [Brt] (0.1743-01) assuming >> H= 101743 xjö 606 0.0003896 <<0.174'3 j = [H₃O+] as ka is small, We can verify pH= -log[ Hy 0+] - -dog (0.0003896) our assumption from the value of a we get . a . -6-06 0.1743 = Bul Ansman