Question

4. Suppose you titrate 50.00 mL of 1.20 M pyridine, C₅H₅N, with 0.634 M HCl.

(a) How many moles of pyridine are present in the original pyridine solution?

(b) What is the initial pH?

(c) What volume (in mL) of HCl solution is required to reach the equivalence point?

(d) What is the pH at the equivalence point?

(e) What is the pH of the solution after the addition of 104 mL of HCl?

Answer 1

pH formula's for salt's weak acid + strong baise salt tho pH at { pk w tp ka +Log e] strong acid + weak base salt tho pH = [ pk wapkb_loge]

one litre of Molarity = no- of moles of solufe present in the solution is called Molarity -number of moles & of solute Molarity = no volume (2) - mass of solute number of moles= molar mass of of solute solute mass of solute — molar mass of .. - Xa мааз 9 volume (2) solute 1000 rebarily a na molarity Mass of solute Molarmass of solute volume (mi)

ÇHEN: Kg = value = 1+2x109 paba-log kb = -10g (17X109) = 8-77 moles of estan X1000 nolarity of - Stan volume Imp saltan 1000 moles of 1.27 - 5o glt, N= o. of moles moles of SH, NHT & of I ÇIHAN + HO o lo I 1-2. al-x X X El 12-X [ gats NHt] LOH ķ= [ghs W] x2 +7x10? +7X107 7.2- 1-7X169 (1.2-x) = x 2004 x109 - 1-28109 CX22 x2 x + 1.7 x16 9CX2 – 204 X10920 ax²+ bx tcro

크622ac 2 . - 이키 나2세에)를u)(-2-04×169) 2 C) | -- xot O-000 ioss 3 아s | -- xo + - Coo lossa 22 Xe CoHS = o. oooo LS2761 M PoFe ACG DOES - -log Co-oooo LS LICA) p얘스 4344 얘=14- pH =lu-4344= 2~656 음

SH-N molarity = 12 M volume soml=0-054 Mat no of moles = 12X0-05 20.06. at equivalence point moles of moles of onof mole ç HN - Hel - Molarity of moles of Hel Hal . volume (2) 0.6345 0.06 volume 0.06 20.094641 0.634 volume of Hola = 94.64 ms Che NHC salt ett Nt Hal - weakbase strong acid Il 0.06 l 0-0 6 el-006 1-0001 1-006 1 0 1 +0-06 arob Total volume = 0.05to.09464 20.144644 = 0.414823 Mze & { selts Whe] =1= salts No v 0.14464 Elkos Cute

For this type of salt pt formula pH = t[ pk w pkb_logo pkwaly pkb = 8.77 c=0.414823 M pHaţ[14-8-77-log (9.414823)] pH= 2.80607

SHSN: molarity = R2 M volumea soml=0.052 M= 2 no of moles=kz X0-05 20-06 | Hel: Molarity = 0.634M volume = 104m) 20.1042 no ot moles a 0.634 xoiloy =0.065936 به طلا برای - )W + را D 0.06 el-o-or El o 0.065936 -0.06 10.00 5936 Total volume = 0-057 0.104 = 0.1545 0-00 5936 [H413 = [H +3 = -0.0385454 M 0.154 plata -log[H+3 it louco.0385454 ) = 1.414 -