Question

8) A 25.0 mL sample of 0.150 M benzoic acid is titrated with a 0.150 M NaOH solution. What is the pH at the eq. point? pka= 4.20

Answer 1

Benzoic acid =%h cool. sodium hyd rouide = NOH CH₂Cool + Nach Ghs coo Na + tho. At equivalence point kumber of moles of GH_COOH = number of Moley of Naon Benzoic acid moles: V=25mi, M = 0.15M M = X looo v (m) n = MV(mi) 1000 ne 0.15) (25) = 3.75 x103mol. 7000 CHCOOH + Nach Initiat: 3 - 75x163 3.75X163 Che Coonattheo 0 - -3-758103 -3.75810 +3.75 X 103 Finally: 0 0 3:75 xro? At equivalence point Chs cooNa is produced with 3.75 X16 "mol. CS Scanned with a CamScanner

Salt is produced from weak acid and strong base. Co ty coot the GH SCOOH + oH 3.75X10 mal 3-75X15 mol 0.075M 0.07EM Total Solution Volume = 25m) + 25 ml At equivalence point Mili Movie → 6.15) (as) = (0:15) * » R = 25ml Total solution volume = 50ml Che cos concentration = 3.75x163 xlo = 0.0751 G Hg coo concentration = 0.075m cs Scanned with CamScanner

Salt produced from strong base To Find so it contains basic nature [or] = knic - We nave pra value c = 0.075M / pra = 4.2; pta + pky = 14 - pky = 14-4-2 =9.8 Apply both side logo - log [or] =- - logcut.c) P = -109 (.c) pot = -1/2 { log ke + log c] por = {-logt & logc] poh = b [ pt - 10g (0,075)] zá [.8- (-. 125)] polt = 5.46 we know that pH + pot = 14 pH = 14-poh → pH = 14 - 5.46 pH = 8.54 Scanned with CamScanner