Question

Name 1 (25 pts) Consider the titration of 50 mL of 0.100 M acetic acid (CH-COOH, K. -1.8 x 10) with 0.200 M NaOH solution. Show all calculations for full credit. a) Write the titration reaction: b) Calculate the pH after 5.00 mL of NaOH: c) Calculate the pH after 12.5 mL of NaOH: d) Calculate the pH after 25 mL of NaOH:

Answer 1

The titration of Acetic Acid (CH₂(00H) and Naou is of the lipe of necak and The titration Reaction is as follows:- Shoug buree lype tibration; D The tibration of some 0.9 60 ML Na 0 H 50 CH₂cool (aq) + oH-Cag) - CH3C.00-(ao) +H20(4) of a 10om Acetic acid and 5.00ml of Lets calculate the number of moles of CH₂ cool to neutralized 0.050 LX0.100M = '0.005 moles. (1000 ml =12 The number of moles of Nach required 005L" x 0.980m = 0.003 woles (1000 ml 02.0051) How to calculate the pH we must use an ICE table. CH₃COoh + OH CH₃Cool + H2O Initial 0.005 mol Add. 0.001 mol Change 0.001 mol -0.001 mol +0.001wol Equilibrium 0.004 mot 0.oolwol. when NaOH and CH₃COOH were wroed the volume 50ml + sme = 55ml Fo.055L Now, concentration of CH₃COOH 0.004 mot 0.073 mol/L 0.053L Concentration of cll3Coo- ooool mol 0.018 mol/L. O 055 L 8 x 10 ka= Therefore, wee the Henderson-Hasselbalch equations, [CH3cool = 1-8810-5 2. pkas 4.74 e-boyli8x10-5). hel have we continere by using pH = pkat Log [CH3 Cook (0.018) 4.74 tlog C0.073). 4.74 +60.61) pH 4.13

pH of the solution when 19.5 mL of NaOH (0.200M) was added The number of miles of Nach required. (1000mL = 1L L=0.01254. So, 12.5ML= : 0.0125LX0.200M 0.0025 mol To calculate pH was murst cons freet l.(of table Initial Add CH2COOH(-) + OH (9) CH3600-609) + H2011) 0.005 mols 0.002smols change 0.0025 mots -0.002smots +0.0025 merly Equilibrium 0.002swols 0.0025 mole. Now total volume aflir mixing - 50ml+12.5mL = 62.5ml = 0.0625L So [CH3 Cook] = 0.002smols F 0.04 mol/L 0.0625 L [CH3600] = 0.0025 mots 0.04 mol/ 0.0625L co pH = pka + [CH3600"] log [CH3Cooke 4.7.4+ 0:04 4 +log 0:04 4.74+logi 4.74ito. pH = 4.74 | This describes that we have found the Half-neutralization point. (d pH of the solutions when 25 ml the solution when 25 mh l o. 200m Nagh is added of The number of moles of Nach required. 0.025LX0.200M =0.005 mols. I.C.E table, get get. so from CH3COOH Coq) + OH (99) CH3 Coo Taq) + H2014 Initial 0.005 wols Add. 0.005 smols change. -0.005mols conos woes to-ooswole Equilibrium owols 0.005 mols is So this, the equivalence point of titration where acid and base are both neuhalized and neither of them Here total roluene 25 mlot som 2 75ml = 0.075L (1000ml=AL) o mole are excess.

So, concennation of [CH₂ coor] = a.cosmot 0.06 moth 0.075L So to Calculate the pH at this point [choo] must From ICE tabloi- hydrolyge + Hoot om + X M хи CH3COOH H₂O t. C.H3Coo- Initial 0.067m OM Change -X +XM Equilibrium (0.067-8) m XM So here Cl₂cool acts as base. Therefore we must determine kb. kb= kw 1.0xioly ka 1.8x10-5 K6=5.56*16-10 The expression of to from 1.C.E table reeget. K6= x x x² (0.067-x) (0-067-x) x 5.58 X20-10 0.067-X. x?+ 5:56X16x -3.73x10"=0. Now solving the quadratic equation weget x= -(5.58x1096) IN (5:56 X1070)-4.1.6-3.73 x1054) 2.1 2 = 76.1x1076.fory X=6.1 XV0-6. The negative a value is ruled out because cone. Cornot be negative. [ou]=xo 6.1X/66. poro-logoH = -log(6-141056) 25.21. , pH = 14-poh =14-5-21 pH=8.79