Question

2. (5 Points) 25.00 mL of 0.0750 M sodium benzoate (NaC6H3CO2) is titrated with 0.100 M HCl. Find the pH of the solution for the following volumes of acid added: 0 mL, 1 mL, 5 mL, 10 mL, 15 mL, 17 mL, 18 mL, 18.75 mL, 20 mL, 22 mL, 25 ml, and 30 mL. Create a titration curve.

Answer 1

NaC6Hs- CO or Colts coo wa is a salt of weak and (Phcoot) and strong base coach). Ph coowat the Phcoot & wooh Phcoot xat the shcoolt wat oH Effective eam, phcoo- + tho Phoot tot 12 cha) caca very Kn J a phwo- where be the ential concentration. & be the degree of dinouation. Eqm constant keq = Aphcoot Lot- attoo ID lydbolysis constant, a phool OH [or, acho = 1) 1. (PL thcoo ) tон) оюн 26- TPh wo) tphcool DOH) thoon I on we assume that for del. soln activity] coefficients (8) are 1. YS ca ca c(ra) P regrotton = - CA2 on L ine si-d = 1 case ac = √knc = OH-

Agam K, = &ph сон сон thcoo- Phwoh Con-cut phoo-cht - Now Phoot phwort ut Katthooth t Phloot/ Ht Coll- phcoo- CHA phoon - Kw Kw = CH + Lot- = 1x10-14 . Col- = kne Ka homepat of Walter COH Git - kw = kw (ka k. = (kukwyk - Wg Cht = Y Wg ka + yow kw Inloge - wg Ct = -1/ log kw - Ywgka tyloge. pH =/pkw ty, pka + bloge 1 H = 7+ Yz pka + k log c) 0 4 for om addition of Hu. the son contain only Ph word. then the pH of the satris due to hydroly's not the salt phoona. Now, pka of benzoic and = 4.20 (by literature, I value conc. of the Coona & (c) = 0.0750 m

Thus, from ege we have, &H - + + 474.20 + / log (0.0750) = 7+2.10+ ½ wg (0.0750 8.54 (2) Cottr word Addition of imL the Cette coonat tu u Colts coolt t way. 25 ml of dotrom getoona = 20mly o.ottom of – 25 only 0.07to mo Looo m ' thom = 1:875X10-3 mdf phwona Im of alm Hu = 1mlxo.im of Hu = Tank yo. / mál of the 1000 m = 1x10 4 me of Hu. the site contain Now, 1xcol mol of the react with 18104 mol of phevona to form 1x10 4 malit PhcoolH . Therefore the son contorn, (1-875x183 – 18109) mo ll, 1775x103 mol of exen Phuona. 1x10-4 mol [Phcoot) = (2+1) mL 1x004 mol 264103 L = tx104m Iwona) = _1.97-X03 mal [th Coona) = (28+)mL ou I es = 14 16 x 103 ['m a 1775 m 17txo3 mol

song by Hendersm egn, pH = pka + log t th loona) (11 tphcool 0.1 = 4.20+ wg _1.725/26 10-1/16 = 4.20 + wg 1.775 ~ 5.45 3) Addeturn of sml of Hi 5 ml of aim Hu = 5mLX a.im Hu. m of wona Woona conva ht fun toxthus Decoool, mo - 5x107 mil of Hu. 5x10 Y mol of Hu react with 5x154 mol of phworry to form 5x04 milf phoot. Thin the solu contorn (1-675X10–3–5 x10 Y mat 1.375X10–3 milit i excom Phcoona 5x10-4 mil - [ Phcooth) = - (25+5 m2 5x10y mil 30X703 I SXOLM thcoona) = _134X103 mot (20+5) mL 1.375X03 moh 3ox ist = 1.375 m from her torsom en [eR ] 1 378/20 pH = 4.20+ wg 5XU.) = 420 + wg 135 4.64

ме (4 Addetion of wom of Hu. tom of ar MHU = lomexo.1 m hun - lomlxv. ir tu 1000 m - 1xco- milattu 1x 103 mol of the react with 1x mit th worra to form 1x 103 mil of phevot. theretter for the the contern (1.875 0103_1x100 mo e 8.75 4104 milof exam phcooNa. - Theooh) = __ 1x10–3 met (25+10m 18163 m) = 85 XII mo 8.75 80 mol [ph wong = -35x632 = 9.375 mg this fome e pH = 4.20thg 6 1/35 = 4.20 + wg 0.875 0.875 31 (5) Addition of 15 m2 of Hel. is mest alm HU = 15m24 0.1 m Hy - 1540.1 mol Hy = 1058 10-3 mil Hu. 1.5x103 mol of Hu react with 1.5 800-3 mol of ph coora to form 1.5x 10-3 mol of phwol.

moto there for the son contorn (1.875 x 103 104 101 is u. 375103 mol of exem phcoona. tphcool] 15 x103 moh (21+ 15810L - 15 m tphcoona Unt 18 mo (25+5) Xull -0.37 m this froment ③ we have, 0.375/40 L- pH = 4.2thg 11/40 = 4.2+ log 0.37 ~ 3.60 (6) Addetun of 17 ml of Hu: (similar way) .. on before [phcooth = the nam = 42 m (Ph cooNa] = 1 (1-875-17) 42 = - 0:175 42 m ". this from eap 6, 0.179/12 pH = 4.20 + leg 117/42 = 4.20 + wg oltre ~ 3.4

(1) Additum of if mit HU (Stomilar way or before) tphcooh) = 169603 = lem 43 Phoonal (1-875-1.8) 43 43 apH - 4.20 tlog = 0.01 m. 0-071/43 1.8/43 =4.201 wg of rest ~ 2.82 (8) Additumn of 18-75 mlhu: 18.75 mL of alm HU = 18.75 ml 0.1 mol of the 1000 m antalone = 1.875 X 1-3 mol of Hu. therefore the son contron only Phool. - tphoH) = 1.875x105 moh. (25 + 18.7t) x L = 0.043 m Ph wollt f Ph wortht CCF 4a orca where is the mitral concentratus & be the degnut dnocention

ka=- 1 Phloo 7172 trh woh) Ca.ca - - car on a 41 1-dei a = kg the = ca = c skole =skał - Light) = 2 kagkat Y loge - wgTHI] = - / Logka - Yloge pH = { pka – ½ loge = 1/2 X 4.20 - ½ wg 0.043) = 2.10 - Ylog (0-043) ~ 2.78 (8) Addition of zome Hu. 20 mL of 0.1 m HU = 2000. 1 molof Hu = 2x10-3 mol of hu. in the sola contorn (2x 10-3 – 1.87 x 103 e, 0.125x103 mol of exum He. the] = (20+2+) x 1 = 2.78 x 103 m 1000 D.125x43 moth (20+2+) Xi Then the pH of the sot is due to presenu exus Hu(strong and). treglecting the contributun from weak and)

pH =- wg thi - 256 calculatin e = log (728x4) (7 Additund at me of the (smelor calce 3 [HU] - (22x0.1 -1.875) nom un (25+22) = 6.91x103 -pH = -wg (6.91x10) ~ 2.16 (10) Addition of 25 mL of Hu (similar in o) (HU) = Huy (25X0-1-.871) M (25+25) = 0.0125 - pH =- wg (0.0125) 1.90. ② Addeturn of 30 mL of Hu (same on they = (300--1-875) m (28+30) – 0.020 Oras otras wheel pH = - wy (0.020) s 1.69xclus Titretun arren, oh with the gradeel caddetion. of Hul, pH of the soln decreases gradually. this the titration curve is as follow- h not of Hali ? decreany, my no as follow- antiche alla