Question

Assume you dissolve 0.367 g of the weak acid benzoic acid, CH3CO2H, in enough water to make 1.00 x 102 mL of solution and then titrate the solution with 0.101 M NaOH. K, for benzoic acid = 6.3 x 10-5) CH, CO,H(aq) + OH- (aq) = CH,CO, - (aq) + H20(0) a. What was the pH of the original benzoic acid solution? pt pl pH = pt b. What are the concentrations of all of the following ions at the equivalence point: Na+ H30+ OH, and CH3CO,? [Nat] = [120+)- [OH-]- м C.H.CO,-) = c. What is the pH of the solution at the equivalence point? pH =

Answer 1

> Amount of benzoic acid = m = 0.3677 - Molar mass of benzoic acid = M = 122-12g/mole - Noo of moles (n) of benzoic acid = m = 0.367g. - M 1 22-12g/mol n = 3x10 mole n = 0.003 mole (O IL = 1000mL) Volume of solution = v= 1.00x 10²mL Concentration of benzoic acid = n = 0.00amole. O.IL 0.03M a = [CoMoCoom] = S ety

Cohgcoolag OM 0.034 + H₂ot cap OM 0.034 C6Hgcoon (aq) + Holl) - Initial 0.0m - final_ 0.03(1-«) E x → Degree of dissociation [Cohs coo] [ Hoot Ka - [CH₂COOH (0-03a) 2 0.03 (1-d) ka= 0.03.2 1-21 [oo weak A cid] x = 6.3x105 O-03 4.58 Xlo 0.0458 x = [Hot] = 0.034 = 0-03 X 0.0458 [H₃ ot] = 1.378103 PH = -log[H3 ot] = -log(1-37X16) (PH = 2086 Answer

(6) At equivalence point No of equivalents of cotts COOH = Now of equivalents of Naoh CGHS CooHlaq) + NaOH(aq) - Cotts Coona & Hot cap No of m-moles of ConscooNa formed = No of m-moles of Che cool reacted civi = C₂ V2 a = Concentration of NaoH = 0.10 M V = volume of wooh required Cosm XoolL = o.lolm X V₂ V₂ = googt 0.02972 Total volume of solution (V) = vt = 0.1 +0-0297) V = 0.1 297L → No of endles of Coh Scoona formed 1 CGH Coona = 0.003 mole 0-1297 L altaat sest = 0.003mole ] = [ Cola Coona C6HCoona is 0.0231M a salt. (ivat] = [CGHS CooNa] = 0.0231 M 40+= [Cong CooNa] = 0-0231M ars Anew en 1 L-on = 'het = 0.0231M = 4.33 X 10 13 An- - [COMşcoo] = [Cousco oma] = 0-023\M Aneu a

G It CGHS Coona is salt of weak Acid will Hydrolyze in presence of water. and stoong Basc PH = √ { pkw + pka tlog pka = -logka = -log (6.3x105) pka = 4.20 pH = {(14 4-20 + logo.021] PH = 8.28 Answer