Question

Question 2 (20pts): A 20 mL sample of 0.01 M propionic acid (CHsCH2COOH; Pk 4.87) is titrated with 0.05 M NaOH A) Write out the chemical reaction for this titration. B) Calculate the initial pH of the sample. C) Calculate the volume of NaOH required to reach the equivalence point. D) Calculate the pH of the solution at the equivalence point. E) Sketch a titration curve for this titration (pH versus volume NaOH added). Note the Jocation of the equivalence point on the curve

Answer 1

AD Balanced CH CH Cool (aq) chemical Reaction + NaOH(aq) - CH₂ CH₂ Counalap + Hold) CB - = o-olm. Given Concentration of CH₂ CH, Cool Given pka = 4.87 - - logka = 4.87 Ka = 1.-4.87 Ka = 1.3sxlos a = Degree of dissociation - Ka Coo weak Acid] = 1.35 x 16 0.01 3 - 61 Klo = 0. 0363 - CH₂ CH₃COONalag) + ] Initial CH₃CH, CooHlaq + Nach Cags o.olm oool(-a) inal 0.014 0.04 [H₂ot - ooola = o.olxo.0367 = 0.000367 M pH = -log[H20t] -109 (0.000367M) PH = 3.43 Answer @ At Equivalence point → No of equivalents of CH CHCOOH = No of equivalents of - NaoH Volume of Propionic acid =V, = 2006 Given Concentrar 1aOH = G= 0.05M V = volume of Naot required to reach equivalence Point Civ, - SV2

0.olm X ZomL = 0.05M X V2 v = 4 mL Answer To The salt. addition of Nooh will result in formation of No of m-moles of salt formed = Now of m-mole of Acid reacted = 0.0 MX 20mL = 0.2m.nde Total volume of solution (V) = (20+4) mL = 24mL Concentration of salt formed = 0.2m.mol - 24mL CH₂ CH₂COONa will undergo = 8.33x103 Hydrolysis PH = Il pkw + pka + loge / [oo pH of salt of weak acid/strong Bar = $(14 + 4.84 +log(0.00lsa] PH = 8.391 Answer