Question

Given the following titration information, what was the molar concentration of t-butyl chloride at 88 minutes? Round to 1 significant digit and put no units. Hint: Calculating the other missing values of the table may be necessary to compute the answer. run minutes mL 0.04 M NaOH (HCI) [t-butyl chloride) 3 88 17.6 7 infinity 26.1

Answer 1

Given run 3 minutes 88 mL 0.04 M Nooh [HQ] [t-buty & 12.6 . chloride] 7 infinity abol For sun 7 at infinite time, a6.8 m2 of 0.04M NAOH 26.1 mL 1000 m2), * 0.04 M Nao 11 = 0.001 05 mol -> 0.001058 moles of NOOH will neutralise 0.001059 moles of Hal that will come from o.o0los moles of tert butyl chloride. At times the reaction is complete. 7. Do, 0.0olosa moles of test butyl chloride also corresponds to initial number of tert butyl chloride. *For run 3, after 88 minutes, 13.9 ml of 0.04 M Nash then, 17.6 m2 -X0.04 M 1000 ml NAO H = 0. 00 010+. mol . 0.000704 moles of NaOH will neutralise 0.900704 of Hel that will come from 0.000704 mol of text butyl chloride nce, the number of moles of text butyl chloride remaining after 88 min. 0.001052-0.000704 -0.000 348 mo xif val of aliquot was used the molar for the titration then Concentration of tart butyl chloride is cs Scann.o 00347 mobs x 10oo mike Do, E CamStarmi