Question

dont know how to calculate moles of HoAc and the ratio and the bottom table. please help.

50 Postlaboratory Exercise ratory Exercise Used Mtor poldle on graph Have this page available for your class session after completing your report. (Do not attach it to your report.) the pH, ml, and moles NaOH columns only, before class (moles - M Calculate the equivalence point values first. note about significant figures: When a decimal number is converted to a log, the number of c ant figures in the decimal number becomes the number of decimal places in the log quantity ne reverse holds, so a 2 decimal place pH reading produces 2 significant figures in a concentration or inak. However, we will use a convention, based on experimental uncertainties, that unless specified all pH (and PK) values have 2 decimal places A Stoichiometry in a titration (use your actual data) Theoretical initial moles HOAC = 0.20 mol/L x 0.01000 L -0.0020 mole Actual initial moles HOA - moles of NaOH at ep. = MNOH VNO - (0.100) (012) 001995_mole pH mL (from graph) moles NaOH added moles OAc formed moles HOAc remaining ratio of OÀc" /HOÁC 4.00 3.84 000334 000384 Halfway point pH = Sob 1.92 000192 .00192 000192 6.00 19.33 .001933 1.001933 equivalence point, 19.95 .001995 .001955 infinare pH = 7.29 = [H+). [OAC B. Calculation of the acid dissociation constant for acetic acid using concentrations (You will not need to convert the moles to molarity since in the ratio of [OAcy [HOAc) the volumes will cancel. [H + ][OAC) Ka = [HOÀc] pH (H') from pH ka 1x 10-9 [HOAc] 4.00 2,19x104 Halfway point, 105.66 2-19 x 16 9 2.19x10a pH = 3.61 6.00 1.0x10 1.7x10-5 CLEMA: 163 Sp20

Answer 1

Solution Here, I will help you for the calculations of pH = 400 and Halfway point. The remaining two you try by following the steps shown. Part-A: At pH = 4:00: Güven molarity of Naon or on = 0.100 M. - volume of Nach or out from graph = 3.89 me 30.00389 L : Moles of ou" - 0.1*0.00389 = 3.89x10-4 moles Now, consider the reaction Naon + HOAC NODAC + H₂O OR ou + HOAC - OAC + M20. — 4) It is clear that, one mole of out react wi gives Ofe a Imole. 4 3.89x10-4 moles of on give OAC = 3.89x10-4 moles → Moles of OAC 'formed - 3.89 xl0-4motes Inittal No.of moles of HOAC = 0.001995 moles. 2. Moles of HOAC remaining = Initial No. of moles of HOAE . - Moles of - OAC = 0.001995--3.89x1074 = 0.001606 moles. :. Ratio of - OAC = 3.89410-4 = 0.2422 0.001606 но Ас

At Halfway point pH = 3.66: - Cliven molarity of Nool or ou = 0.100 m. Volume of NaOH orou from graph = 1192ml -0.001924 ... Moles of on = 0.100%0.20192=1.92x10-4 moles Now, from ep. (1), it is clear that 1-92x1004 moles of on" give OAC = 1.92x10-4 moles → Moles of Ole formed = 192x10-4 moles Initial No. of moles of Hote= 0.001995 moles . Moles of Hole remaining = & Initial No. of moles of HOAC - Moles of rote = 0.001995 - 1.92 410-4 = 0.001803 moles Ratio of one = 1.92x1044 = 0.1064 restos И0Ae 0.001803 part-B : At pH=4.00: pH = -log[41] - » [H] - 10 P4 , 10-4 [ht] = 10-4 Coac] = 0.2422 So, Kaz (H+).(0 Ac] [Ho ac] hoac] Also

Ka= 10-4x0.2422 | Ką= 2.422x10-5 At Half way Point PM = 366 : [hº] - 10P* = [+) = 10*366 = 2.187X1074 an Also Coac] - 0.1064 (hone] !. : 2:187x|0-4 x 0.15 64 Ko - [R1]. (TOÁc (hoac = Ka = 2.326X10-S