Question

Section DATA SHEET: EXPERIMENT 16 Date A. Determination of NaOH Concentration 1. Molarity of HCl solution M Trial #1 Volume of acid 0.0831 Trial #2 Trial #3(...) 25.00 ml 25.00_ml 25.00 ml 160.40 ml 5.55_ml 18.75 ml 1.30_mi 0.25_ml. 3.80 ml 3. Final buret reading Initial buret reading 5. Volume of NaOH - 6. Molarity of NaOH (Show set-up) 7. Average molarity of NaOH

Answer 1

④ Determination of NaOH Concentration Values Provided in the Question are - 0 Molarity of HCO = 0.083) M. @ Volume of Acid in 3 Trials Trial # 1 Trial #2 Tral #3 - 25 ml 25ml 25ml ③ Final Buret Reading = 16.4ome is 55ml 18.75 ml 9 Initial Buret Reading = 130me 0.25 ml 3.8ome Now, we need to calculate> Trial #1 Trial#2 Trial #3 ③ volume of NaOH = 15.10 ml (15-30 me 14.95 ml ① molarity of NaOH [0.13761m [0-13501m [0.1389 m So, To calculate volume of NaOH formula applied is - 6 = Volume of Nach for Trial #1= (Final Buret Reading) of Trial #1 - Fritial Buret Reading of Trial #1 Vol. of NaOH = 16.40 - 1.30me (Trial #1) vol. of NaOH = 15-10ml Trial # 1 Volume of NaOH for Trial #2 = (Final Buret Reading) - Initial Buret Reading for Trial #2 = 15.55-0.25 vol. of NaOH = 15.30 me Trial #2

= Volume are of NaOH of NaOH for 180 for Terial #3 = (Final Buret Reading - (Onitial Burst Reading beb of Trial #3 P = 18.75– 3.80 Vol. of NaOH = 14.95 me Trial #3 s on Now, To calculate Molarity of NaOH 6 Formula Used is for 3 y of NaOH trials for s. - Molarity of Acid x volume of Aud) = (Molarity of Basex - Volume of Base) ů.e. My = M₂V2 a (Acid) (Base) l a autor Given molarity Molarity of Acid = M = 0.0831M. . Volume of Aled = V, = 25 ml (same for 3 trials) So, now - nolarity of NaOH (M2) for Trial #1 e t Trial #1) - M, V = M₂ V2 [ already 0.083) X 25 = M calculated X 15-10 Lin 5th part) M2 = 0.0831 x 25 [V2 = 15.101 15.10 it to M₂=0.1376 M Trial #1 > Molarity of NaOH (M2) for Trial #27 My = M₂ V₂ (Trial #2) I already M₂ = MV = 0.0831 x 25 M2 = 0.1350M Trial #2 15.30 in 5th part) 2 15.30

#2= - Molarity of NaOH (M2) for Trial M, V = M2 V2 M = M. Vi (Trial#3) V₂ abready calculated in 5th part [V2 = 14.95 =0.0831 x 25 14.95 M2 = 0.1389 m) Trial #3 & Average molarity of NaOH = Sum of Molarities of NaOH of all 3 Trials = 0.1376 +0-1358+0.1389 [Avg. Molarity of NaOH = 0.1374M]