Question

Need help with a Titration Lab calculations, as we just shifted to online classes, so I don't know exactly how to do what is asked of us, since this data was simply provided to us. If I could simply get the formula / equations, I can do the work from there. Also unsure if my first calculations are correct.


Standardization of NaOH Name Part I Titrant Preparation Calculate the volume of 6 M stock NaOH required to prepare 500 mL of 0.1 M NaOH M1 x V1 = M2 x V2 M1 = 0.1 M V1 = 500 mL (0.5 L) M2 = 6.0 MV2 = ? 0.1 x 0.5 = 6 xV2 V2 = (0.1 x 0.5) / 6 = 8.3 x 10-3L (8.3 mL) Part II Standardization Show the calculation you would do to determine the mass of oxalic acid required for a 30 ml NaOH titration Coarse mass of oxalic acid initial volume final volume volume added molarity 0.188 g 0.00 mL 31.50 mL 31.50 mL 1 0.190 g 1.54 mL 37.30 mL 35.76 mL 2 0.187 g 0.54 mL 34.94 mL 34.40 ml 0.186 g 2.50 mL 39.11 mL 36.61 mL Average molarity - Sample calculations (show your calculations for a single fine titration including units):

Answer 1

o my=ma V2 - Vezmiw - OLX 500 - 8.3 me molwrity - Wt of oxalic x1000 mw of oxalicx vol. (me) mw of oxalic = go molqu'k coarse = 0188 X1000 90X 31.5 0.066M trial = 0.190 x1000 = 0,059 m. 90 x 35.76 trial 22 0.187 x 1000 90X 34.4 o oboum tial3 = 0.186 x 1000 - 0.056um 904 3661 average molaritya 0-066 +0.059 to 06ou to wszy u = 0.0 boys M