Question

answer the question based on data available for each trial

Table I. Gravimetric and Titration Data for the Determination of Ba(OH). Concentration Trial 1 Trial 2 Trial 3 Volume at Equivalence Point (mL) 7.857 10.867 8.821 Mass of filter paper + precipitate (g) 0.4171 0.8905 0.8724 Mass of filter paper (g) 0.2577 0.6387 0.6294 Mass of precipitate (g) Molarity if HSO (M) 0.100 0.100 0.100 Moles of H.SO, used to reach equivalence point Average M Molarity of Ba(OH) (titration) d d Moles of BaSO from mass of ppt. Average M Molarity of Ba(OH) (gravimetric) d d

Answer 1

Reaction between Bacott 2 999) & Hy soy is as fellows Bas04 (29) + H2 504 (aq) BaSO4 (3) + 2 H2O(l) #1 #2 #3 8.821mL 7.857mL (0.867mL Vol. q Ha sou at Eq.pt mass of Preeipitation (9) 0.4171-0.2594 Basoa = 0.15949 0.8905-0-6387 0.25189 0.2439 0.10 0.10 Molaurity q Ha 504 (m) 0.10 inder of Hq soq used = Mclarityx Volume 14 O IOM X 7.857ml 1000 = 0.0007857 mde 01x10.867 1000 -0.0010867 ndes QI*8.821 1000 - 0000882 males mees of Ra Cottl2 = mees of Has04 0.0007857 mdy 0.0010867 00008824 Mdacity of Ba (OH)2 moles of Ra COH)2 val. 0.0007857-mdy 7.857 ML 0.000867 10.867 0.0008821 9.821mu - Om -OM Avelagem 0.OM 0.1 +0.1 +0.) 3 OM standard deviation d = O d efference Cortortor OIMIO 2 mdes q Ba 2042 mals mdar mass 0.25189 233.38 907 0.2439 233.38 0.15949 233.38 molt 000 108 males -0.00104 moles - 0000683 miles

Please note that the red colored box is incomplete since you have not provided the volume of Ba(OH)2.

So, please provide the volume of Ba(OH)2, and I will help out sure.

Thanks.