{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ChemLab113-0060

# ChemLab113-0060 - What the Computer Does The computer...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: What the Computer Does The computer begins by obtaining from you all the data exempliﬁed in the following table. Aspirin Analysis NaOH molarity 0.1004 M Mass of aspirin 0.2080 g Initial buret reading 0.50 mL Initial pH 2.81 Data for Large Additions Buret reading pH 3.59 mL 3.20 6.60 mL 3.55 9.52 mL 4.20 1 1.62 ml. 5.30 Data for Dropwise Addition pH after ﬁrst drop 5.38 pH after second drop 5.49 pH after third drop 5.62 pH aﬁer fourth drop 5.79 pH after ﬁﬁh drop 6.06 pH after sixth drop 6.52 pH after seventh drop 7.24 pH aﬁer eighth drop 8.52 pH aﬁer ninth drop 9.17 pH after tenth drop 9.51 Final buret reading 12.16 mL The ﬁrst result the computer calculates is the molar mass of the aspirin. It begins doing so exactly as in Block I: by calculating the drop volume, then using the largest pH change to ascertain which drop caused the equivalence point to be passed, and ﬁnally calculating the volume of NaOH solution used to reach the equivalence point. In this case the drop volume is 0.054 mL/drop, the equivalence point was passed with the eighth drop (equivalence point is 7.5 and the counting of drops starts with one), and 11.52 n1L of NaOH was used to reach the equivalence point (Volume of base before dropwise addition is 11.12 mL and the volume of drops to equivalence point is 0.405 mL). That volume, the mass of the aspirin sample, and the NaOI-l molarity are used to calculate the molar mass of the earth product. IL 1000 mL 1 1 mo] NaOH 0.1004 mo] NaOHH [L )(11.52 mL 1 mol aspirin Molar mass = 0.2080 g aspirin ( = 179.8 gfmol 83 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online