{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ChemLab113-0032 - What the Computer Does The computer will...

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

View Full Document Right Arrow Icon
Background image of page 1
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 will have you enter all the data exemplified in the following sample data table. Acid Molar'ity Mass of N32C03 0.2018 g Initial buret reading 0.19 mL Buret reading before first drop 37.24 mL pH before first drop 4.66 pH before second drop 4.60 pH before third drop 4.54 pH before fourth drop 4.47 pH before fifth drop 4.39 pH before sixth drop 4.29 pH before seventh drop 4.18 pH before eighth drop 4.00 pH before ninth drop 3.88 pH before tenth drop 3.78 Final pH 3.69 Final buret reading 37.77 mL The calculations begin with the computer finding which drop caused the equivalence point to be passed. This is the drop that Causes the largest pH change and is the seventh drop in this case. The change in pH between "before first drop" and "before second drop" is 0.06; and the other differences are 0.06, 0.07, 0.08, 0.10, 0.11, 0.18, 0.12, 0.10, and 0.09 respectively. The change of 0.18 occurs after the seventh drop (which reads “before the eighth drop”). Note that the counting of drops begins at zero. Next the computer calculates the volume of a drop of acid by using the number of drops added, together with the buret readings before and after dropwise addition, the final buret reading. Drop volume = (37.77 mL - 37.24 mL)/10 = 0.053 nil/drop The equivalence point is taken to occur at half the drop that caused the largest pH change or at a total of 6.5 drops added in this case. The computer uses that value, the drop volume, and the volume of acid used before dropwise addition to calculate the volume of acid used to reach the equivalence point. Volume of acid : V of acid before dropwise addition + V of drops to equivalence point. Volume ofaeid = (37.24 mL - 0.19 mL) + 6.5 drops (0.053 mL/drop) = 37.39 mL 44 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online