Pre-lab assignment PDF

Title Pre-lab assignment
Author Rafael Dador
Course General Chemistry For Engineers (Laboratory)
Institution Mapua University
Pages 2
File Size 65.6 KB
File Type PDF
Total Downloads 5
Total Views 146

Summary

Questions asked before attempting the experiment ...


Description

1. Assuming the density of a 5% acetic acid by mass solution is 1.0 g/mL, determine the volume of the acetic acid solution necessary to neutralize 25.0 mL of 0.10 M NaOH. Also record the calculation on your Report Sheet. A) 1.0 L * 0.05 * 1000g/1L = 50 g V1=M2V2/M1 0.10 M * 0.025 L/50g *60.05 = V1 = 0.003 L = 3 mL 2. a. A chemist often uses a white card with a black mark to aid in reading the meniscus of a clear liquid. How does this technique make the reading more accurate? Explain. A) The white gives a more accurate reading of the substance because it can be used as a part of comparison. The card will be visible in the meniscus against the black mark, they'll see the white very clearly, giving them an accurate the reading. b. A chemist should wait 10-15 seconds after dispensing a volume of titrant before a reading is made. Explain why the wait is a good laboratory technique. A) Waiting for 10-15 allows the acid and base to react we stir to make sure no bubbles are left. c. The color change at the endpoint should persist for 30 seconds. Explain why the time lapse is a good titration technique. A) 30 seconds is a good wait time because the longer we wait the more CO 2 is absorbed. 3. Lemon juice has a pH of about 2.5. Assuming that the acidity of lemon juice is due solely to citric acid, that citric acid is a monoprotic acid, and that the density of lemon juice is 1.0 g/mL, then the citric acid concentration calculates to 0.5% by mass. Estimate the volume of 0.01000 M NaOH required to neutralize a 3.71-g sample of lemon juice. The molar mass of citric acid is 190.12 g/mol.

0.5 A) ¿ 3.71 x 100 ¿ 0.01855 g 0.01855 g ¿ g 190.12 mol ¿ 9.76 x 10−5 mol ( NaOH ) 9.76 x 10−5 mol ¿ mol 0.0100 L

Computation zone 1) Find the mass of citric acid by multiplying the 0.5% to the sample mass of lemon juice ( ¿ 3.71 x

0.5 ). 100

2) You then find the number of moles of citric acid by dividing the mass by its molar mass (

¿

0.01855 g g 190.12 mol

). Since it’s a 1:1

ratio, the number of moles for citric acid will be the same for NaOH. 3) Finally, you use the number of moles that you have received for citric acid to divide it by the molarity in order to find the

¿ 9.76 x 10−3 L...


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