Question

Name Section Date Lab Partner PRE-LAB ASSIGNMENT: CONDUCTIMETRIC TITRATION AND GRAVIMETRIC DETERMINATION OF A PRECIPITATE (Show work for All Problems) 1. Define the analyte and titrant in this experiment. How can you differentiate between them? 2. The following data below was gathered during a conductimetric titration involvinga standard solution of 0.100 M sulfuric acid, H.,SO, and 20.0 mL of barium hydroxide, Ba(OUsing LoggerPro, plot these data, perform a linear regression (refer to the Data Analysis section), and answer the questions below. Conductivity (uS/cm) 55.8 10.3 64.1 159.3 259.6 349.6 451.9 537.8 625.7 714.6 805.6 887.3 968 Volume (mL) Conductivity Volume (mL) (HS/cm) Haso, 13 14 15 16 17 18 19 20 21 H,so, 0 834.06 747.7 678.4 607.1 532.6 462.3 388.9 325.8 251.3 184.1 23 24 25 10 12 a) Write a balanced chemical equation for the reaction that occurred during the titration. b) Print out your graph and indicate on the graph where the equivalence point was reached c) What was the equivalence point volume of H,SO,? d) How many moles of H,SO were titrated? e) Calculate the number of moles of Ba(OH), originally present in the solution. f Calculate the molarity of Ba(OH), 1331 SUFFOLK COUNTY COMMUNITY COLLEGE 3. In the same experiment, the following data was gathered for gravimetric determination of the precipitate, BasO,. Using the data below, answer the following questions. Equivalence point (mL) Mass of filter paper + precipitate (9) Mass of filter paper (9) 13.87 1.601 1.310 a) Calculate the mass of BaSO, produced in the reaction. b) Calculate the number of moles Baso, c) Calculate the number of moles of Ba(OH), d) Calculate the molarity of Ba(OH),

Answer 1

1) The analyte is the solution that is titled, in this case the Ba (OH) 2, and the titrant is the solution with known concentration, in this case H2SO4.

2)

Vol H2S04 Conductivity 969 908 Conductivity 2834,06 1200 678.4 1000 532,6 800 600 325,8 400 184.1 200 55.8 0 3747.7 607,1 462.3 388,9 Conductivity 10 251.3 12 117,9 0 5 10 15 2025 30 10,3 64,1 159.3 259.6 349,6 451.9 537.8 625,7 714,6 805,6 887.3 968 14 16 18 20 21 23 24 25

a) Ba (OH) 2 + H2SO4 = BaSO4 + 2 H2O

b) It is the inflection point of the conductivity values.

c) V H2SO4 = 14 mL

d) n H2SO4 = M * V = 0.1 mol / L * 0.014 L = 0.0014 mol

e) Because the stoichiometric ratio between the reactants is 1: 1, the moles of H2SO4 are equal to the moles of Ba (OH) 2 = 0.0014 mol

f) The concentration of Ba (OH) 2 is calculated:

[Ba (OH) 2] = mol / V = ​​0.0014 mol / 0.02 L = 0.07 M

3) a) The difference between the weight of the filter paper and the weight of the filter paper plus the analyte, then:

Mass BaSO4 = 1.601 - 1.310 = 0.291 g

b) With the molar weight, the moles are calculated:

n BaSO4 = 0.291 g * (1 mol / 233.38 g) = 1.25x10 ^ -3 mol

c) The moles of BaSO4 are the same as Ba (OH) 2, because their ratio is 1: 1, n Ba (OH) 2 = 0.00125 mol.

d) The concentration of Ba (OH) 2 is calculated:

[Ba (OH) 2] = 0.00125 mol / 0.02 L = 0.0625 M

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