7/16/2019 Prelab Acid Base

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Aditya MahadevanAP Chemistry

2nd Period January 14, 2013

Prelab

1.Equivalent Mass = Molar Mass/(# of H+ ions per molecule)

a.) HC2H3O2 Equivalent Mass = (60.05 g/mol) / 1 H+ per molecule = 60.05g/equivalent

b.) KHCO3 Equivalent Mass = (100.12 g/mol) / 1 H+ per molecule = 100.12g/equivalent

c.) H2SO3 Equivalent Mass = (82.08 g/mol) / 2 H+ ions / molecule = 41.04g/equivalent

2. KHC8H4O4 + NaOH -> KC8H4O4- + Na+ + H

2O

(.5632g KHC8H4O4) (204.23 g/mol KHC8H4O4)(1 mol OH-/ 1 mol KHC8H4O4) (1 mol OH-/1 mol NaOH)

= 2.758 x 10-3 mol NaOH

(2.758 x 10-3 mol NaOH)/(23.64mL) = 1.167 x 10-1 M

3. (24.68 mL NaOH) x (0.1165 M) = 2.875 x 10-3 moles

(2.875 x 10-3 moles/1 mole OH- per molecule) = 2.875 x 10-3 equivalent

(0.2931 g Acid) / 2.875 x 10-3equivalent = 101.9 g/equivalent

4. a.) see graph paper

b.)The pH at the equivalence point is approximately 8.73. This was determined by finding the point of inflection in the graph

of the titration curve and measuring the pH at that point.

 

c.)The pK a is about 4.74 . This was determined by finding the pH at the half equivalence point (12.5 mL).

pKa = 4.74 Ka = 10-4.74= 1.82 x 10-5

 

d.) The most appropriate indicator would be Thymol blue because the equivalence point is between pH values of 8 and 9.6 (8.73)