Cliffs AP Chemistry, 3rd Edition

Samples: Free-Response Questions

1. In an experiment to determine the equivalent mass of an unknown acid, a student
   measured out a 0.250-gram sample of an unknown solid acid and then used 45.77 mL
   of 0.150 M NaOH solution for neutralization to a phenolphthalein end point.
   Phenolphthalein is colorless in acid solutions but becomes pink when the pH of the
   solution reaches 9 or higher. During the course of the experiment, a back-titration was
   further required using 1.50 mL of 0.010 M HCl.

(a) How many moles of OH–were used in the titration?

(b) How many moles of H+were used in the back-titration?

(c) How many moles of H+are there in the solid acid?

(d) What is the equivalent mass of the unknown acid?

Answer

1. Given: 0.250 g solid acid
   45.77 mL of 0.150 M NaOH required for phenolphthalein end point
   Back-titration: 1.50 mL of 0.010 M HCl

(a) Restatement: Moles of OH–used in the titration.

moles of OH–= moles of NaOH

..
1

0 04577

1

liter 0 150

liter

= # mole NaOH

= 6.87 × 10 –3mole OH–

(b) Restatement: Moles of H+used in back-titration.

moles of H+= moles of HCl

..
1

0 00150

1

liter 0 010

liter

= # mole HCl

= 1.5 × 10 –5mole H+

(c) Restatement: Moles of H+in solid acid.

moles H+in solid acid = moles OH–−moles H+

= (MNaOH×VNaOH) ×(MHCl×VHCl)

= 6.87 × 10 –3mole OH–×1.5 × 10 –5mole H+

= 6.86 × 10 –3mole H+

Acids and Bases