Question

2. A student is given an antacid tablet that weighed 5.6832 g. The tablet was crushed and 4.3628 g of the antacid was added to 200 mL of simulated stomach acid. This was allowed to react and then filtered. It was found that 25.00 mL of this partially neutralized stomach acid required 8.50 mL of a NaOH solution to titrate it to a methyl red endpoint. It took 21.54 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the "stomach acid" had been neutralized in the 25.00 mL sample which was titrated? (Show work.) a. b. How much stomach acid was neutralized by the 4.3628 g sample used? (Show work.) c. How much stomach acid would have been neutralized by the original 5.6832 g tablet? (Show work.)

Answer 1

Ans

Part a

Volume of NaOH solution added V1 = 21.54 mL

Volume of stomach acid neutralized V2 = 25 mL

Volume of NaOH solution titrated V3 = 8.50 mL

Volume of stomach acid neutralized in 25 mL sample

= V3*V2/V1

= 8.5 * (25.00 / 21.54 )

= 9.87 mL

Part b   

Volume of 4.3628 g sample = 200 mL

Stomach acid neutralized = Volume of 4.3628 g sample - Volume of stomach acid neutralized in 25 mL sample

= 200 - 9.87

= 190.13 mL

Part c

Stomach acid neutralized =

5.6832 g x (190.13 mL / 4.3628 g )

= 247.67 mL