Question

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 a imulated "stomach acid." Thiswas allowed to react and then filtered. It was found that 25.oo ml of this partiallt nuetralized stomach acid required 8.50 ml of a NaOH solution to titrate it to amethyl red endpoint. If it took 21.5 mL of this naOH solution to neutralize 35.00mL of the original stomach acid

a. How Much of the stomach acid had been neutralized in the 25.00 ml sample which was titrated?

b. how much stomach acid was neutralized by the 4.3629 g tablet?

c. How much stomach acid would have been neutralized by the original 5.6832 g tablet?

Answer 1

Answer (a): 25.5 ml of NaOH solution is equivalent to 25.00 ml original stomach acid

Therefore,

8.5 ml NaOH X (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid

Answer (b): Since It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (from question 1)

Therefore, the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml

Answer (c): 4.3628 g antacid is equivalent to 191.67 ml acid

5.6832 g antacid x (191.67 ml acid / 4.3628 g antacid) = 249.67 ml acid