Question

A student is given an antacid tablet that weighed 5.6832 g. The tablet was crushed and 4.3628 g of he antacid was added to 200 mL fo simulated stomah acid. This was allowed to react and the filtered. It was found that 25 mL fo this partially neutralized stomach acid required 8.5 mL of a NaOH solution to titrate it to a methyl red endpoint. If it took 25.5 mL of this NaOH solution to neutralize 25 mL of the original stomach acid.

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

b) how much stomach acid was neutralized y the 4.3628 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