A student is given an antacid tablet that weighs 5.4630 g. The tablet is crushed and 4.3620 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 13.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.7 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the stomach acid (in mL) has been neutralized in the 25.00 mL sample that was titrated

Solution :

It is given that :

Weight of the antacid tablet = 5.4630 g

4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.

25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.

27.7 mL of solution is equivalent to of the original stomach acid. Therefore, 13.6 mL of NaOH will take x

                                                             = 12.27 ml of the original stomach acid.

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890