Question

(1) A student is given an antacid tablet that weighs 5.4320 g. The tablet is crushed and 4.1220 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 14.4 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.8 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?

(2) A student is given an antacid tablet that weighs 5.3690 g. The tablet is crushed and 4.1160 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 11.4 mL of a NaOH solution to titrate it to a methyl red end point. It takes 27.1 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid.

How much of the stomach acid (in mL) is neutralized by the 4.1160 g crushed sample of the tablet?

(3) A student is given an antacid tablet that weighs 5.8330 g. The tablet is crushed and 4.2890 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 11.1 mL of a NaOH solution to titrate it to a methyl red end point. It takes 28.4 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid.

How much of the stomach acid (in mL) would be neutralized by the full 5.8330 g tablet?

******The answers I got were 1. 12.949

2. 189.483

3. 258.709

But these were apparently all incorrect! Please help!!!

Answer 1

1.

Without the antacid 27.4 mL NaOH would neutralise 25mL stomach acid.

With the antacid 14.4 mL NaOH would neutralise 25mL stomach acid. Which means some part of stomach acid was neutralised by the antacid.

Without the antacid 14.4 mL NaOH can neutralise ((14.4 * 25) / 27.8) mL stomach acid.

That is, without the antacid 14.4 mL NaOH can neutralise 12.949 mL stomach acid out of the 25mL.

So the remaining volume(mL) will be neutralised by antacid.

Volume(mL) neutralised by antacid in the 25.00 mL sample = 25 - 12.949 = 12.050 mL

2.

Without the antacid 27.1 mL NaOH would neutralise 25mL stomach acid.

With the antacid 11.4 mL NaOH would neutralise 25mL stomach acid. Which means some part of stomach acid was neutralised by the antacid.

Without the antacid 11.4 mL NaOH can neutralise ((11.4 * 25) / 27.1) mL stomach acid.

That is, without the antacid 11.4 mL NaOH can neutralise 10.517 mL stomach acid out of the 25mL.

So the remaining volume(mL) will be neutralised by antacid.

Volume(mL) neutralised by antacid in the 25.00 mL sample = 25 - 10.517 = 14.483 mL

But this for 25mL sample. The atacid was added to 200mL stomach acid. So we multiply it by (200/25 = ) 8 to get the total acid neutralized by the 4.1160 g crushed sample of the tablet

stomach acid (in mL) neutralized by the 4.1160 g crushed sample of the tablet = 14.483*8 = 115.86 mL

3.

Without the antacid 28.4 mL NaOH would neutralise 25mL stomach acid.

With the antacid 11.1 mL NaOH would neutralise 25mL stomach acid. Which means some part of stomach acid was neutralised by the antacid.

Without the antacid 11.4 mL NaOH can neutralise ((11.1 * 25) / 28.4) mL stomach acid.

That is, without the antacid 11.1 mL NaOH can neutralise 9.771 mL stomach acid out of the 25mL.

So the remaining volume(mL) will be neutralised by antacid.

Volume(mL) neutralised by antacid in the 25.00 mL sample = 25 - 9.771 = 15.229 mL

This for 25mL sample.

The antacid was added to 200mL stomach acid. So we multiply it by (200/25 = ) 8 to get the total acid neutralized by the 4.2890 g crushed sample of the tablet

stomach acid (in mL) neutralized by the 4.2890 g crushed sample of the tablet = 15.229 * 8 = 121.83 mL

to get  stomach acid (in mL) neutralized by the full 5.8330 g tablet we multiply this volume by (5.8330/

4.2890 =) 1.356

Stomach acid (in mL) neutralized by the full 5.8330 g tablet = 121.83 mL * 1.356 = 165.69 mL

So your answers should be:

1. 12.050 mL

2. 115.86 mL

3. 165.69 mL