January 30, 2014Peter Fenger

Experiment 10.2Acid/Base Titration

I. Purpose

The purpose of this experiment is to better understand titration. Titration is a process of slowly reactinga base of unknown concentration with an acid of known concentration until enough has been added to react with all of the base. Basically, by reacting these two special types of elements we try to figure outhow much acid it will take to react perfectly with a base (or vice versa). This is very similar to the stoichiometry experiment we did in module six this time dealing with only acids and bases.

II. Hypothesis

If a scientist takes a base with unknown concentration, and titrates it with an indicator and an acid having a known concentration, then that scientist will be able to determine the concentration of the base.

III.Materials and Supplies

• Eye dropper• Mass scale• Distilled Water (you will need at least 100 mL of it)• Totally blank white sheet of paper• Stirring rod• 6-10 cabbage leaves• 2 Large Beakers• flame or stove for heating• graduated cylinder• clear ammonia solution (cannot be colored)• clear vinegar (cannot be colored)• safety goggles• a stopwatch

IV. Procedure

1. Rinse your beakers, graduated cylinders, spoons, medicine droppers, and anything else you hope to use for the experiment with distilled water to avoid contamination.

2. Put your cabbage leaves in the small beaker and fill it with about 70 mL of distilled water, if you are using a pot add three times more distilled water.

3. Boil the cabbage and water for about 3 minutes4. Allow the beaker to cool and then carefully remove the leaves from the beaker5. Use your graduated cylinder to add 10.0 mL of the clear ammonia to the beaker6. Add 90.0 mL of distilled water to the beaker as well and stir the solution.7. Add a little less than half of the indicator (the cabbage) to the beaker

January 30, 2014Peter Fenger

8. Measure the mass of the graduated cylinder9. Fill the graduated cylinder with 50.0 mL of vinegar and measure the new mass of the cylinder

plus the vinegar.10. Determine the mass of the vinegar by difference (subtraction)11. Calculate the density of the vinegar using the mass and the volume (50.0 mL)12. Place the beaker full of ammonia on the white piece of paper so that it is easy to see the color of

solution13. Take your medicine dropper and fill it up with vinegar14. Squirt the vinegar in drop by drop and stir after a few drops until you see a color change15. Try to determine how much vinegar you added by determining the mL of vinegar left in the

graduated cylinder.16. What we just did is referred to as a rough titration because the titration process was done fairly

quickly. Now start over and do a careful titration17. Pour the old solution down the drain and rinse the beaker out thoroughly with distilled water.18. Fill it back up with 10.0 mL of ammonia and 90.0 mL of distilled water19. Stir the solution and then add what is left of the indicator (cabbage) to the solution20. Place the beaker back on the white piece of paper and once again fill up the graduated cylinder

with 50.0 mL of vinegar.21. Begin the careful titration, by first adding vinegar to the medicine dropper as before.22. Now carefully add vinegar drop by drop but this time after every drop make sure you stir the

solution thoroughly looking for color change after color change until the solution turns pink.23. Squirt the remainder of the vinegar back into graduated cylinder.24. Read the graduated cylinder and by difference calculate the number of mL you added to reach

the true endpoint of the titration

V. Observation/Data

1. We made sure to rinse out and clean everything with distilled vinegar so that it was all clean andsparkly.

2. We placed our cabbage leaves into the beaker and threw 70.0 mL of water in with it.3. Our Cabbage leaves boiled into purple liquid after a few minutes.4. We removed the leaf parts from the indicator leaving just the purple cabbage juice and water in

there.5. Using a different cylinder we poured our ammonia very carefully into the clean beaker

6-7. We dumped in 10.0 ml of ammonia, 90.0 mL of distilled water, and then when we added the indicator (cabbage) the solution immediately turned blue indicating that we had a base in the solution. The base of course was the ammonia (NH3).

8. The weight of our cylinder was 71.7 grams.9. The mass of 50.0 mL of vinegar+cylinder was 121.3 10. So the mass of the vinegar calculated out to 49.6 grams.11. Now we calculated the density. The equation for density is Mass divided by volume. So our

density is 0.992 grams/mL because 49.6 grams divided by 50.0 mL is what that amounts to. 12-14. This first time we kept on adding drop after drop after drop of vinegar without stirring the solution, eventually we did stir it and the solution turned pink. However, we did not see the process of the color change that we did the second time and we also were not sure we had used more vinegar than needed.

January 30, 2014Peter Fenger

15. We calculated that we had used about 2 mL of vinegar this first time.16. Because of how we poured in the vinegar we had no idea if we used too much vinegar or not

that is why this is called rough titration. We roughly threw in a bunch of vinegar and didn't do itvery carefully, so this second time we set out to try and do it very carefully.

17-20. We poured the old solution down the drain and started over repeating the process up to the point when the vinegar should be added.

21-22. This time when adding the vinegar we carefully stirred the solution after every drop verythoroughly. The solution went through an amazing array of changes! It started out as purple then it turned green, then blue, then light blue, and then pink! As soon as we saw pink we knewthat the solution had reacted completely and we were done.

23-24. We calculated the amount of vinegar we had used to 2.08 mL (I think). I am not sure if that is the number that was calculated or not due to the lack of clarity in our group, but that is what I was told the amount of vinegar used was. This part of our experiment was extraordinarily confusing for thegroup due to the rush to end the class since we where running so late on time (Long experiments kill time).

VI. Conclusion

Overall, we took an ammonia solution (the base) with an unknown concentration and titrated it with a vinegar solution (the acid) whose concentration was known. By using an indicator (cabbage dye) and the adding of vinegar drop by drop we where able to determine exactly when we had added enough acid to neutralize our base because of the pinkish color the solution would change to. This is what titration is! We then did a rough titration and a careful titration, the careful titration proved to the exact amount how much vinegar (acid) was needed to neutralize the base (ammonia). Now that we have accomplished doing rough titration and careful titration I will attempt to prove my hypothesis by solving for the amount of ammonia concentration (base).

The reaction that occurred in this experiment was: C2 H4 O2 + NH3 ----> NH4 + C2 H3 O2 -. Our acid clearly is C2 H4 O2 because it gives an H+ to the NH3 and thus NH3 is our base. This means that for every molecule or mole of acid added every molecule or mole of the base reacted with that. So the first step we need to take is to convert our acid (vinegar) from grams to moles, but in order to do this we need to first convert from milliliters to grams since our vinegar is currently shown in mL. We do this by multiplying our amount of milliliters by the density of the vinegar. We calculated our density to be 0.992 grams/milliliters and we used about 2.08 mL of vinegar, so we multiply the density by the mL used: 2.08 mL x 0.992 grams/mL = 2.06 grams. Now that we have our acid in grams we need to multiply this by the percentage of the acid within the vinegar to find the mass of acid: 2.06g (mass of vinegar) x .05 = 0.103 grams (mass of acid within the vinegar). Finally now that we have found the amount of acid in grams we can convert from grams to moles. Now we have to find the chemical formula of the acid. We do that by finding the total amu of C2 H4 O2 the amu of C is 12.0, H is 1.01, O is 16.0 so 12x2 = 24, 1.01 x 4 = 4.04, and 16 x 2 = 32. Wethen collect the amu of the whole molecule: 16 + 24 + 4.04 + 32 = 60.04 amu, and amu converts to grams so the chemical formula of C2 H4 O2 is 60.04 grams. We know also that 60.04 grams = 1 mole of this entire formula, so we can now convert to moles: 0.103 grams (C2 H4 O2) x 1 mole (C2 H4 O2)/ 60.04 grams (C2 H4 O2) = 0.0017 moles. Now that we have converted to moles for our acid we need tofigure out how many moles of the base originally was present to find the mole to mole relationship

January 30, 2014Peter Fenger

between the acid and the base. The good thing is that it's a 1 to 1 ratio, because every single one of the moles of the acid reacted with the moles of the base so the moles are the same NH3 is also .0017 moles.Finally we can now find our base concentration. All we have to do is use the moles to liters equation tofind the concentration, but in order to do this we have to convert from milliliters to liters for the ammonia. We used 10.0 mL of ammonia so that is 0.01 Liters. Now we just plug our moles and liters in the concentration equation and we are done: 0.0017 moles (NH3)/ 0.01 Liters = 0.17 M and we have found the concentration of the ammonia!

So it turns out that all we needed to find an unknown concentration of a base was to have an acid with aknown concentration and an indicator so that we can see color change and know when we have fully reacted the acid with the base, thus my hypothesis has been proven correct! I personally find it amazing that it is possible to calculate not just the amount of molecules in something but also just the amount of concentration of molecules within something doing just a little math and lab work! I would love to know what a molecule looks like.