Ml Hcl L Molarity Naoh Now example essay topic

Titration of an Acid with a Base Jeff Barker Chem. A-1 Wednesday, May 17, 2000 Introduction: When you combine strong acids and bases, perfect ionization is achieved. This means that all that is left after the reactions are water and the spectator ion. The definition of an acid is a substance that produces hydronium ions when dissolved in water. Acids have a pH range from 1-6.

Bases release hydroxide ions when dissolved in water, with a pH level of 8-14. There are different segments of acids and bases, weak, and strong. In this lab, we used strong acids and bases, to make it easier for us to find the levels of concentration (it is easier to calculate too! ). Also, since the acids and bases are strong, they ionize completely, which develops a neutralization reaction. This allows the indicators to work.

When the acids neutralize, they produce water and a salt. The production of water induces the indicator to lose an acidic hydrogen, causing a change in the chemical structure of the phenolphthalein, changing it from clear to a pinkish color. This happens because light travels through the structure at a different angle, absorbing everything but some red of the visible spectrum! To find the concentration of the HCL, follow these procedures: 1) Fill a burette with NaOH until it is full to the 0 ml mark. 2) Carefully measure out 10.0 ml of your unknown acid, and pour into a beaker. Add a few drops of phenolphthalein.

3) Place the beaker onto a stirring plate, with the stirring capsule inside the beaker, and set to a medium level. (Note- this should be enough to thoroughly mix the solution, but none of the solution should spill out of the beaker.) 4) Titrate the unknown acid with NaOH until the solution in the container turns a very pink color; it may almost appear to be purple. (Note- This color should last for at least 30 seconds when the solution i neutralized.) 5) Record the amount of NaOH used to complete the titration. 6) Repeat steps 1-5 until you have 3 pair data. Results: Balanced Equation: NaOH (aq) + HCL (aq) = H 20 (l) + NaCl (aq) As stated in the last part of the introduction above, the 0.1 M of NaOH was titrated into the unknown concentration of the 10 ml of HCl. Well, in our 3 observations, Steven Koch and I titrated 78, 80.4, and 76.7 ml of NaOH to neutralize the 10 ml of HCl.

We averaged these numbers to get 78.4 ml on average to neutralize the acid. To find the concentration of the acid, we get to use stoichiometry (yippie!) For the equation, look at this: ml HCl / L molarity NaOH Now take that answer and multiply it by the L of NaOH the result is the molarity of the HCl solution. Now, if there were reacting ratios, then you would apply those (this makes it much more difficult to do!) Now, lets do that same equation, but now we will use the actual data. 10 ml HCL / L = . 01 L HCl |. 01 L HCl.

1 M NaOH = . 001 LM |. 001 LM /. 078.4 L = . 0127 M HCl Discussion & Conclusion In thinking our results, I think that my information might be a little illogical, because of the jump of the data from the test data to our unknown data.

But with the information given, I feel that my calculations are accurate and precise. As I said before, the possible errors could be from bad titration data, wrong markings on the molarity, maybe even some of the math, but not too much! As described in the book, possible real world applications could be to find the molarity of the poorly marked flask of acid. Maybe it could be used to neutralize a lake, and then find out how acidic the lake was, and the concentration has affected the surrounding terrain.

To use this technique for a research project might to, like above use in environmental biology to find approximate acidic, or basic levels of a water gathering.

Bibliography

Scott, Ran nah: Titration of an acid with a base lab report rubric Phillips, Strozak & Wi strom Chemistry: Concepts and Applications (1997) pgs. 540-545.