# Acid Base Titrations Essay

Acid and Base Titrations: Preparing Standardized Solutions Introduction:
This experiment focuses on titrations of acids and bases. A titration depends on addition of a known volume of solution and is a type of volumetric analysis. Many titrations involve either acid-base reactions or oxidation-reduction reactions. In this experiment we do one of each. We monitor the pH of the reaction with the use of a color indicator. We also learn about the standardization of bases (NaOH) and acids (HCl) which is basically making a dilution to change the molarity.

The first reaction consists of titrating sodium hydroxide (NaOH) into potassium acid phthalate (KHP or K[HC8H4O4]): K+[HC8H4O4]- + Na+OH- => K+Na+[HC8H4O4]- + H2O

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The second titration we did was hydrochloric acid (HCl) with sodium hydroxide (NaOH): HCl(aq) + NaOH(aq) => NaCl(aq) + H2O(l)
Procedure:

You need to calculate the volume of 3 M NaOH needed to make 500 mL of a 0.

As you approach the end point, the solution will turn pink very briefly. Add very small amounts until the solution stays very light pink for more than ten seconds. Record the final volume then and calculate the volume of base necessary to reach the end point of the reaction. Titrate sample two in the same exact way. (This titration should proceed a little faster since we now know the approximate volume needed to reach the endpoint). Just remember to record the initial volume, take it slow towards the end point, and record final volume. After you calculated the molarity and taken the average, take your NaOH and place it in a bottle to save for the next experiment. Clean out the 250 flasks and the 500 mL flask for use in the next titration.

Molecular Weight of KHP: 204.22 g/mol
I. Standardization of the NaOH solution
SAMPLE
1
2
Wt. KHP
0.4628 g
0.4926 g
mmol of KHP
2.266 mmol
2.412 mmol
24.3 mL
30.2 mL
0.7 mL
5.65 mL
Volume of NaOH used
23.6 mL
24.55 mL
Molarity of NaOH
0.096 M
0.098 M
Average Molarity
0.097 M

II. Standardization of the HCl solution
SAMPLE
1
2
43.75 mL
39.8 mL
8.6 mL
4.5 mL
35.15 mL
35.3 mL
31.9 mL
42.1 mL
1.4 mL
8.2 mL
Volume NaOH used in titration
30.5 mL
33.9 mL
Calculated Molarity of HCl
0.0842 M
0.0932 M
Average
0.0887 M

Analysis and Discussion:
Since NaOH solution is used for the titration of the unknown and also in the standardization of HCl, its standardization is important as well. In order to measure the molarity of the NaOH, I reacted it with the KHP with is a primary standard. I made two very precise measurements to make the molarity a much more accurate number. With molarity, a 1 M solution has one mole per liter of solution. It’s easier to work with smaller numbers so I converted them to millimoles and milliliters: Molarity = (moles / liters) = (10-3 x moles / 10-3 x liters) = (mmol / mL) In the titration, the mmol of KHP equaled the mmol of NaOH added to the solution. At the end of the first titration, I had the information needed to calculate the molarity of the NaOH solution using this equation: M base = (g KHP x 103) / (mL base)(mw KHP)

Conclusion:
The acid base titrations provided key insights into the behavior of both the acid and the base with each other and with water. It was observed that when a strong acid and a strong base react with each other in equivalent concentrations, the pH of the solution is neutral, reaching an end point. It was also shown that the molarity of the bases and the acids could be determined if a few facts were known about the acid-base system. Finally, using all the information and data we obtained from the experiment, we put it all together. These observations and accomplishments demonstrate important qualities of the acid-base equilibrium. These qualities have real effects on real systems in the world. The ability to predict pH, molecular weight, molarity, or any of the other values that were used in this experiment gives great power and understanding to understanding the chemical
system.

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