Titration Research Paper

My results yielded a high Average Molarity .270M . The ideal would be around 1.000M . Deviation was ± 1.20

We know that that the end point of the titration is reached when, after drop after careful drop of NaOH, the solution in the flask retains its pale pink color while swirling for about 30

1. To titrate a hydrochloric acid solution of “unknown” concentration with standardized 0.5M sodium hydroxide.

(0.074 mol HCl x 1 mol NaOH) / 1 mol HCl = 0.074 mol NaOH

The results showed the molarity of the NaOH solution. This experiment was completed twice and a new average molarity

After the twenty minutes elapsed, the flask was cooled to room temperature and then titrated with the remaining NaOH until the colorless solution remained pink. The final volume was then recorded. While solution #1 was heating the same process was repeated with solution#2 and the second burette

2. Calculate the molarity of the Hydrochloric acid in the flask. You may refer to the Titration demo at the beginning of the honors lesson, just above the Virtual Lab to see sample calculations.

9. How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?

A higher volume of NaOH will result in more moles of NaOH being added to the HCl, which results in more HCl reacting. This makes the calculated molarity of the HCl be smaller than the actual molarity of the HCl.

Chemistry 102 is the study of kinetics – equilibrium constant. When it comes to the study of acid-base, equilibrium constant plays an important role that tells how much of the H+ ion will be released into the solution. In this lab, the method of titrimetry was performed to determine the equivalent mass and dissociation constant of an unknown weak monoprotic acid. For a monoprotic acid, it is known that pH = pKa + log (Base/Acid). When a solution has the same amount of conjugate base and bronsted lowry acid, log (Base/Acid) = 0 and pH = pKa. By recording the pH value throughout the titration process and determining the pH at half- equivalence point, the value of Ka can be easily calculated. In this experiment, the standardized NaOH solution has a concentration of 0.09834 M. The satisfactory sample size of known B was 0.2117 g. The average equivalent mass of the unknown sample was found to be 85.01 g, pKa was found to be 4.69, which was also its pH at half-equivalence point and Ka was found to be 2.0439×〖10〗^(-5). The error was 1.255% for equivalent mass and 0.11% for Ka. In other word, the experiment was very precise and accurate; the identity of the unknown sample was determined to be trans-crotonic by the method of titrimetry.

3. How could a conductometric titration be used to determine the molarity of either reacting solutions, assuming the concentration of one solution was known? HINT: Consider the variables needed to calculate molarity, and how can these values be obtained from the titration. The symbol M stands for molarity with units of mol/L. Molarity is a measure of the concentration of a solution. If you knew the concentration of one of the solutions, the molarity could be found in this way. The concept plan would go as follows: Volume of Titrant (in liters) x Moles of Titrant (mol) = moles of the unknown, and then take Moles of the unknown / Liters of the unknown to get the Molarity of the unknown in mol/L. 4

ii. The second part of the titration series involves titration of NaOH with Hydrochloric acid (HCL). Again, three reps of titration and a blank titration have to be completed. A volumetric pipet is used to measure 10.00mL of HCL into three labeled conical flasks. Then the flasks are filled with deionized water until about the 50mL mark. A buret is

During a titration the pH of the solution will be monitored using a pH meter from that we get a titration curve. The titration curve is then used to determine the equivalent molecular weight and Ka value of the unknown weak acid, from that we are

One milliliter of 6.00-M phosphoric acid was placed into a 125-mL Erlenmeyer flask using a volumetric pipette. Using a slightly larger pipette, six milliliters of 3.00-M sodium hydroxide was transferred into a 50-mL beaker. Then a disposable pipette was used to slowly mix the sodium hydroxide into the phosphoric acid while the solution was swirled around. Then both the beaker and flask were rinsed with 2-mL of deionized water and set aside. A clean and dry evaporating dish was weighed with watch glass on a scale. Then the solution was poured into the dish and the watch glass was placed on top. The solution was then heated with a Bunsen burner to allow for the water to boil off to reveal a dry white solid. After the dish cooled to room temperature it was once again weighed and the new mass was recorded.

5. In reaction three, the number of moles of NaOH can be calculated from the concentration of the solution (1.0M = 1.0mole/L) and the volume used. The calculation is below. Enter the result into Data Table 2.