Phenolphthalein Chemistry Lab Report
C20H14O4+ NaOH → NaKC8H4O4 (aq)+ H2O(l)
Phenolphthalein is used as an indicator in this reaction because it changes color when a solution reaches a pH of around eight, or in other words the endpoint of a solution. It also allows us to see a physical change, thus resulting in a more accurate measure of how much was titrated. Citric acid is a weak acid, making it best titrated by using a strong base, such as sodium hydroxide. The combination of a weak acid and a strong base would result in an endpoint that corresponds to the range of phenolphthalein indicator approximately a pH of eight.
Phenolphthalein is composed of C14H20O4, as shown in the image above and can …show more content…
To continue, a primary standard is a solution prepared by direct measurements of the mass of solute and the volume of solution, it has known quantitative data. However, a secondary standard solution is a solution whose concentration cannot be determined directly by weight of solute and volume of solution, the concentration must be determined by analysis or experimentation of the solution itself, as presented in ChemWiki. The purpose of titrating KHP with NaOH was to neutralize our experiment, or reach an endpoint as indicated by the phenolphthalein. Titration is used in order to find the concentration of a basic solution. The amount of reactants that have been mixed at the equivalence point depends on the stoichiometry of the reaction.
In order to calculate the molarity of NaOH, My partner and I, were able to take the amount of grams of KHP used and divide it by the mL of NaOH. We also had to ensure that we took into account the different units and conversions needed to do so. After plugging in g KHP as well as mL NaOH into this equation provided:
M NaOH = g KHP x 1mol KHP x 1 mol NaOH x 1000 …show more content…
Both the grams of KHP as well as the mL of NaOH varied for the two trials. For trial one, we had .50 g of KHP and 24.5 ml NaOH, which lead to the following calculations. .50 hKHP divided by 24.5 mL NaOH,multiplied by 1 mol KHP over 204.23 g KHP, in order to cancel out the grams as a unit. Then we multiplied that by 1 mol of NaOH over 1 mol KHP due to the 1:1 ratio and cancel the unit mol KHP. Lastly, we multiplied by 1000 ml over 1 L because molarity is moles over liter of solution. These calculations resulted in the M NaH as 0.099927 or approximately 0.1M. For trial two, we used .50 g KHP and 24.4 ml NaOH, resulting in a molarity of 0.1003 or approximately 0.1M. We applied the same steps mentioned above, with different substitutions for g KHP and mL NaOH. In addition, using the above results, we were able to find the average value of molarity of NaOH by adding to two molarity values and dividing it by two. Hence the .1mol/L average value of molarity of NaOH.
The chemical equation of Citric acid reacting with NaOH is :
H3C6H5O7(aq) + 3NaOH(aq) -----> 3H20(l)+