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Uni lab report on titration
Titration practical investigation
Uni lab report on titration
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In the titration experiment, the endpoint was recorded in the experimental data to be at 21.30 mL of NaOH and at a pH of 10.44. However, when all of the data from the table was graphed, the observed endpoint was too high up and on a part of the upper concave down portion of the graph. To ensure that the proper equivalence point was used, a new point had to be extrapolated that was roughly the point at which the graph went from concave up to concave down. This point was at 21.28 mL of NaOH added and pH of 9.20. Dividing both of these points by two, the half equivalence point was found to be at a pH of 5.30 and 10.64 mL of NaOH added. The pH is equal to the pKa here, so the pKa was found to be 5.30. Using data from the equivalence point extrapolated from the graph, the molar mass of the unknown was calculated to by 148 grams per mole. Lastly, because there was only one region of …show more content…
Error analysis calculations showed a 1.4 % error. A possible source of error is extrapolating a value for the freezing point that was too high. The Y-axis of the graph was done in intervals of two degrees Celsius, so precisely finding the value from the graph has the possibility to be estimated incorrectly. Extrapolating too high of a value would cause ΔT to be too low. This would then cause molality to be too low, moles of solute to be too low and the molar mass to be falsely high. After doing an IR spectrum for unknown 149, the only information that could be determined about the structure was that it contained a hydroxide group, a carbonyl and a C-C aromatic bond somewhere in the structure. There was a lot of trouble getting a readable IR spectrum from unknown 149 but after numerous pellets made, it was later found that the printed spectrum would suffice for comparison to trans-Cinnamic and o-Chlorobenzoic acid
Compress the safety bulb, hold it firmly against the end of the pipette. Then release the bulb and allow it to draw the liquid into the pipette.
Thorough analysis of the graph displayed enough evidence suggesting that an increase in substrate concentration will increase the height of bubbles until it reaches the optimum amount of substrate concentration, resulting in a plateau in the graphs (figure 2). Hence; supported the hypothesis.
We were then to make a base solution of 0.7 M NaOH. In order to standardize
The equation shows how 1 mol of Na2CO3 reacts with 1 mol of H2SO4, so
For part C, the concentration of was determined to be 1.01 mol/L, 0.973 mol/L, and 1.158 mol/L. These results show a relatively closed to the accepted 1.00mol/L of NaOH. The differences of these results are understandable since the concentration of NaOH would changes over time because during the transfer of NaOH powder in part A, it was exposed to the air, thus it could reacts with CO2 in the atmosphere to produce Na2CO3 and water, therefore, changing the concentration of NaOH. Furthermore, the NaOH could also react with the glass thus it wills also reducing its concentration. However, all of the concentration of NaOH that was determine are maximum of 0.158mol/L differences compare to the standard 1.00 mol/L, therefore, it can be concluded that the result are accurate.
The purpose of the experiment is to determine the ID of an unknown diprotic acid by establishing its pKa values. The first phase is to determine the unknown diprotic acid by titration, which is a technique where a solution of known concentration is used to determine the molecular weight. While the second phase involved seeing how much NaOH needed to standardize diprotic acid.
The Iceman is kept in a glass vitrine with regulated temperature of -6℃ and humidity of 98% at glacier-like conditions.
This lab has a typical percentage error usually around 30%-40%. By looking at the percentage error of the calculated enthalpy and entropy, the percentage error is large but small when compared to the typical percentage error that is usually found in the lab. This typically large percentage error shows that there is an error in this lab that causes the experimental data to be less than the theoretical data.
Also, it had provided enough trials to find an accurate volume of Ca(OH)₂. The data collected gave enough information to find the mol of H⁺ ions which equaled the mol of OH⁻ ions because of neutralization. Furthermore, this helped to find the concentration of OH⁻, which helped find the concentration of Ca²⁺ because every 2 OH⁻ ions have one Ca²⁺ ion. That is why the concentration of Ca²⁺ was half of the concentration of OH⁻. With these two concentrations the Ksp was found which was 2.92 x 10^⁻⁵. The actual Ksp is 5.5 x 10^⁻⁶, therefore, there is a big difference. In fact, the percentage error is 431%. This could of happened because of some sources of errors in the
A precipitation reaction can occur when two ionic compounds react and produce an insoluble solid. A precipitate is the result of this reaction. This experiment demonstrates how different compounds, react with each other; specifically relating to the solubility of the compounds involved. The independent variable, will be the changing of the various chemical solutions that were mixed in order to produce different results. Conversely the dependent variable will be the result of the independent variable, these include the precipitates formed, and the changes that can be observed after the experiment has been conducted. The controlled variable will be the measurement of ten droplets per test tube.
During our lab we found our mass percentages for our sample to consist of 20.5% of NH4CL, 48.5% of NaCl, and 25.6% of SiO2. Being that the actual percentages of our mixture were 20% of NH4CL, 50% of NaCl, and 30% of SiO2. After using the percent error equation that was provided to us in class we found there to be a 5.4% error which is relatively accurate.
From looking at the results I can conclude that when the pH was 3 and
In this, the amount of moles in the sodium hydroxide solution after it has been reacted with the aspirin is found using titration, and then compared with the amount of moles it had without the aspirin being added. The difference in moles is the number of moles of sodium hydroxide that reacted with the aspirin, and therefore the number of moles of
borate) and 1.0 g. of sodium hydroxide in 20 mL of warm water. It may
In this experiment three different equations were used and they are the Stoichiometry of Titration Reaction, Converting mL to L, and Calculating the Molarity of NaOH and HCl (Lab Guide pg. 142 and 143).