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Naoh Solution Lab Report

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For Experiment 9, the NaOH solution was standardized with the primary standard acid KHC8H4O4 in order to determine the molar concentration of the NaOH solution. As seen in Table 1, the known mass of KHC8H4O4 was 0.2037 g. The moles of KHC8H4O4, could then be calculated to get 9.97E-04 mol. Based on the balanced equation between the sodium hydroxide solution and primary standard acid, the mole ratio of 1:1 and the molar mass of NaOH was applied to calculate the molar concentration of NaOH, which was 5.42E-02 mol/L (Table 2). Since there was only one trial conducted for this part, the average molar concentration, standard deviation, and relative standard deviation calculations were not applicable. The molar concentration of the standardized …show more content…

The calculated average percent by mass of acetic acid in this experiment was 1.72%. This indicates that there was an error during the experiment, regarding the lower than expected amount of acetic acid in the vinegar sample. A possible error may be the addition of too much NaOH during Experiment 10. This would have caused inaccurate data, since the calculations would account for a higher amount of sodium hydroxide solution added. This can be a source of human error, in which it may be difficult to obtain the same color of pink for the different trails. Also, it was fairly easy to go over the endpoint with the addition of a few extra drops of the NaOH solution, causing a dark pink color. A possible improvement to the lab would be to include more trials during Experiment 9 and 10. Performing multiple trials allows for more consistent and precise results. In reference to Experiment 9, only one trial was conducted, which increases the likelihood of an error within the results. An error in the data from Experiment 9 would cause the calculated molar concentration to be off, thus affecting the calculations for Experiment 10. Another error would be the lack of waiting 10-15 seconds before reading the buret. This step is important since it the solution may have an overall light pink color upon the addition of a few drops of NaOH, in which some may automatically assume that the endpoint has been

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There is one mole of OH- in the solution since NaOH goes to Na+ and OH-. Trial 1: 25.65mL NaOH x 0.100mol/1000mL = 2.57 x 10-3 mol NaOH = 2.57 x 10-3 mol HA = 2.57 x 10-3 mol H+. The equivalent mass is 0.356g Acid / 2.57 x 10-3 mol H+ = 139g/mol H+ Trial 2: 49.57mL NaOH x 0.100 mol / 1000 mL = 4.96 x 10-3 mol NaOH = 4.96 x 10-3 mol HA = 4.96 x 10-3 mol H+. The equivalent mass is 0.644g Acid / 4.96 x 10-3 mol H+ = 130.g/mol H+ Average =

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Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029

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The data from the first experiment concluded that pH level of 4 and pH level 7 had the most activity. Thanks to the chemistry lab the pH levels were created and data was collected. In both experiments, there were procedures that were done differently that could have had an effect on the data. The

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3. In this experiment, the percent yield was 90%. This number implies that there was little error in this experiment. However, this result could have been caused by certain external factors.

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When the sodium hydroxide pellet was added, the test tube became warm and it took awhile to dissolve. The dissolution of NaOH in water resulted in a basic solution (When tested with the red litmus paper, the paper turned a blue color and when tested with the blue litmus paper, the paper remained a blue color). The net ionic equation for the dissolution of NaOH in water is displayed below: NaOH(s) → Na+(aq) + OH-(aq) The addition of magnesium hydroxide to water resulted in a cloudy slurry. There was no change in the color of the blue or the red litmus paper.

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In this lab, human error could have possibly been that the salt wasn’t fully dissolved or even the Kool-Aid wasn’t fully dissolved. To fix this next time, both mixtures can be stirred a little longer. A third human error could have been when putting 20 drops into the test tubes, some drops were bigger than others causing there to be more than mL of mixture in the test tube. At the end of the lab, a red and yellow M&M were used to do a home material test. I dissolved the color off the shell with warm water and placed a dot of each color onto a strip of chromatograph and placed them in two individual

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I am not sure how the lab could be modified to improve the results. For most of the lab, I was not present, so I am not aware of any inherent weaknesses. Overall, the results were pretty similar to the intended conclusions, with the exception of purple and green / .6 M and .8

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The materials used during the experiment included Sodium Hydroxide, Copper Nitrate, a beaker, a pipette, a test tube, a graduated cylinder, a digital scale, gloves, and goggles. The scientist put the goggles and gloves on for safety and the scientist began the experiment. First, the scientist set the graduated cylinder on the digital scale. Then, the scientist pressed zero on the digital scale. While the graduated cylinder is still on the digital scale, the scientist measured ten milliliters of Sodium Hydroxide into the graduated cylinder.

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Some of the solution had multiple nutrients in them and we were able to figure them out by identifying which test the unknown resulted positively. As we enjoyed every part of this lab we recommend to have a side activity where we have to investigate what might be the unknown solutions be using our results, this will make the lab more intresting and challenging. It will influence us to reseach more about chemical makeups of foods in order to determine what the unkowns solutions where based upon the results aquirred from the

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Otherwise the experiment had been done improperly, therefore giving out abnormal results. The source of error lies in the fact that the experiment hadn’t been done several times to find its maximum accuracy. Another error made may be human error, such as mixing up the chemicals on accident or over dosing more than the experiment needed. There are other errors such as systematic errors that could have had an impact on this experiment. Refinement to make this experiment’s data more accurate could be done by cleansing all the equipment that is going to be used thoroughly and use a consistent quality of water.

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Procedure A. Preparation of NaOH solution The molarity of a solution is the ratio of the number of solutes dissolved in a liter of solution. To figure out the needed mass (in grams) of NaOH pellets to be dissolved in a 0.25 L of water, remember that a mole is equivalent to the quotient of mass over the molar mass of the substance. This was used to rearrange the base formula and to derive the mathematical equation of mass in terms of molarity. mass (g) =

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Commercial vinegar, Yamaha brand 0.1 mol/dm3, NaOH soloution Phenolpthalein indicator soloution (50.00 ± 0.5 cm3 ) cm3 burrete (250.00 ± 0.5 cm3) volumetric flask a (250 cm3± 0.5 cm3)

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Practical I: Acid-base equilibrium & pH of solutions Aims/Objectives: 1. To determine the pH range where the indicator changes colour. 2. To identify the suitable indicators for different titrations. 3.

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That caused a new initial reading of NaOH on the burette (see Table1 & 2). The drops were caused because the burette was not tightened enough at the bottom to avoid it from being hard to release the basic solution for titrating the acid. The volume of the acid used for each titration was 25ml. The volume of the solution was then calculated by subtracting the initial volume from the final volume. We then calculated the average volume at each temperature.

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Explain how the molarity of the standard solution (the alkali) was calculated in the experiment (equation explained)- 0.1M of NaOH is required, this equation will be used: Concentration = moles volume This will be rearranged to find the moles needed to carry out the experiment. The concentration of the experiment using NaOH is 0.1M so we just need to rearrange the equation to find the molarity. 0.1 x 0.250 = 0.0250 moles Number of moles =

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