temperature was obtained by using the MeasureNet. Using a graduated cylinder‚ 50 mL of 0.901M NaOH was added into the cup‚ and the final temperature was obtained. The solution was disposed‚ and the coffee cup calorimeter was rinsed with distilled water. The next experiments were performed. The acid was always added before the base. Experiment 2 consisted of 100 mL of 0.862M HCl and 100 mL of 0.901M NaOH. Experiment 3
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analysis‚ and advanced chemical equilibrium. In solutions that used in the experiment were two species are present reaction between CuSO4 and NaOH. In this method of experiment‚ the total molar concentration of the two solutions are held constant‚ but their mole fractions are varied. In the second experiment‚ it was observed the reaction of acid-base in solution of NaOH with HCl in different condition of volume for each. An observable that is proportional to complex information is plotted against the mole
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[ (2 mole NaOH) / (1 mole CaCl2) ]*[ .005 mol CaCl2 ] = .010 moles NaOH needed to completely react CaCl2‚ only .005 moles available‚ therefore NaOH limits the reaction Theoretical Yield of Ca(OH)2: [ .005 moles NaOH ] * [ (1 mole Ca(OH)2 ) / (2 moles NaOH) ] = .0025 moles Ca(OH)2 formed [ .0025 moles Ca(OH)2 ] * [ (74.0926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] =
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Diagram 13G: N/A Diagram 13C: N/A Observation 13G: Mass of NaOH = 10.836g Table 1:oxalic acid Calculated mass of oxalic acid dihydrate 3.150g Mass of weigh boat (g) .500g Mass of weigh boat + oxalic acid (g) 3.662g Mass of oxalic acid (g) 3.162g Concentration of 100.00ml .2510M Table 2: NaOH Mass of NaOH + weigh boat 11.336g Mass of weigh boat .500g Calculated mass of NaOH 10.836g Moles of NaOH .2709mol Concentration of 500.0ml NaOH .5418M Table 3: HCl Volume of Concentrated HCl 10.00mL Concentration
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(the endpoint). You are allowed three trials‚ and will be graded on accuracy. Lab Questions: 1. HA(aq) + NaOH(aq) NaA(aq) + H2O 2. Trial 1: .1010 M of NaOH = moles of NaOH / 0.0299 L = 0.0030199 mols of NaOH Trial 2: .1010 M of NaOH = moles of NaOH / 0.0227 L = 0.0022927 mols of NaOH Trial 3: .1010 M of NaOH = moles of NaOH / 0.0158 L = 0.0015958 mols of NaOH After we find the moles of NaOH‚ we can use that to find the molar mass of each acid. Trial 1: 0.261g / 0.0030199 = 86.43 mm of acid Trial
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result. Part A: Standardization of 2M NaOH(aq) Data Collection : Weight of substances Uncertainty Weight of watch glass + salt (potassium hydrogen phthalate) w1= 23.4380g ± 0.001g Weight of watch glass ( after emptying the salt)w2 = 21.5430g ± 0.001g Weight of salt (potassium hydrogen phthalate) w1 –w2= (23.4380± 0.001g) – (21.5430g ± .001g) = 1.8950± 0.002g Titration Table - Volume of distilled water Burette Reading NaOH/ml Volume of NaOH solution for neutralization End point
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scale Procedure: Step 1: Standardization of the NaOH solution using Standard 0.1M PHP 1. Take 10ml of the given NaOH solution and dilute with 90ml of water. Fill up the burette with this solution. Take 10ml of 0.1M potassium hydrogen phthalate (PHP) in the conical flask‚ add 1-2 drops of phenolphthalein and titrate with the diluted solution of the NaOH till the solution turns light pink. 2. Calculate the strength of the original NaOH solution. Step 2: Poultry eggshell 1. Carefully
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as well as showing that the moles of the acid in the solution are equal to the moles of the base (Survey of Chemistry lab manual). Using a dilution equation of M1xV1=M2xV2‚ the concentration of the hydrochloric acid can be calculated. M1=NaOH(molarity)‚ V1=NaOH(volume)‚ M2=HCl(molarity)‚ V2=HCl(volume). Procedure: Instructions for laboratory were found on page 91 of Survey of Chemistry lab manual. All calculations were rounded off by 2 decimal places for accuracy. Preparation of the indicator
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Worked solutions to student book questions Chapter 4 Analysing acids and bases Q1. Antacid tablets should normally be chewed before they are swallowed. Why? A1. Antacid tablets are normally chewed to provide a larger surface area for faster reaction with stomach acids. Q2. A laboratory test to determine how much hydrochloric acid is neutralised by a brand of antacid does not give a complete picture of its effectiveness in the stomach. What other factors might be important? A2. Other factors to
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Data Tables: Data Table 1: Quantity of NaOH needed to Neutralize 5 mL of Acetic Acid Brand of Vinegar Used: heinz Label Notes: none | | Initial NaOH reading(Interpolate to 0.1 mL) | Final NaOH reading(Interpolate to 0.1 mL) | Volume of NaOH used | Trial 1 | 9.5 | 1.3 | 8.2 | Trial 2 | 9.6 | 1.4 | 8.2 | Trial 3 | 9.7 | 1.4 | 8.3 | Average volume of NaOH used: | 8.23 | Observations: A. Calculate the average number of mL of NaOH used for the 3 trials and record. 8.23
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