stoichiometric amounts of the acid and base have combined is the equivalence point. An example of this is shown in the equation: HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l). The number of moles is given by knowing the exact concentration and volume added of the titrant. The latter‚ in turn‚ is related by stoichiometry to the number of moles of acid initially present in the unknown. To detect the equivalence point‚ indicators are usually added to acid-base titrations. The point at which the indicator changes color
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N/A N/A Moles 6.75x10^-4 .032 Theoretical: 3.29x10^-4 N/A Final Mass N/A N/A Theoretical: 99mg (.099g) N/A Actual: Melting pt (or bp) 81-83 C N/A Literature: 312-315 C N/A Experimental: Percent Yield Calculation: Limiting Reagent: 1 mg= .001 g‚ 100 mg= .1 g vanillin Moles vanillin = .1g/152.15 g/mol= 6.57x10^-4 moles Density H2O2= 1.45 g/mL Mass H2O2= 1.45 g/mL x .75mL = 1.088g Moles H2O2= 1.088g/34.015 g/mol =.032 moles Limiting
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of the other solution. The following steps tell how to calculate the unknown concentration: 1. Write the balanced equation for the reaction. From the coefficients‚ determine how many moles of acid reacts with 1 mole of base (or vice versa). Use the coefficients to form a mole ratio. 2. If the mole ratio is 1:1‚ the following relationship can be used to calculate the unknown concentration: Ma x Va = Mb x Vb Where Ma =
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=-mH2O×cH2O×△TH2O As the water has gained the heat produced by the reaction‚ the heat change of reaction is negative when the temperature of the water increases. As the heat change observed depends on the amount of reaction‚ for example the number of moles of fuel burned‚ enthalpy change reaction are usually expressed in kJmol-1. Chemicals Name of chemicals concentration Volume Water used for diluting Materials Name of materials Measuring cylinder Measuring cylinder Beaker Stirring rod Thermometer Balance
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|28.00 |34.00 | Volume of acid = 50 cm3 ± 0.25 Acid concentration = 1.0 M Volume of base = 50cm3 ± 0.25 Base concentration = 1.0 M Number of mole of solution used = [pic] = [pic] = 0.05 mole ± 0.5 % Data Processing: Calculation: Temperature Change‚ [pic] = Final temperature – Average temperature of Acid and Base |Reaction |Average Temperature of Acid and Base‚ °C|Final
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Based on the molecular weight calculated which was 151 g/mole and the melting point of 108°C obtained from the experiment‚ the identity of the product was 4-ethylbenzoic acid. When comparing this to the correct identity of benzoic acid‚ its molecular weight was 122 g/mole and its melting point was 122°C. Therefore‚ there were errors that occurred within the experiment. One source of error was from the separation
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The Ka and Molar Mass of a Monoprotic Weak Acid Chemistry Lab 152 Professor: James Giles November 7‚ 2012 Abstract: The purpose of this experiment was to determine the pKa‚ Ka‚ and molar mass of an unknown acid (#14). The pKa was found to be 3.88‚ the Ka was found to be 1.318 x 10 -4‚ and the molar mass was found to be 171.9 g/mol. Introduction Acids differ considerable as to their strength. The difference between weak and
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reacted‚ The concentration of sodium thiosulphate. Then we can know the mass of vitamin C = No. of moles of vitamin C × Molar mass of vitamin C = No. of moles of I2 reacted with vitamin C ×10 × Molar mass of vitamin C = ( No. of moles of I2 produced - No. of moles of I2 reacted with Na2S2O3 ) ×10 × Molar mass of vitamin C = ( No. of moles of KIO3 used ×3 - No. of moles Na2S2O3 ÷2) ×10 × Molar mass of vitamin C Results: Part A: Standardization of sodium thiosulphate solution
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chemical compound which is in violet liquid form; potassium dichromate is an orange inorganic chemical reagent and methylene blue is a chemical compound commonly used for staining because of its color blue. These three substances have 158 g/mole‚ 294 g/mole and 374 g/mole‚ respectively. A jelly-like substance‚ agar-water gel in a petri dish was used as a medium for diffusion since water and air‚ as a medium‚ are not controllable
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the molar mass of the metal that was present in the sample. The concentration of metal that was present in the TAP water was solved by the following calculations: 150 mL EDTA * (1 L/ 1000 mL) * (0.005 M) = 0.00075 moles of metal (0.00075 moles/ 0.050 L) = 0.015 moles/ L 0.00075 moles of metal * (32.192 g/ 1 mol) * (1000 mg/ 1 g) = 24.144 mg The concentration of metal that was present in the DI water was solved by the following calculations: 75 mL EDTA * (1L/1000 mL) * 0.005 M = 0.000375
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