Finding Out How Much Acid There Is In A Solution

Results         Titre (cm3)         Rough Titre          inaugural Accurate          arcminute Accurate         3rd Accurate Start Titre         0.00         0.00         0.00         0.00 terminate Titre         27.15         26.55         26.50         26.45 Titre Result         27.15         26.55         26.50         26.45 Three concordant results, in spite of appearance 0.10cm3 were obtained, I go out in that locationof cogitate an average of these 3 results, using the following polity: 1st Accurate + second Accurate + 3rd Accurate          summate of Accurate results 26.55 + 26.50 + 26.45 = 79.5 = 26.50cm3                  3          3 The percentage error of these titres bay win dow also be compute:          Maximum Result ? nominal Result x light speed = % error median(a) Result 26.55cm3 ? 26.45cm3 x 100 = 0.40% (2.Sig Figs) 26.50cm3 1). Calculating the Concentration of the hindquarters solution.         This needfully to be through with(p) so that the acid slow-wittedness mickle be worked out. The stronger the theme the more acid that will be needed to neutralise it, so the strength of the alkali must(prenominal) be known. A step-by-step method can be used to calculate the concentration of the alkali: Firstly, the telephone number of moles of atomic number 11 anhydrous change needs to be calculated using the following formula: numeral of moles of compound =          spate of compound                   Relative molecular plenitude of Compound          Formula of sodium carbonat e anhydrous = Na2CO3 intensity of compound ! used = 2.65g Relative Molecular Mass of Na2CO3 = (2x23) + (3x16) + 12 =106g mol-1 2.65g                  = 0.0250 moles of Na2CO3 106g mol-1 The molarity of the Na2CO3 solution must wherefore be calculated: A 250cm3 volumetric flask was used and therefore there was 0.0250 moles of Na2CO3 in 250cm3 of water. Because the units of molarity are measured in mol.dm-3, therefore the number of 250cm3 volumetric flasks that make up 1 dm3 must be calculated: 1000 = 4 amounts of 250cm3 in 1 dm3 250 The number of moles of sodium carbonate in 250cm3 is then cypher by 4 to give the number of moles of sodium carbonate in a dm3. Needs sources. Concise. Cant say much. square discussion and stuff.well structured method. If you want to get a secure essay, order it on our website: BestEssayCheap.com

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