A 2.0 g sample of juice from a fresh lime was diluted with water and filtered to

Question

A 2.0 g sample of juice from a fresh lime was diluted with water and filtered to remove suspended matter. The clear liquid required 40.39 ml of 0.04022 M NaOH for titration to the phenolphthalein end point. Calculate the acidity as % citric acid. (b)The carbon dioxide evolved on heating a 2.407 g sample of steel in a closed system was swept into a flask containing 50 ml of 0.05081 M barium hydroxide. After filtering the solution to remove the precipitated barium carbonate, the remaining base consumed 14.87 ml of 0.1125 M hydrochloric acid for titration to the phenolphthalein end point. Calculate the percentage of carbon in the steel A 2.0 g sample of juice from a fresh lime was diluted with water and filtered to remove suspended matter. The clear liquid required 40.39 ml of 0.04022 M NaOH for titration to the phenolphthalein end point. Calculate the acidity as % citric acid. (b)The carbon dioxide evolved on heating a 2.407 g sample of steel in a closed system was swept into a flask containing 50 ml of 0.05081 M barium hydroxide. After filtering the solution to remove the precipitated barium carbonate, the remaining base consumed 14.87 ml of 0.1125 M hydrochloric acid for titration to the phenolphthalein end point. Calculate the percentage of carbon in the steel A 2.0 g sample of juice from a fresh lime was diluted with water and filtered to remove suspended matter. The clear liquid required 40.39 ml of 0.04022 M NaOH for titration to the phenolphthalein end point. Calculate the acidity as % citric acid. (b)The carbon dioxide evolved on heating a 2.407 g sample of steel in a closed system was swept into a flask containing 50 ml of 0.05081 M barium hydroxide. After filtering the solution to remove the precipitated barium carbonate, the remaining base consumed 14.87 ml of 0.1125 M hydrochloric acid for titration to the phenolphthalein end point. Calculate the percentage of carbon in the steel

Explanation / Answer

(a) sample of juice = 2.0 g

1 mole citric acid need 3 moles of NaOH to neutralize

moles of NaOH needed to neutralize = 0.04022 M x 40.39 ml

                                                           = 1.6245 mmol

moles of citric acid present = 1.6245/3 = 0.5415 mol

mass of citric acid in sample = 0.5415 mol x 192.124 g/mol/1000

                                              = 0.104 g

acidity % citric acid = 0.104 x 100/2.0 = 5.202%

(b) moles of Ba(OH)2 added = 0.05081 M x 50 ml

                                              = 2.5405 mmol

moles of HCl needed to neutralize excess Ba(OH)2 = 0.1125 M x 14.87 ml

                                                                                   = 1.6729 mmol

moles of excess Ba(OH)2 present = 1.6729/2

                                                       = 0.8364 mmol

moles of Ba(OH)2 reacted with steel = 2.5405 - 0.8364

                                                           = 1.7041 mmol

moles of carbon present in steel = 1.7041 mmol

mass of Carbon in steel = 1.7041 mmol x 12 g/mol/1000

                                       = 0.02045 g

Percentage of carbon in steel = 0.02045 x 100/2.407

                                                = 0.85%

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