Dilutions, calculations etc Flashcards
A purchased standard solution of sodium hydroxide had a concentration of 1.0 mol/dm3. How would you prepare 100 cm3 of a 0.1 mol/dm3 solution to do a titration of an acid?
M1 x V1 =M2 x V2 1 x V1=0.1 x 100V1 = 0.1 X 100 / 1 =10so 10 cm3 (ml) of standard soln + 100-10=90cm3 waterSimples !!!orThe required concentration is 1/10th of the original solution.To make 1dm3 (1000 cm3) of the diluted solution you would take 100 cm3 of the original solution and mix with 900 cm3 of water.The total volume is 1dm3 but only 1/10th as much sodium hydroxide in this diluted solution, so the concentration is 1/10th, 0.1 mol/dm3.To make only 100 cm3 of the diluted solution you would dilute 10cm3 by mixing it with 90 cm3 of water.
Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?Why might there be inaccuracies?
M1 x V1 =M2 x V22.0 x V1= 0.5 x 250V1 =0.5 x 250/2 = 62.5 cm3 of stock + (250-62.5)=187.5 water ORGiven a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0 mol/dm3.To make 1dm3 (1000 cm3) of a 0.5 mol/dm3 solution you would take 250 cm3 of the stock solution and add 750 cm3 of water.Therefore to make only 250 cm3 of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm3 of the stock solution plus 187.5 cm3 of pure water.This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.It can be done much more accurately by using a burette or a pipette to measure out the stock solution directly into a 250 cm3 graduated-volumetric flask.Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time!
Given a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?Why might there be inaccuracies?
M1 x V1 =M2 x V2M=”molarity” or concentrationV=volume2.0 x V1= 0.5 x 250V1 =0.5 x 250/2 = 62.5 cm3 of stock + (250-62.5)=187.5 water ORGiven a stock solution of sodium chloride of 2.0 mol/dm3, how would you prepare 250cm3 of a 0.5 mol/dm3 solution?The required 0.5 mol/dm3 concentration is 1/4 of the original concentration of 2.0 mol/dm3.To make 1dm3 (1000 cm3) of a 0.5 mol/dm3 solution you would take 250 cm3 of the stock solution and add 750 cm3 of water.Therefore to make only 250 cm3 of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm3 of the stock solution plus 187.5 cm3 of pure water.This can be done, but rather inaccurately, using measuring cylinders and stirring to mix the two liquids in a beaker.It can be done much more accurately by using a burette or a pipette to measure out the stock solution directly into a 250 cm3 graduated-volumetric flask.Topping up the flask to the calibration mark (meniscus should rest on it). Then putting on the stopper and thoroughly mixing it by carefully shaking the flask holding the stopper on at the same time!Measuring errorsSpillageInaccurate apparatus
You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm-3 (conc. ammonia! ~18M!)What volume of the conc. ammonia is needed to make up 1dm3 of 1.0 molar ammonia solution?
M1 x V1 =M2 x V218x V1 = 1 x 1000V1 = 1000/18=60 CM3 OF CONC AMMONIA ADD (1000-60) =940 CM3 OF WATER TO THE 60 CM3 OF CONC NH3 TO MAKE 1DM3 9 ie a litre of 1.0 molar amonnia solnMethod (i) using simple ratio argument.The conc. ammonia must be diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.Therefore you need (1.0/17.9) x 1000 cm3 = 55.9 cm3 of the conc. ammonia.If the 55.9 cm3 of conc. ammonia is diluted to 1000 cm3 (1 dm3) you will have a 1.0 mol dm-3 (1M) solution.Method (ii) using molar concentration equation - a much better method that suits any kind of dilution calculation involving molarity.molarity = mol / volume (dm3), therefore mol = molarity x volume in dm3Therefore you need 1.0 x 1.0 = 1.0 moles of ammonia to make 1 dm3 of 1.0M dilute ammonia.Volume = mol / molarityVolume of conc. ammonia needed = 1.0 / 17.9 = 0.0559 dm3 (55.9 cm3) of the conc. ammonia is required,and, if this is diluted to 1 dm3, it will give you a 1.0 mol dm-3 dilute ammonia solution.
You are given a stock solution of concentrated ammonia with a concentration of 17.9 mol dm-3 (conc. ammonia! ~18M!)What volume of conc. ammonia is needed to make 5 dm3 of a 1.5 molar solution?
M1 x V1 =M2 x V218 x V1= 1.5 X 5000V1= 1.5 X 5000 /18= 417 ML OF STOCK AMMONIA + 5000-417=4583 CM3 WATERTO DILUTEOR 407 CM3 STOCK TOPPED UP TO 5000 CM3 WITN WATER
