2. Aqueous 50.5 % (w/w) sodium hydroxide has a density of 1.53 g cm-3 (a) Show t
ID: 1026593 • Letter: 2
Question
2. Aqueous 50.5 % (w/w) sodium hydroxide has a density of 1.53 g cm-3 (a) Show that 50.5 % (w/w) sodium hydroxide is equivalent to a molar concentration of 19.3 mol dm3 4 marks] (b) Determine the mole fraction of sodium hydroxide in this solution. mar (c) Calculate the volume of 19.3 mol dm 3 aqueous sodium hydroxide required to prepare 100 cm* of 0.200 mol dm3 aqueous sodium hydroxide. Describe fully how you would carry out this dilution in the laboratory [4 marks] (d) The amount of phosphoric acid (H3PO4) in a sample of a cola drink was determined by titration with 0.200 mol dm NaOH (i) Write a balanced chemical equation to show the complete neutralisation of phosphoric acid by sodium hydroxide [2 marks] if 4 marks] [3 marks] (ii) Determine the concentration of phosphoric acid in the cola drink in mol dm 25.0 cm' portions of the drink required a mean titre of 7.68 cm* of0.200 mol dm NaOH for complete neutralisation. (iii) Determine the concentration of phosphoric acid in the cola drink in ppm.Explanation / Answer
a) weight of NaOH in 1cm3=50.5 % of 1.53g/cm3=0.505*1.53 g/cm3=0.773 g/cm3
mole of NaOH in 1cm3=0.773g/MW [,MW of NaOH=39.997 g/mol]
mole of NaOH in 1cm3=0.773g/39.997 g/mol=0.0193mol
molar concentration=0.0193mol/cm3=0.0193mol/10^-3L=19.3 mol/L=19.3M
b)mass of water in 1cm3=1.53g -0.773 g=0.757 g
mol of water in 1cm3=0.757g/MW =0.757g/18.015g/mol=0.0420 mol[,MW of water=18.015g/mol]
mol fracton of NaOH =mol of NaOH in 1cm3/total mol in 1cm3=0.0193mol/(0.0193mol+0.0420mol)=0.315
c)Using equation ,C1*V1=C2*V2
C1=19.3mol/dm3
V1=?
C2=0.2mol/dm3
V2=100cm3
V1=C2V2/C1=(0.2mol/dm3)*(100cm3)/(19.3mol/dm3)=1.036 cm3
d)i)H3PO4(aq)+3NaOH --->Na3PO4 +3H2O
b)V1=25.0cm3
V2=7.68cm3
C2=0.200 mol/dm3
mol H3PO4/mol NaOH=1/3
3*mol H3PO4=mol NaOH
or,3*C1V1=C2V2
C1=C2V2/3V1=(0.2mol/dm3)(7.68cm3)/3(25.0cm3)=0.0205 mol/dm3
C(H3PO4)=0.0205 mol/dm3
iii) MW of phosphoric acid=97.994 g/mol
C(H3PO4)=0.0205 mol/dm3=(0.0205 mol/dm3)(97.994g/mol)=2.007 g/dm3=2007 mg/dm3=2007mg/L=2007ppm
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