This is a distinction for the assignment of Level 3 Applied Science 2A Titration & Colorimetry and should only be used as an example and not to be copied.
I certify that the work submitted for this
assignment is my own. I have referenced any
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declaration is a form of malpractice.
DATE: ________________
,Determining the concentration of Sodium Hydroxide by Titration
Risk Assessment
Risk Nature Precaution
Na 2 CO 3 solution Eye irritant Safety glasses
≥ 0.8 mol/dm
−3 (CLEAPPS, 2017) (CLEAPPS, 2017)
Hazcard 95a
HCl<2.7 mol /dm−3 Skin and eye irritant Safety glasses
(CLEAPPS, 2017) Lab coat
Hazcard 47A Wash skin (if coming
into contact)
(CLEAPPS, 2017)
Methyl Orange Indicator Flammable (including No naked flames
vapours) Safety glasses
Hazcard 32 Eye irritant Small quantities
Toxic if inhaled Ventilation
Skin burns Wash skin (if coming
(CLEAPPS, 2017) into contact
(CLEAPPS, 2017)
Sodium hydroxide Eye damage Eye protection
−3
≥ 0.125 mol/dm Harmful if infected in (CLEAPPS, 2017)
open skin
(CLEAPPS, 2017)
Results – data and calculations
Preparing a sodium carbonate (primary) standard
Mass of sodium carbonate used was 1.43g
M r of Na 2 CO 3
=( 22.990 ×2 ) +12.011+ ( 15.999 ×3 )
= 105.988
Moles of sodium carbonate used.
mass
moles=
Mr
1.43
¿
105.988
¿ 0.013492 mol
Calculating concentration of sodium carbonate used.
Titration 1 results table:
HCl Titre (cm 3) 1 2 3 4
Initial Reading 13.95 20.00 15.40 11.50
Final Reading 39.80 45.90 41.60 37.90
Volume of HCl 25.85 25.90 26.20 26.40
used (cm 3)
Volume of HCl 0.02585 0.0259 0.0262 0.0264
used (dm 3)
Balanced equation for reaction between HCl and Na 2 CO 3:
Na 2 CO 3 +2 HCl → 2 NaCl+ H 2 O+ CO2
Reaction ratio: 1 : 2
Calculating the concentration of HCl :
Na 2 CO 3 concentration is 0.054mol /dm−3
3 ÷1000
Volume of solution made is 25.00 cm →
0.02500 dm 3
3 ÷1000
HCl volume is 26.00 cm →
0.26000 dm
3
number of moles of Na2 CO 3=concentration of Na2 CO3 × volume of Na 2 CO 3
¿ 0.054 × 0.02
¿ 1.35 ×10−3 mol
number of moles of HCl=number of moles of Na 2 CO 3 × reactionratio
−3
¿ 1.35 ×10 ×2
−3
¿ 2.7 ×10 mol
number of moles of HCl
concentration of HCl=
volume of HCl used
−3
2.7 ×10
¿
0.026
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