msc part 1 analytical unit 2 chemometrics

1. A SEMINAR ON
BY
Miss. NEHA MILIND DHANSEKAR
MSc I ANALYTICAL CHEMISTRY
THE INSTITUTE OF SCIENCE
15, MADAM CAMA ROAD

2. Concentration of solution
Definition : The concentration is defined as amount of solute
dissolved in a specific amount of solvent.
Concentration of solution expressed in different ways :
1. Percentage by mass : [ w/w ]
2. Percentage by volume : [ v/v ]
3. Molefraction
4. Molarity
5. Molality
6. Normality
7. Formality , etc.

3. Percentage by mass & volume
Percentage by mass [w/w] =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
×100
Percentage by volume [ v/v ]=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100
Percentage by weight by volume
[w/v ]=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
(𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛)
× 100

4. Molarity & molality
Molarity : no of moles of solute present in 1 dm³ volume of
solution.
Molarity =
𝑛𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑑𝑚³
or
1g = 1000ml = 1 molar solution
molality : no of moles of solutes dissolved in 1 kg of
solvent .
Molality =
𝑛𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒𝑠
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑘𝑔

5. Mole concept

7. Concept of millimole (mmol)=
A millimole is one-
thousandth of a mole.
Moles=
Millimoles=
Millimole=Molarity X milliliter or mmol=M . mL
g/mol=mg/mmol
Weight of solute(g)
Formula weight(g/mol)
Weight of solute(mg)
Formula weight(mg/mmol)

8. Calculation conversions

9. Concept of ppm/ppb (wt/vol)
ppm is an abbreviation for "parts per million" and
it also can be expressed
as milligrams per liter (mg/L).
Parts per billion (ppb) is the number of units of
mass of a contaminant per 1000 million units of
total mass. Also µg/L or micrograms per liter
ppm(wt/vol)= X106 ppmWeight of solute(g)
Volume of solution [ L ]
Weight of solute(g)
Volume of sample(mL)
ppb(wt/vol)= ×10⁹

10. Concept of ppm/ppb (wt/wt)
ppm(wt/wt)= X106 ppm
ppb(wt/wt)= X109 ppb
Weight of solute(g)
Weight of sample(g)
Weight of solute(g)
Weight of sample(g)

11. Common units for expressing trace
concentration
Units Abbreviation Wt/Wt Wt/Vol Vol/Vol
Parts per
million
ppm mg/kg
µg/g
mg/L
µg/mL
µL/L
nL/mL
Parts per
billion
ppb µg/kg
ng/g
µg/L
ng/mL
nL/L
pL/mL
1 ppm=1000 ppb, pL=picoliter=10-12 Liter, ng=nanogram=10-
9g,

12. Dilutions:-Preparing the right
concentration
Mstockx mLstock=Mdilutedx mLdiluted
M1.V1=M2.V2

13. Problems on ppm,ppb and millimole
Q1-How many grams and millimoles of Potassium Sulphate are
present in its 50 cm3 of 0.2M solution?
Ans-
1mol____________1000mL____________________1M
X mol_____________50ml_____________________0.2M
X=(0.2x50)/1000=0.01mol of Potassium Sulphate=0.01x1000
mmol=10 mmol
Weight of Potassium Sulphate= no.of mol x
M.W.=0.01x174.25= 1.7425g

14. Q2-Using AgNO3, how will you prepare 1.2 dm3 of a solution
containing 25 ppm of Ag+?
Given- Required conc of Ag+=25ppm Final vol of
solution=1.2L=1200 mL
1 mg_______________1L__________________1ppm
Xmg________________1.2L________________25ppm
X= 1.2x25=30 mg=0.03 g of Ag+
Moles of Ag+=0.03/107.86=2.78x10-4 mol
1 mole of Ag+= 1 mole of AgNO3,
2.78x10-4 mole of Ag+= 2.78x10-4mole of AgNO3,
Weight of AgNO3=no. Of moles of AgNO3x M.W.=2.78x10-
4x169.87=0.0472 g in 1.2L

15. Q3-How much of potassium nitrate must be weighed to prepare a
solution containing 7 mg of K+ per liter?
Given- Conc of K+ = 7mg/L=7ppm=7x10-3g/L
Weight of K+=7x10-3g moles of K+=7x10-3/39=1.795x10-4
moles
1 mole of K+=1 mole of potassium nitrate
Hence 1.795x10-4 mole of K+=1.795x10-4 mole of
potassium nitrate
Weight of potassium nitrate required=no. Of moles of
potassium nitrate x M.W. of potassium nitrate
weight=1.795x10-4x101.1=0.0181g

16. Q4-Calculate the number of millimoles of chloride ions are
required to react with AgNO3 to produce 4.5g of AgCl. what will be
the mass of CaCl2 required to generate the required amount of Cl-
.
AgNO3 + Cl- = AgCl + NO3
-
1mol 1mol 1mol
169.87g 35.5g 143.3g
Xg 4.5g
X=(35.5x4.5)/ 143.5 =1.113g=1113mg , millimoles of Cl-
=1113/35.5=31.35mmol=0.031mol
CaCl2 = Ca2+ + 2Cl- Weight of CaCl2= n
of moles x M.W
1mol 1mol 2mol Weight of
CaCl2=0.0155x 110.98=1.72g
0.031/2=0.0155mol 0.031mol

17. Q5-A solution (100mL) containing 325 ppm K+ is analyzed by precipitating it as the
tetraphenyl borate K(C6H5)4B, dissolving the precipitate in acetone solution, and
measuring the concentration of tetraphenyl borate ion,(C6H5)4B- , in the solution. If
the acetone solution volume is 250 mL,what is the concentration of tetraphenyl
borate in ppm?
1 mg_____________1000 mL_______________1ppm M.W. of (C6H5)4B-
=319.2 g/mol
Xmg_____________100mL________________325ppm M.W of K+=39g/mol
X=325x100/1000=32.5mg of K+=0.032g Moles of K+.=0.032/39=8.2x10-4mol
K(C6H5)4B = K+ hence number of moles of K(C6H5)4B=8.2x10-
4mol
Since K(C6H5)4B=(C6H5)4B- hence no. of moles of (C6H5)4B- =8.2x10-4mol
Weight of (C6H5)4B-= no. of mo lx M.W.=8.2x10-4 mol x 319.2=0.262g =262mg
1mg_________________1000 mL________________1 ppm
262mg________________250mL________________Yppm
Y=(262x250)/1000=65.5 ppm

18. Q-6 A solution contains 6.0 pmol Na2SO4 in 250 mL,how many ppm
sodium does it contain?
Ans- Given
Number of moles of Na2SO4=6.0x10-12mol=6.0x10-9.mmol
=2x6.0x10-9 mmol of Na+=12x10-9 mmol of Na+
M.W.of Na=23 mg/mmol
Volume of solution=250mL=0.25L
Weight of Na+=mmol of Na+ x M.W of Na+ =12x10-9x23=
2.76x10-7mg
1mg___________ _____1L_________1ppm
0.000000276mg_______0.25L________xppm
x=0.000000276/0.25=1.104x10-6ppm

19. Q8- Calculate the molar concentration of 1 ppm solution of AgNO3.
Given- conc. of solution=1ppm
1mg____________1L________________1ppm
Weight of AgNO3=1mg=1x10-3g
M.W. of AgNO3.=169.87g/mol
169.87g____________1L_______________1M
1x10-3g_____________1L_______________xM
x=1x10-3/169.87=5.88x10-6M

20. Q10-You want to prepare 1 L of a solution containing 1 ppm Fe2+.
How many gram ferrous ammonium sulphate FeSO4
.(NH4)2SO4.6H2O must be dissolved and diluted in 1L?what would
be the molarity of this solution.
Given-Conc of Fe2+=1ppm
M.W. of Fe2+=55.84g/mol Volume of solution=1L
1 mole of FeSO4 .(NH4)2SO4.6H2O= 1 mole of Fe2+
1mg________________1L___________________1ppm
Hence weight of Fe2+=1mg=0.001g moles of
Fe2+=0.001/55.84=1.79x10-5mol
Hence moles of FeSO4 .(NH4)2SO4.6H2O=1.79x10-5mol
1mol_____________1L___________________1M
1.79x10-5mole_______1L___________________yM
y=1.79x10-5M

21. Q11-A 0.456 g sample of an ore is analyzed for chromium and
found to contain 0.560 mg Cr2O3 express the conc. Of Cr2O3 in the
sample as ppm and ppb.
Given- Weight of ore=0.456g=0.000456kg
Weight of Cr2O3 =0.560mg
1mg________________1kg__________________1ppm
0.560mg_____________0.000456kg___________xppm
x=0.560/0.000456=1228ppm

22. Q12- How many grams NaCl should be weighed out to prepare 1L
of 100 ppm solution of Na+?
Given- Volume of solution=1L Conc. Of NaCl=100ppm
Weight of NaCl=? M.W of NaCl=58g/mol M.W. of
Na+=23g/mol
1mg___________________1L_______________________1
ppm
xmg____________________1L_______________________
100ppm
X=100mg=0.1g of Na+ moles of Na+=0.1/23=0.00434mol
1 mol of NaCl=1 mol of Na+ hence moles of
NaCl=0.00434mole

23. Q13- You have a 250 ppm solution of K+ as KCl. You wish to
prepare from this a 0.00100 M solution of Cl-. How many milliliters
must be diluted to 1L?
Given- Conc of K+=250 ppm
1mg____________1L______________1ppm
xmg____________1L_________________250ppm
x=250mg=0.250g of K+
number of moles of K+=0.250/39=6.41x10-3mol
1 mol of K+=1 mol of KCl,1 mol of KCl=1 mol of Cl-
6.41x10-3mol of K+ = 6.41x10-3 mol of Cl-
1mol_________1L_______________1M
6.41x10-3mol____1L________________yM
y=6.41x10-3M
M1V1=M2V2
Given-Final vol. M2=1L=1000mL
Final conc. M2=0.001M
Initial conc.M1=6.41x10-3M
V1=?
Now,
6.41x10-3xV(mL)=0.001x1000
V=156.25mL

24. Q14-One liter of 500 ppm solution of KClO3 contains how many
grams K+?
Given- Conc. of KClO3 =500 ppm weight of K+=?
M.W. of K+=39g/mol, M.W. of KClO3 =122.55g/mol
1mg___________________1L__________________________1ppm
xmg____________________1L_________________________500ppm
x=500mg=0.5g of KClO3 moles of KClO3=0.5/122.55=0.00408 mol
1mol of KClO3= 1mol of K+ hence 0.00408 mol of
KClO3=0.00408mol 0f K+
Weight of K+=mole x M.W.=0.00408x39=0.16g of K+

25. Q15- A 12.5mL portion of a solution is diluted to 500mL,and its
molarity is determined to be 0.125. Whats is the molarity of the
original solution?
Given- let final conc. M2=0.125M Initial conc. M1=?
Final vol. V2=500mL initial volume V1,=12.5mL
We know that
Mstockx mLstock=Mdilutedx mLdiluted
M1.V1=M2.V2
M1x12.5=0.125x500
M1=5M

26. Q19-What weight of Pb(NO3)2,will have to be dissolved in 1 L of
water to prepare a 100 ppb Pb2+ solution?
Given-Conc.Of Pb2+ solution=100 ppb=100microgram/L=10-
4g/L
M.W. of Pb2+=207.2g/mol M.W. of Pb(NO3)2,=331.2g/mol
Moles of Pb2+=10-4g/207.2(g/mol)=4.82x10-7 mol
1 mol of Pb2+=1 mol of Pb(NO3)2,
Hence 4.82x10-7 mol of Pb2+= 4.82x10-7 mol of Pb(NO3)2
Weight of Pb(NO3)2,= no. of mole x M.W=4.82x10-
7x331.2=1.59x10-4g

27. Q20-You are required to prepare 100mL working std. Solution of
1.00x10-4 glucose from stock solution of 0.01M?Calculate the
volume of stock solution required for dilution?
Given- let final conc. M2=1.0x10-4M Initial conc. M1=0.01M
Final vol. V2=100 mL initial volume V1,=? mL
We know that
Mstockx mLstock=Mdilutedx mLdiluted
M1.V1=M2.V2
V1x0.01=1.0x10-4x100
V1=1mL

28. References
Hsc text book ncert and state board
Greenfacts.org
Scienceing.com
https://courses.lumenlearning.com/boundless-
chemistry/chapter/solution-concentration/
Analytical Chemistry by-G.D.Cristin

29. Thank you
I express my sincere thanks to all my professors, seniors
and my parents.