Preparation of the LAB reagents

Preparation of Reagents
Md. Imran Nur Manik ; Lecturer Page 0
Department of Pharmacy; Northern University Bangladesh (NUB).
Prepared By
Md. Imran Nur Manik
Lecturer
Department of Pharmacy
Northern University Bangladesh

How to prepare:
For the Qualitative analysis (identification) of anions and cations from inorganic salt
solutions.
1. For sulphate (SO42
¯) identification
a. Dilute (6M) Hydrochloric acid (HCl)
b. 0.1M Barium chloride (BaCl2) or Barium nitrate {Ba(NO3)2} solution
2. For halide ions identification
a. Dilute nitric acid (2M)
b. 5% Silver nitrate (AgNO3) solution
c. Dilute ammonium hydroxide solution (2M)
d. Concentrated ammonium hydroxide solution (~9M)
3. For nitrate ion identification
a. Concentrated Hydrochloric acid (37%) {÷Conc. Sulphuric acid }
b. 5% Ferrous sulphate (FeSO4) solution
4. For Acetate ion identification
a. Dilute Hydrochloric acid
b. 0.2M Ferric chloride (FeCl3) solution
A. For group I cations
a. Dilute hydrochloric acid (6M)
b. Ammonia solution (9M)
c. Potassium chromate (1M)
B. For group II cations
a. Nitric acid (6M)
b. Stannous chloride (0.1M)
c. Concentrated Ammonia solution (9M)
d. Dilute Ammonia solution (2M)
e. Dilute Sodium hydroxide solution (2M)
C. For group III cations
a. Ammonia solution (9M)
b. Ammonia solution (2M)
c. Potassium ferricyanide (K3[Fe(CN)6]) [250 mg in 10 mL]
d. Potassium thiocyanate [250 mg in 10 mL]
e. Sodium hydroxide (2M)
f. Sodium hydroxide (10M)
Md.
Imran
Nur
manik

Reagents and their preparations including their shelf life:
01. Dilute (6M) Hydrochloric Acid (25mL) solution. Purity 37% and Density 1.19g/mL:
Method of Preparation:
Take 12.435 mL hydrochloric acid in 100 mL volumetric flask.Now completely dissolve it with a small
volume of D.W. Finally adjust the volume up to mark by adding distilled water Q.S.
Calculation: We know, W=SMV÷1000
Here, strength of HCl, S=6M, Volume,V=25 mL,
Molecular weight of HCl, M= 1+35.5=36.5 gm/mol,
Required amount W=?
Now, W={(636.525)÷1000} gm =5.475 gm
Since, ρ=M÷V; Here, ρ=1.9 gm/mL and M=5.475gm, Thus, V=(5.475÷1.19) mL =4.6008 mL.
On the other hand, purity of HCl=37%,
That means, 37 mL is present in=100 mL
∴ 4.6008 mL is present in= {(1004.6008)÷37} mL =12.435 mL of HCl
Shelf Life: 03(Three) months from the date of preparation.
Question: Calculate the volume of concentrated HCl, having a density of 1181 g/L and containing 36.5 %
HCl by weight, needed to prepare 500 mL of a 0.1 mol/L hydrochloric acid solution?
=======================
To prepare 500 mL of a 0.1 mol/L solution of hydrochloric acid we will need to dilute 4.22917 mL of
36.5 % HCl to a final volume of 500 mL with deionized (distilled) water.
SOURCE:
-------------
Name: Hydrochloric acid
Formula: HCl
Formula weight: 36.461 g/mol
CAS Number: 7647-01-0
NFPA: Health 3, Flammability 0, Instability 1, Special
CALCULATION:
-------------
The key concept is that the amount of solute in the desired solution must be equal to the amount of
solute in the source solution. Remember, the concentration is the amount of a solute divided by the
volume of the solution.
Before we make any calculations we have to make sure that we only use one system and one unit of
measurement. DO NOT mix measurement systems and units.
Desired solution:
V0 = (500 mL×1÷1000)L = 0.5 L
c0 = 0.1 mol/L
Source solutions:
w1 = 36.5 % = (36.5× 1)÷100 = 0.365
d1 = 1181 g/L
--------------------------------
Then, we determine the concentration of the source (stock) solution
c1 = d(HCl)× w(HCl) / M(HCl)
c1 =( 1181 g/L× 0.365 )÷36.4609 g/mol
c1 = 11.8227 mol/L
Md.
Imran
Nur
manik

Since the total amount of solute is the same before and after dilution, the volume of stock
solution needed is
V1 = (V0 × c0 )÷ c1
V1 = (0.5 L × 0.1 mol/L )÷ 11.8227 mol/L
V1 = 0.00422917 L
To convert the result into a desired unit we will use dimensional analysis again
V(36.5 % HCl) = 0.00422917 L= (0.00422917×1000) mL = 4.22917 mL
Method:02- Calculation:
We know, W=SMV÷1000
Here, strength of HCl, S=0.1M,
Volume, V=500 mL,
Molecular weight of HCl, M= 36.4609 g/mol
Required amount W=?
Now, W={(0.136.4609500)÷1000} gm =1.823045 gm
As the purity of the solution is 36.5% (w/w) thus , 36.5 gram is present in 100 gm
Since, the density is 1181 g/L=1.181 g/mL
Thus , 1.181gm=1 mL So, 100 gm= {(100×1)÷1.181} mL= 84.67400508 mL
That means, 36.5 gm is present in=84.67400508 mL
∴ 1.823045 gm is present in= {(1004.6008)÷37} mL =4.229165011 mL of HCl
Method-03: Calculation:
Here, strength of HCl, S=0.1M,
Volume, V=500 mL,
Molecular weight of HCl, M= 36.4609 g/mol
Required amount W=?
Now, W={(0.136.4609500)÷1000} gm =1.823045 gm
As the purity of the solution is 36.5% (w/w) thus , 36.5 gram is present in 100 gm
Since, the density is 1181 g/L=1.181 g/mL
Thus, 1.181gm=1 mL So, 100 gm= {(100×1)÷1.181} mL= 84.67400508 mL
We know that, molarity is the moles of solute per litre of solution.
Here the moles of HCl is (36.5÷36.5) = 1 mole
And the volume is (84.67400508÷1000)L=0.084674005 L
Thus the molarity =(1÷0.084674005)=11.8099999M
Md.
Imran
Nur
manik

Now
We know that,
M1V1 = M2V2
11.8099999
5001.0
M
VM
VOr,
1
22
1


= 4.23370029 mL
Here, Volume of diluted HCl Solution (Final volume), V2 = 500 mL
Strength of diluted solution (Final Strength), M2 = 0.1 M
Volume of Stock Solution (Volume to be taken), V1 = ? mL
Strength of Stock solution, M1 = 11.8099999M
02. 0.1M Barium chloride solution (20 mL). Purity 99%
Completely dissolve 0.493494 gm of BaCl2 in 100 mL volumetric flask with a small volume of DW.
Finally adjust the volume up to the calibration mark by adding distilled water Q.S.
Calculation:
Here, strength of BaCl2, S=0.1M,
Volume, V=20 mL,
Molecular weight of BaCl2, M= 244.28gm/mol,
Required amount W=?
Now, W={(0.1244.2820)÷1000} gm=0.48856 gm
On the other hand, purity of BaCl2=99%,
That means, 99 mg is present in=100 mg
∴ 0.48856 gm is present in= {(1000.48856)÷99} gm =0.493494 mg of BaCl2
03. Dilute (2M) Nitric Acid solution (25 mL). Purity 65% and Density 1.39g/mL:
Take 3.4869 mL of Nitric acid in 100 mL volumetric flask. Now completely dissolve it with a small
volume of D.W. Finally adjust the volume up to the calibration mark by adding distilled water Q.S.
Calculation:
Here, strength of HNO3, S=2M,
Volume, V=25 mL,
Molecular weight of HNO3, M= 63.01 gm/mol,
Required amount W=?
Now, W={(263.0125)÷1000} gm =3.1505 gm
Since, ρ=M÷V; Here, ρ=1.39 gm/mL and M=3.1505 gm, so, V= (3.1505÷1.39) mL=2.2665 mL.
On the other hand, purity of HNO3=65%,
That means, 65 mL is present in=100mL
∴ 2.2665 mL is present in= {(1002.2665)÷65} mL =3.4869 mL of HNO3
Md.
Imran
Nur
manik

04. Conc. (6M) Nitric Acid solution (25 mL). Purity 65% and Density 1.39g/mL:
Take 10.4609 mL of Nitric acid and completely dissolve it in 25 mL volumetric flask with a small volume
of DW. Finally adjust the volume up to the 25 mL calibration mark by adding distilled water Q.S.
Calculation:
Here, strength of HNO3, S=6M,
Volume, V=25 mL,
Molecular weight of HNO3, M= 63.01 gm/mol,
Required amount W=?
Now, W={(663.0125)÷1000} gm = 9.4515 gm
On the other hand, purity of HNO3=65%,
That means, 65 mL is present in=100mL
∴ 6.7996 mL is present in= {(1006.7996)÷65} mL =10.4609 mL of HNO3
Question 01: Method 02 and Method 03 will be applicable if the reagent percentage is
an indicative of percentage by Weight (W/W).
05. 5% Silver nitrate (AgNO3) (25 mL). Purity 99.5%:
Take 1.25628 gm of AgNO3 and completely dissolve it in 25 mL volumetric flask with a small volume of
DW. Finally adjust the volume up to 25 mL calibration mark by adding distilled water Q.S.
Calculation:
100 mL contains= 5 gm AgNO3
∴ 25 mL contain= {(525)÷100} gm= 1.25 gm AgNO3
On the other hand, purity of AgNO3 =99.5%,
That means, 99.5 gm is present in=100 gm
∴ 1.25 gm is present in= {(1001.25)÷99.5} gm=1.25628 gm of AgNO3
06. Dilute (2M) Ammonia (aq.) solution (25 mL). Purity 25% and Density 0.9g/mL:
Take 3.78464 mL of NH3 completely dissolves it in 25 mL volumetric flask with a small volume of DW.
Finally adjust the volume up to 25 mL calibration mark by adding distilled water Q.S.
Md.
Imran
Nur
manik

Calculation:
Here, strength of NH3, S=2M,
Volume, V=25 mL,
Molecular weight of NH3, M= 17.031 gm/mol,
Required amount W=?
Now, W={(217.03125)÷1000} gm =0.85155 gm
On the other hand, purity of NH3=25%,
That means, 25mL is present in=100mL
∴ 0.94616 mL is present in= {(1000.94616)÷25} mL =3.78464 mL of NH3
07. Conc. (9M) Ammonia (aq.) solution (25 mL). Purity 25% and Density=0.9g/mL:
Take 17.031 mL of NH3 and completely dissolve it in 25 mL volumetric flask with a small volume of DW.
Finally adjust the volume up to 25 mL calibration mark by adding distilled water Q.S.
Calculation:
Here, strength of NH3, S=9M,
Volume, V=25 mL,
Molecular weight of NH3, M= 17.031 gm/mol,
Required amount W=?
Now, W={(917.03125)÷1000} gm =3.831975 gm
Since, ρ=M÷V; Here, ρ=0.9 gm/mL and M=3.831975 gm, so, V= (3.831975 ÷0.9) mL=4.25775 mL.
On the other hand, purity of NH3=25%,
That means, 25mL is present in=100mL
∴ 4.25775 mL is present in= {(1004.25775)÷25} mL =17.031 mL of NH3
08. 5% Ferrous sulphate solution (25 mL). Purity 99.5%:
Take 2.29919 gm of FeSO4.7H2O and completely dissolve it in 100 mL volumetric flask with a small
volume of DW. Finally adjust the volume up to 25 mL by adding distilled water Q.S.
Md.
Imran
Nur
manik

Calculation:
100 mL contains= 5gm FeSO4
∴ 25 mL contain= {(525)÷100} gm= 1.25 gm FeSO4
But the FeSO4 has 7 molecules of H2O with its structure and its molecular weight is 278.02 gm/mole
Therefore, 151.908 gm FeSO4 is present in =278.02 gm FeSO4.7H2O
∴ 1.25 gm FeSO4 ,, ,, ,, = {(278.021.25)÷151.908} gm =2.2877 gm FeSO4.7H2O
On the other hand, purity of FeSO4.7H2O =99.5%,
That means, 99.5 gm is present in=100 gm
∴ 2.2877 gm is present in= {(1002.2877)÷99.5} gm=2.29919 gm of FeSO4.7H2O
09. 0.2M Ferric chloride solution (25 mL). Purity 98%:
Method of preparation:
Take 0.8276 gm of FeCl3 and completely dissolve it in 25 mL volumetric flask with a small volume of
Calculation:
Here, strength of FeCl3, S=0.2M,
Volume, V=25 mL,
Molecular weight of FeCl3, M= 162.21 gm/mol,
Required amount W=?
Now, W={(0.2162.2125)÷1000} gm=0.81105 gm
On the other hand, purity of FeCl3=98%,
That means, 98 gm is present in=100 gm
∴ 0.81105 gm is present in= {(1000.81105)÷98} gm=0.8276 gm of FeCl3
10. 1M Potassium chromate (K2CrO4) solution (25 mL). Purity 99%:
Take 4.904 gm of K2CrO4 and completely dissolve it in 25 mL volumetric flask with a small volume of
Calculation:
Here, strength of K2CrO4, S=1M,
Volume, V=25 mL,
Molecular weight of K2CrO4, M= 194.20 gm/mol,
Required amount W=?
Now, W={(1194.2025)÷1000} gm=4.855 gm
Md.
Imran
Nur
manik

On the other hand, purity of K2CrO4=99%,
∴ 4.855 gm is present in= {(1004.855)÷99} gm=4.904 gm of K2CrO4
11. 0.1M Stannous chloride (SnCl2.2H2O) solution (25 mL). Purity 97%:
Take 0.58152 gm of SnCl2.2H2O and completely dissolve it in 25 mL volumetric flask with a small volume
of DW. Finally adjust the volume up to 25 mL calibration mark by adding distilled water Q.S.
Calculation:
Here, strength of SnCl2.2H2O, S=0.1M,
Volume, V=25 mL,
Molecular weight of SnCl2.2H2O, M= 225.63 gm/mol,
Required amount W=?
Now, W={(0.1225.6325)÷1000} gm=0.564075 gm
On the other hand, purity of SnCl2.2H2O =97%,
∴ 0.564075 gm is present in= {(1000.564075)÷97} gm=0.58152 gm of SnCl2.2H2O
Water of hydration may be taken into account as per Question no. 08 ( Although the
result of both will be the same.)
12. 2M Sodium Hydroxide (NaOH) solution (25 mL). Purity 98%:
Take 2.0408 gm of NaOH and completely dissolve it in 25 mL volumetric flask with a small volume of
Calculation:
Here, strength of NaOH, S=2M,
Volume, V=25 mL,
Molecular weight of NaOH, M= 40 gm/mol,
Required amount W=?
Now, W={(24025)÷1000} gm=2 gm
On the other hand, purity of NaOH =98%,
∴ 2 gm is present in= {(1002)÷98} gm=2.0408 gm of NaOH
Md.
Imran
Nur
manik

13. 10M Sodium Hydroxide (NaOH) solution (25 mL). Purity 98%:
Take 10.2041 gm of NaOH and completely dissolve it in 25 mL volumetric flask with a small volume of
Calculation:
Here, strength of NaOH, S=10M,
Volume, V=25 mL,
Molecular weight of NaOH, M= 40 gm/mol,
Required amount W=?
Now, W={(24025)÷1000} gm=10 gm
On the other hand, purity of NaOH =98%,
∴ 2 gm is present in= {(10010)÷98} gm=10.2041 gm of NaOH
14. Potassium Ferricyanide (K3[Fe(CN)6]) [250 mg in 10 mL]
Shelf Life: N/A. Should be instantly prepared.
15. Potassium thiocyanate (KSCN) [250 mg in 10 mL]
Shelf Life: N/A. Should be instantly prepared.
Md.
Imran
Nur
manik

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