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Sodium phosphate [1 M sodium phosphate buffer (pH 6.0–7.2)] -click below:
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SALINE SOLUTION [Sodium Chloride, 0.9% (w/v), Isotonic Saline] :
Saline solution is a mixture of sodium chloride salt (NaCl) and water (H2O). Normal saline solution contains 0.9 percent (%) of sodium chloride (salt) in waterl; which is similar to the sodium concentration in the human blood and tears. Saline solution is usually called normal saline, but it's sometimes referred to as physiological or isotonic saline.
More Information on normal saline:
• Normal saline is 0.9% saline. This means that there is 0.9 G of salt (NaCl) per 100 ml of solution, or 9 G per liter. This solution has 154 mEq of Na per liter.
• This solution is used for correction of hypovolemia.
• This solution is used for maintenance IV fluids in all pediatric patients greater than 1 month old due to the risk of hyponatremia with hypotonic IV fluids.
• Useful hint: if you ever have to convert grams of salt (NaCl) into mEq of Na, just remember normal saline: 9G of salt = 154 mEq of Na. You can apply this conversion factor to any other amount.
• 0.9% NaCl solution = Normal Saline= ~ 0.154 M NaCl= ~ 154mM NaCl
Reference: https://www.utmb.edu/pedi_ed/corev2/fluids/Fluids6.html (press ctlr+enter)
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Chromatographic Separation:
Converting From Linear Flow (cm/hour) to Volumetric Flow Rates (mL/min) and Vice Versa
Sepharose CL-4B
Click: www.cytivalifesciences.com/en/us/shop/chromatography/resins/size-exclusion/sepharose-cl-4b-p-05602
Preparation and standardization of 0.1N hydrochloric acid (HCl) by using standard sodium carbonate (Na2CO3)
Click : engineering.tiu.edu.iq/petromining/wp-content/uploads/2018/01/3rd-Lab.pptx
Preparation and standardization of 0.1N hydrochloric acid (HCl) by using standard sodium carbonate (Na2CO3)
· Hydrochloric acid (HCl) is not standard substance; therefore it must be standardized by using standard sodium carbonate (Na2CO3).
·
Preparation of solutions
1- Prepare 0.1N Na2CO3 in 250mL of distilled water
· Na2CO3 is a solid substance. Dissolve 1.325 g Na2CO3 in 250 ml of distilled water to obtain 0.1N Na2CO3.
2- Prepare 0.1N HCl in 250mL of distilled water
HCl is a liquid substance.; Sp.gr = 1.19; %HCl = 37%
Take 2.1 ml of concentrated HCl with a pipet, and dilute to 250 ml with distilled water in a 250ml-volumetric flask to obtain approximately 0.1 N
N= [Sp.Gr x % x 1000 ] / Eq.Wt.
N= [1.19 x 37/100 x 1000] / 36.5
N= 12 eq/L
Eq Wt= FWt of HCL / 1
Eq Wt= 36.5/1 = 36.5 g/moL
No. of milli equivalence before dilutions=No. of milli equivalence after dilutions.
(N1 X V1) conc = (N2 X V2)dilution ; 12 X V1 = 0.1 X 250 mL ; V1 = 0.1 x 250 / 12 = 2.08mL = 2.1 mL.
How do you calculate the molarity of HCl in a 37% mass by volume solution?
To calculate the molarity of HCl in a 37% mass by volume solution, you must first convert the mass percentage to grams by multiplying it by the total volume in mL. Then, divide the number of grams by the molar mass of HCl (36.46 g/mol) to get moles. Finally, divide the moles by the volume in liters to get the molarity.
Reference: https://www.physicsforums.com/threads/solved-calculate-molarity-of-hcl-in-37-mass-by-volume.203183/
%of HCL and their Molarity [% HCL= M Molarity]
37% HCL=12.08 M
36% HCL=11.65M
35% HCL= 11.51M
32% HCL= 10.2M
How to Prepare 1.0 N H2SO4 (1000mL).?
Equivalent weight = (molecular weight)/(number of equivalent moles)
The molecular weight of H2SO4 is 98.07 g/mol.
The number of acid hydrogen in the compounds is 2.
or, Equivalent weight = 98.07g/mol/2 = 49.035
Therefore, the Equivalent weight is 49.035 g/mol.
Calculating the mass of sulfuric acid required for the preparation of the desired volume of solution.
To calculate the amount of sulfuric acid we can use the given formula:
Grams of compound needed = (N desired) x (equivalent mass) x (volume in liters)
Desired N is 1,
The equivalent mass is 49.035 g/mol
Grams of compound needed = (N desired) x (equivalent mass) x (volume in liters)
Grams of compound needed = 1N x 49 x 1 litre = 49 grams
Therefore, the grams of compound required is 49 grams.
A 1 N solution necessitates 49 g of pure sulfuric acid powder (if available) diluted to 1 L. However, the acid is a liquid, and it is not 100 percent pure active sulfuric acid. You must determine the amount of concentrated acid that includes 49 grams of sulfuric acid.
Since acid is in the liquid form we need to calculate the volume of the concentrated acid required using the given formula:
Volume of concentrated acid required = (grams of acid needed) / (percent concentration x specific gravity)
Here,
Concentration percent of sulfuric acid H2SO4 = 97%
The volume of concentrated acid required = 49 grams
Specific gravity = 1.84 g/cm3
The volume of concentrated acid required = (grams of acid needed) / (percent concentration x specific gravity)
Volume of concentrated acid required = 49 grams / (0.97 x 1.84) = 27.5 mL.
If you take 27.5 mL of concentrated sulfuric acid and dilute it to 1 L, you will get 1N H2SO4.
Step 1: Take a volumetric flask of 1 L.
Step 2: Add 500 mL of distilled water to the flask
Step 3: Using a pipette take 27.5 mL of conc. H2SO4 and mix it with 500 mL of distilled water in the flask.
Step 4: Mix it properly.
Step 5: Allow it to cool at room temperature.
Step 6: Adjust the volume to 1 litre adding the distilled water to the solution.
AND
How to Prepare 0.5 N H2SO4 (200mL).?
Preparation of 200ml of 0.5 N H2SO4 from concentrated sulphuric acid (purity 96%, specific gravity 1.84, molecular weight 98)
Note: Normality = n factor x Molarity, where n factor = the number of hydrogen ions in acid (or hydroxide ions in the base).
Therefore, 0.5 N = 2 x Molarity. The Molarity of H2SO4 = 0.25 M
Preparation of 200 ml of 0.5 N H2SO4 :
First, we need to know the molarity of the concentrated H2SO4 before we can prepare the diluted H2SO4.
It can be calculated using the formula:
M = (density x % purity x 1000)/molecular mass of H2SO4.
3. Thus, M = (1.84 x 0.96 x 1000)/98) = 18.02 M
4. Next, we will calculate the volume of the concentrated acid required to prepare the solution by using a dilution formula of C1 x V1 = C2 x V2, where C1 = 0.5 N or 0.25 M H2SO4 solution, V1 = 200 ml of the diluted H2SO4 solution, C2 = 18.02 M H2SO4 and V2 = ? ml
5. Therefore, from the notation, 0.25 M x 200 ml = 18.02 M x V2. On solving for V2 = (0.25 x 200/18.02) = 2.77 ml or approx. 2.8 ml
6. So, you will need 2.8 ml of the concentrated H2SO4 to prepare the diluted acid.
7. Next, you will take 2.8 ml of the concentrated H2SO4 with a 3-ml graduated pipette and add carefully into a 200- ml glass beaker containing some 100 ml of distilled or deionised water.
8. Mix or stir the solution well with the help of a glass rod.
9. Allow the solution to cool down to near the room temperature before doing any further dilution.
10. Transfer the solution to a 200 ml volumetric flask with the help of a small glass funnel.
11. Rinse the beaker, glass rod and funnel with about 50 ml of distilled or de-ionized water.
12. Top up the flask to the mark with more distilled or de-ionized water with the help of a wash bottle.
13. Stopper the flask well, hold onto it and carefully invert the flask several times to mix the solution.
14. This then is your 200 ml of 0.5 N H2SO4 solution
15. For an accurate work, you may need to standardise the prepared acid solution with a primary standard and a suitable indicator. But l’ll leave this piece of work to you.