You have the following stock solutions and solid chemicals in your lab:
1.5M NaCL
0.2M Tris buffer, pH 8.0
10 mg/ml Bovine serum albumin (BSA)
MgCl2 (solid, anhydrous) MW = 95.211 g/mole (95% pure)
How would you make 250 ml of the following solution:
0.2M NaCl
0.05M Tris, pH 8.0
0.25 mg/ml BSA
25mM MgCl2
There are just dilutions. You can use the dilution formula for most if not all.
c1v1 = c2v2
c = concn
V = volume
1.5M x v1 = 0.2M x 250 mL
v1 = 0.2 x 250/1.5 = 33.33.
So you place 33.33 mL of the 1.5M stock solution in a 250 mL volumetric flask, add water to the mark on the flask and mix thoroughly.
Someone will be happy to check your answers for the others if you care to post them.
I am confused on the dillution of the MgCl2 do I have to divide by 95 because it is only 95% pure?
Yes, if you divide the right number by the right number. :-)
Calculate the grams MgCl2 you need, then divide that number by 0.95 to find the final grams to use.
To make each of the desired solutions, you would need to calculate the required quantities of each component and then accurately measure and mix them in the appropriate volumes. Here's how you would do it for each solution:
1. 0.2M NaCl:
- The desired concentration is 0.2M, and you want to make 250 ml of this solution.
- The molar mass of NaCl is 58.44 g/mol.
- Use the formula: C1V1 = C2V2 to calculate the volume of the 1.5M NaCl stock solution needed.
- Substitute the given and desired values into the formula:
(1.5M)(V1) = (0.2M)(250 ml)
- Rearrange and solve for V1:
V1 = (0.2M)(250 ml) / 1.5M
V1 ≈ 33.33 ml
- Measure 33.33 ml of the 1.5M NaCl stock solution and add it to a 250 ml container.
- Make up the remaining volume with distilled water (250 ml - 33.33 ml = 216.67 ml).
- Mix thoroughly to obtain a 0.2M NaCl solution.
2. 0.05M Tris, pH 8.0:
- The desired concentration is 0.05M, and you want to make 250 ml of this solution.
- The pKa of the Tris buffer is pKa = 8.06, so at pH 8.0, the concentration of the protonated form of Tris is roughly equal to the concentration of the deprotonated form.
- Calculate the volume of the 0.2M Tris buffer stock solution needed using the formula C1V1 = C2V2.
- Substitute the values into the formula:
(0.2M)(V1) = (0.05M)(250 ml)
- Rearrange and solve for V1:
V1 = (0.05M)(250 ml) / 0.2M
V1 = 62.5 ml
- Measure 62.5 ml of the 0.2M Tris buffer stock solution and add it to a 250 ml container.
- Make up the remaining volume with distilled water (250 ml - 62.5 ml = 187.5 ml).
- Mix thoroughly to obtain a 0.05M Tris, pH 8.0 solution.
3. 0.25 mg/ml BSA:
- The desired concentration is 0.25 mg/ml, and you want to make 250 ml of this solution.
- Measure 250 ml of distilled water and add it to a 250 ml container.
- Weigh out 62.5 mg of BSA (0.25 mg/ml x 250 ml = 62.5 mg) using a balance.
- Dissolve the 62.5 mg of BSA in the 250 ml of distilled water.
- Mix thoroughly to obtain a 0.25 mg/ml BSA solution.
4. 25mM MgCl2:
- The desired concentration is 25mM, and you want to make 250 ml of this solution.
- The molecular weight of anhydrous MgCl2 is given as 95.211 g/mol.
- Calculate the mass of MgCl2 needed using the formula: mass = molar mass x moles.
- First, calculate the moles of MgCl2 required using the formula: moles = concentration x volume.
- moles = 0.025 mol/L x 0.25 L
- moles = 0.00625 mol
- Then, calculate the mass of MgCl2 needed:
- mass = 0.00625 mol x 95.211 g/mol
- mass ≈ 0.595 g
- Weigh out approximately 0.595 g of anhydrous MgCl2 (since it is given as 95% pure) using a balance.
- Add the weighed amount of MgCl2 to a 250 ml container.
- Make up the remaining volume with distilled water (250 ml - negligible volume of MgCl2 = 250 ml).
- Mix thoroughly to obtain a 25mM MgCl2 solution.
Remember to handle all chemicals safely and use appropriate lab techniques when preparing the solutions.