Question

You need to prepare a buffer with the following concentrations: 0.2 M Tris (FW Tris Base = 121.1), 0.2 M NaCl (FW NaCl = 58.44), 0.01 M MgAc (FW MgAc = 214.4) and 5% glycerol. Prepare a table that shows how to prepare 1 L of this solution.

Answer 1

We have a total volume of 1L.

(i) 0.2 M Tris:  Now for Tris, we have molecular weight = 121.1g/mol

Molarity = 0.2M, therefore from the formula given below, we have:

Molarity = moles/volume(in L) .................................................(1)

Therefore, 0.2 = moles / 1L

hence, moles = 0.2 moles

Now, for moles, we know that:

Moles = given weight(g) / molecular weight(g) .......................(2)

0.2 = given weight / 121.1

Therefore, given weight = 24.22 grams

(ii) 0.2 M NaCl: Similarly for NaCl, we know that:

Molarity = moles/volume(L)

0.2 = moles/ 1L

moles = 0.2 moles

Now, as seen above, moles is given by:

moles = given weight(g) / molecular weight(g)

0.2 = given weight / 58.44

Therefore, given weight of NaCl = 11.688 grams

(iii) 0.01 MgAc: For MgAc, we have:

Molarity = moles/volume(L)

Therefore, 0.01 = moles/1L

moles = 0.01 moles of MgAc

Now for moles, again we know:

Moles = given weight / molecular weight

0.01 = given weight / 214.4

Hence weight required of MgAc = 2.144 grams

(iv) 5% Gycerol: Since glycerol is liquid, so the given 5% concentration is volume/volume(v/v).

Now the volume of glycerol required for the given concentration is given by:

(volume of glycerol / volume of solvent) x 100 = 5%

Therefore, (volume of glycerol / 1L) x 100 = 5%

Therefore, volume of glycerol = 0.05L or 50ml

The table is given as below:

Solutions	0.2M Tris(121.1)	0.2M NaCl	0.01M MgAc	5% Glycerol
Concentrations	24.22 grams	11.688 grams	2.144 grams	0.05L or 50ml