Question

2. You weighed 89.9 g of Tris-H+ and put it into a beaker. You dissolved it with 2 L of water to make 2L of a 0.5 M Tris buffer, 7.6 pH. You also weighed 146 g of NaCl and placed into a beaker and added 500 mL of water to make 500mL of 5M NaCl. ~How would you make 1 L of 20 mM Tris, 10mM NaCl from these solutions. How much BSA (bovine serum albumin -- molecular weight 66.5 kDa) would you need to add to the solution to make it 1% BSA?

Answer 1

1) Preparing 1L of 20 mM Tris by dilution:

Solution of tris-HCl prepared=2L of 0.5 M

Let V1=?

M1=0.5

V2=1L

M2=20mM=20*10^-3M

Using equation, M1*V1=M2*V2

V1=M2*V2/M1=(20*10^-3M)*1L/0.5M=0.045L=0.045L*1000ml/L=45ml

So,45 ml of 0.5M tris H is to be taken in a 1L volumetric flask and distilled water to be added to make it 1L.The resulting solution will be 20mM

2)Preparing 10mM NaCl from 500ml of 5M Nacl

Let M1=5M

V1=?

M2=10mM

V2=1L

Using equation, M1*V1=M2*V2

V1=M2*V2/M1=(10*10^-3M)*1L/5M=0.002L=0.002L*1000ml/L=2ml

So,2 ml of 5M NaCl stock solution is to be taken in a 1L volumetric flask and distilled water to be added to make it 1L.The resulting solution will be 10mM

3)preparing 1% (w/V) BSA (bovine serum albumin ) solution: weigh out 1 g BSA in a graduated cylinder ,add water upto 100 ml mark

This solution has 1g/100 ml or 1%(w/V) concentration