Buffer problems with additional solutions I know this is different from normal b

Question

Buffer problems with additional solutions

I know this is different from normal buffer problems in that you have to add additional grams of something at the end, just not sure what to do with all the extra available solutions/solutes listed.

Outline how you would prepare 500 mL of the following solutions:

0.01M Tris, pH 7.5, 100 mM NaCl

0.05 M Tris, pH 7.8, 500 mM NaCl, 20 mM CaCl2

100 mM Tris, pH 8.7, 20 mM EDTA

You have the following available:

Tris base (FW = 121.1 g/n, pKa = 8.21)

NaCl (FW = 58.44 g/n)

100 mM CaCl2

6 M HCl

6 M NaOH

500 mM EDTA

Explanation / Answer

We use following conversions:

1 M = 1000 mM

1L = 1000 mL

0.01M Tris, pH 7.5, 100 mM NaCl

The final volume of the solution is 500 mL or 0.500 L

Find out the moles of Tris in this final solution

n Tris = molarity x volume in L

=0.01*0.5=0.005 mol

Mass of Tris = mol x molar mass = 0.005 x 121.1 = 0.6055 g

Mass of NaCl = moles x molar mass = molarity x volume in L x molar mass

=0.1 M x 0.5 L x 58.44 g/mol

= 2.92 g

So the solution is made using 0.61 g Tris and 2.92 g NaCl and fill the volumetric flask upto the 500 mL mark.

0.05 M Tris, pH 7.8, 500 mM NaCl, 20 mM CaCl2

Mass of Tris = 0.05 * 0.5 * 121.1 = 3.0275 g

Mass of NaCl = 0.500 M x 0.5 L x 58.44 g/mol

=14.61 g

We use dilution law for the CaCl2

M1V1=M2V2

Here M1 is the molarity of the stock solution of CaCl2 , V1 is the volume of stock solution required, M2 is the desired molarity and the V2 is the final volume.

V1 = M2V2/M1= 0.020 M x 0.500 L / 0.100 M

=0.100 L = 100 mL

So the mass of Tris = 3.03 g , Mass of NaCl = 14.61 g and the volume of CaCl2 = 100 mL is used to make solution of 500 mL volume.

100 mM Tris, pH 8.7, 20 mM EDTA

Mass of Tris = 0.100*0.500 *121.1=6.1 g

Volume of EDTA = 0.02*0.5/0.500

=0.020 L

= 20.0 mL EDTA

Sot mass of Tris = 6.1 g and volume of EDTA = 20.0 mL and diluted to 500 mL in the volumetric flask

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.