Question

1. Calculate how to make 500.0 mL of 0.100 M triethanolaime solution.

2. You need to prepare a buffer for biochemistry lab. The required solution is 0.5 M sodium phosphate, pH 7.0. Use the Henderson-Hasselbach equation to calculate the number of moles and grams of monobasic sodium phosphate (NaH₂PO₄) and dibasic sodium phosphate (Na₂HPO₄) necessary to make 1 liter of the solution.

3. Dscribe how you would prepare a 0.1 M glycine buffer of pH 10.0. You have available isoelectric glycine and sodium glycinate. Does it matter if you use sodium glycinate or potassium glycinate for buffer? Why or why not?

4. Describe how you would prepare a 0.20 M Tris-HCl buffer of pH 8.0. The only Tris reagent you have available is Tris base. What other reagent do you need, and how would you use it to prepare the solution?

Please show all work. I appreciate the help!

Answer 1

CALCULATION OF NUMBER OF MOLES OF Na2HPO4 AND Na2HPO4

[ Na2HPO4] = 1.5135 [ NaH2PO4]

This means that if the concentration of NaH2PO4 is 1 mol /L the the concentration of Na2HPO4 is

1.5135 mol /L

Here the concentration is expressed in number of moles /L

To prepare 1L of the buffer solution we have to take 5oomL of Na2HPO4 solution and 500 ml of

NaH2PO4 solution

The Na2HPO4 solution should contain 1.5135/2 mol of Na2HPO4 = 0.7568 mol Na2HPO4 and

The NaH2PO4 solution should contain 1/2 mol of NaH2PO4 = 0.5 mol NaH2PO4

(The ratio of number of moles = 0.7568 mol/0.5 mol = 1.5135)

CALCULATION OF NUMBER OF GRAMS OF Na2HPO4 AND NaH2PO4

Molar mass of Na2HPO4 = 2x23 + 1 + 31+ 4 x16 = 142 g/mol

1 mol Na2HPO4 = 142 g/mol

Therefore 0.7568 mol Na2HPO4 = 0.7568 mol x 142 g/mol = 107.5 g

Molar mass of NaH2PO4 = 1x23 + 2x1 + 31+ 4 x16 = 120 g/mol

1 mol NaH2PO4 = 120 g/mol

Therefore 0.5 mol Na2HPO4 = 0.5 mol x 120 g/mol = 60 .0 g

Number of moles of Na2HPO4 = 0.7568 mol

Number of grams of Na2HPO4 = 107.5 g

Number of moles of NaH2PO4 = 0.5 mol

Number of grams of NaH2PO4 = 60.0 g