Calculate how you will prepare 01M buffer solutions ranging

Calculate how you will prepare 0.1M buffer solutions ranging from pH 1-12 in increments of 1 pH unit. NaOH and HCl droppers will be available to make adjustments to the final pH.

a. pH 1, 2, 3. Phosphoric acid (14.8M, pKa 2.15, 98g/mol) and monobasic potassium phosphate (136.09g/mol) are available.
b. pH 4, 5, 6. Acetic acid (17.48M, pKa 4.75, 60.05g/mol) and potassium acetate (98.15g/mol) are available.
c. pH 7, 8. Dibasic potassium phosphate (pKa 6.82, 174.18g/mol) and monobasic potassium phosphate (136.09g/mol) are available.
d. pH 9 Tris Base (pKa 8.08, 121.14g/mol) and Tris-HCl (151.59g/mol) are available.
e. pH 10, 11, 12. CAPS (3-(Cyclohexylamino)-1-propanesulfonic acid, 221.32g/mol, pKa 10.4) is available.

Solution

Answer:

Prepare buffer

a. pH = 1

H3PO4/H2PO4- (acid/base)

pKa = 2.15

let total volume of buffer = 1 L

Using Hendersen-Hasselbalck equation,

pH = pKa + log(H2PO4-/H3PO4)

1 = 2.15 + log(H2PO4-/H3PO4)

(H2PO4-) = 0.071(H3PO4)

we have,

(H3PO4) + (H2PO4-) = 0.1 M x 1 L = 0.1 mol

(H3PO4) + 0.071(H3PO4) = 0.1 mol

(H3PO4) mol = 0.1/1.071 = 0.093 mol

volume of 14.8 M H3PO4 to be taken = 0.1 mol x 1000/14.8 M = 6.76 ml    

mol (H2PO4-) = 0.1 - 0.093 = 0.007 mol

mass of KH2PO4 = 0.007 mol x 136.09 g/mol = 0.953 g

similarly, pH 2 and 3 buffers can be prepared

b. pH = 4

acetic acid/acetate (acid/base)

pKa = 4.75

Using Hendersen-Hasselbalck equation,

pH = pKa + log(acetate/acetic acid)

1 = 2.15 + log(acetate/acetic acid)

(acetate) = 0.071(acetic acid)

we have,

(acetic acid) + (acetate) = 0.1 M x 1 L = 0.1 mol

(acetic acid) + 0.071(acetic acid) = 0.1 mol

(acetic acid) mol = 0.1/1.071 = 0.093 mol

volume of 14.8 M acetic acid to be taken = 0.1 mol x 1000/17.48 M = 5.72 ml    

mol (acetate) = 0.1 - 0.093 = 0.007 mol

mass of potassium acetate = 0.007 mol x 98.15 g/mol = 0.69 g

similarly pH 5 and 6 buffers can be prepared.

c. pH = 7

pKa = 6.82

H2PO4-/HPO4^2- (acid/base)

Using Hendersen-Hasselbalck equation,

pH = pKa + log(HPO4^2-/H2PO4-)

1 = 2.15 + log(HPO4^2-/H2PO4-)

(HPO4^2-) = 0.071(H2PO4-)

we have,

(H2PO4-) + (HPO4^2-) = 0.1 M x 1 L = 0.1 mol

(H2PO4-) + 0.071(H2PO4-) = 0.1 mol

(H2PO4-) mol = 0.1/1.071 = 0.093 mol

mass of KH2PO4 = 0.1 mol x 136.09 g/mol = 1.3609 g    

mol (HPO4^2-) = 0.1 - 0.093 = 0.007 mol

mass K2HPO4 taken = 0.007 mol x 174.18 g/mol = 1.22 g

similarly pH 8 buffers can be prepared.