Question

Show how to make:

A) 1.000 L of 0.0500 M potassium phosphate buffer (pH=7.00)

Both salts must to be used. The pK_a is 7.21.

K₂HPO₄•3H₂O = 228.22 g/mol

KH₂PO₄ = 136.086 g/mol

B) 50 mL of 1 mg/ml myoglobin in the buffer.

Myoglobin = 17 kDa (17,000 g/mol)

C) 1.00 L of 6 M guanidinium hydrochloride in the buffer

Guanidinium hydrochloride = 95.53 g/mol

Answer 1

V = 1 L

M = 0.05 Molar

total mol = MV = 1*0.05 = 0.05 mol in solution

K2HPO4•3H2O --> 2K+ + HPO4-2 + 3H2O

KH2PO4 --> K+ + H2PO4-

pH = pKa2 + log(K2HPO4•3H2O/KH2PO4 )

7 = 7.21 + log(ratio)

ratio = 10^(7-7.21) = 0.6165

K2HPO4•3H2O = 0.6165*KH2PO4

mol of K2HPO4•3H2O + mol of KH2PO4 = 0.05

substitute and solve

0.6165*KH2PO4 + 1*KH2PO4 = 0.05

total KH2PO4 = 0.05/ (1.6165) = 0.030 mol of KH2PO4   

mass of KH2PO4 = mol*MW = 0.03*136.086 = 4.08258 g of KH2PO4

for K2HPO4•3H2O

K2HPO4•3H2O = 0.05-0.030 = 0.020 mol

mass = mol*MW = 0.020 *228.22 = 4.5644 g of K2HPO4•3H2O required

B)

V = 50 mL

1 mg/mL

so...

total hemoglobyn mass --< 50*1 = 50 mg of hemoglobyn

add 50 mL and mix

C)

M = mol/V

so..

mol = MV = 1*6 = 6 mol required

mass = mol*MW = 6*95.53 = 573.18 g of Guanidinium hydrochloride