Phosphate buffer is a commonly used buffer for biological experiments. The buffe
ID: 1072317 • Letter: P
Question
Phosphate buffer is a commonly used buffer for biological experiments. The buffer is made by combining H2PO4- (pKa = 7.21) and HPO42-.
a) How much KH2PO4 (MW= 136.1 g/mol) and K2HPO4 (174.2 g/mol) would you need to make 300 ml of 50 mM phosphate buffer at pH 7.35?
b) You then add 1 ml of 1 M HCl to the buffer you made in a), what is the final pH after the addition of the acid?
Phosphate buffer is a commonly used buffer for biological experiments. The buffer is made by combining H2PO4 (pKa 7.21) and HPO4 a) How much KH2PO4 MW 136.1 g/mol) and K2HPO4 (174.2 g/mol would you need to make 300 ml of 50 mM phosphate buffer at pH 7.35? b) You then add 1 ml of 1 M HCl to the buffer you made in a), what is the final pH after the addition of the acid? c) You are interested in making IL of 75 mM phosphate buffer pH 7.4. You have a starting solution of 1 M H2PO4 and a 0.5 M solution of HCl. Describe how you would make this buffer.Explanation / Answer
a)
first, identify the equation of buffer
pH = pKa + log(K2HPO4 / KH2PO4)
7.35 = 7.21 + log(ratio)
ratio = 10^(7.35-7.21) = 1.3803
[K2HPO4] = 1.3803* [KH2PO4]
these are concentrations, but we require mol
[K2HPO4] + [KH2PO4] = 50*10^-3 M
solve for concentrations
1.3803* [KH2PO4] + [KH2PO4] = 50*10^-3
2.3803*[KH2PO4] = 50*10^-3
[KH2PO4+ = (50*10^-3)/(2.38) = 0.0210
so
[K2HPO4] = 1.3803* [KH2PO4] = 1.3803*0.0210= 0.02898
for mol:
mol = MV
K2HPO4 = MV = 0.3*0.02898 = 0.008694
K H2PO4 = MV = 0.3*0.0210 = 0.0063
mass:
mol of K2HPO4 = mol*MW = 173.2*0.008694 = 1.5058 g of K2HPO4
mol of KH2PO4 = mol*MV = 0.0063*136.1 = 0.85743 g of KH2PO4
b)
mmol of acid = MV = 1*1 = 1 mmol
mmol of H2PO4 = 0.0063*1000 = 6.3
mmol of H2PO4- after addition = 1+6.3 = 7.3
mmol of H2PO4- = 0.008694*1000 = 8.694
mmol of HPO4- after addition = 8.694 -1 = 7.694
then
pH = pKa + log(HPO4-H2PO4-)
pH = 7.21 + log(7.694/8.694) = 7.156
c)
V = 1 Liter, M = 75*10^-3 pH = 7.4
pH = pKa + log(conjugate/acid)
7.4 = 7.21 + logratio)
ratio = 10^(7.4-7.21) = 1.54881
conjugate = acid*1.54881
conjguate + acid = 75*10^-3
(1.54881+1)*acid = 75*10^-3
acid = (75*10^-3)/(2.54881) = 0.029425
conjugate = acid*1.54881 = 1.54881*0.029425 = 0.0455
Volume of conjugate = mmol/M = 0.0455/1 = 0.0455 L = 45.5 mL
Volume of HCL = 0.029425/0.5 = 58.85 mL
finally, add up water up to V = 1 L