A) You need to make 900. mL of a buffer that has a pH of 7.50 in which the least
ID: 228789 • Letter: A
Question
A) You need to make 900. mL of a buffer that has a pH of 7.50 in which the least concentrated buffer component has a concentration of 0.20M. How would you make this buffer if the available compounds are K2HP04(s), 3.00M HCl(aq), and 3.00M NaOH(aq)?
B) You need to make 900. mL of a buffer that has a pH of 7.50 in which the least concentrated buffer component has a concentration of 0.20M. How would you make this buffer if the available compounds are KH2P04(s), 3.00M HCl(aq), and 3.00M NaOH(aq)?
H3PO4 Ka1 = 7.5x10^-3 Ka2 = 6.2x10^-8 Ka3 = 4.2x10^-13
Explanation / Answer
Solution:
a. To prepare a buffer HPO4^2-/H2PO4- (base/acid) components
pH of buffer = 7.50
pKa = 7.20
buffer concentration = 0.20 M x 0.900 L = 0.18 mol
0.18 mol = (H2PO4-) + (HPO4^2-) ---- (1)
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
7.50 = 7.20 + log(HPO4^2-/H2PO4-)
(HPO4^2-) = 2(H2PO4-)
feed in (1),
(H2PO4-) + 2(H2PO4-) = 0.18 mol
(H2PO4-) = 0.18 mol/3 = 0.06 mol
Volume 3.0 M HCl needed to add = 0.06 mol x 1000/3 M = 20 ml
(HPO4^2-) = 0.18 - 0.06 = 0.12 mol
mass K2HPO4 needed = 0.12 mol x 174.2 g/mol = 20.904 g
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b. To prepare a buffer HPO4^2-/H2PO4- (base/acid) components
pH of buffer = 7.50
pKa = 7.20
buffer concentration = 0.20 M x 0.900 L = 0.18 mol
0.18 mol = (H2PO4-) + (HPO4^2-) ---- (1)
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
7.50 = 7.20 + log(HPO4^2-/H2PO4-)
(HPO4^2-) = 2(H2PO4-)
feed in (1),
(H2PO4-) + 2(H2PO4-) = 0.18 mol
(H2PO4-) = 0.18 mol/3 = 0.06 mol
mass of KH2PO4 needed = 0.06 mol x 136.086 g/mol = 8.165 g
(HPO4^2-) = 0.18 - 0.06 = 0.12 mol
Volume 3.0 M NaOH needed to add = 0.12 mol x 1000/3 M = 40.0 ml
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