In the study of biochemical processes, a common buffering agent is the weak base
ID: 869029 • Letter: I
Question
In the study of biochemical processes, a common buffering agent is the weak base trishydroxymethylaminomethane, (HOCH2)3CNH2, often abbreviated as Tris. At 25 ?C, Tris has a pKbof 5.91. The hydrochloride of Tris is (HOCH2)3CNH3Cl, which can be abbreviated as TrisHCl. Question 1.) What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl? ANSWER IS 6.67 Question 2.) The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0100mol of hydrogen ions without changing the volume. What is the pH of the final solution? Answer IS 7.58 Question 3.) What additional volume of 10.0 M HCl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?
Explanation / Answer
TrisHCl + NaOH ------> Tris + H2O + Na+ + Cl-
According to Henderson equation,
pOH = pKb + log (Salt / Base)
In our case, it becomes,
pOH = pKb + log (TrisHCl / Tris)
Given, pH required = 7.79. => pOH = 14 - 7.79 = 6.21 , pKb = 5.91
=> 6.21 = 5.91 + log (TrisHCl / Tris)
=> log (TrisHCl / Tris) = 0.3
=> (TrisHCl / Tris) = 1.995
Mass of TrisHCl = 31.52 g
Molar mass of TrisHCl = 157.6
=> Moles of TrisHCl = 31.52 / 157.6 = 0.2 moles
Suppose X moles of TrisHCl reacts with NaOH
TrisHCl + NaOH ------> Tris + H2O + Na+ + Cl-
0.2 - X.........................X
We calculated above that, (TrisHCl / Tris) = 1.995
=> 0.2 - X / X = 1.995
=> X = 0.06678 moles
Therefore, Moles of TrisHCl reacted = 0.06678
According to the stoichiometry of the reaction 1 mole of NaOH reacts with 1 mole of TrisHCl
Moles of NaOH required = 0.06678
Moles = Molarity x Volume (L)
=> 0.06678 = 10 x V
=> V = 6.67 x 10^-3 L = 6.67 mL
2) pH after adding H+
pOH = pKb + log (Salt + X / Base - X)
X = H+ added = 0.01 mol
Salt = 0.2 - 0.06678 = 0.1332 moles in 1 L
In 0.5 L salt = 0.1332 / 2 = 0.0666 moles
Base = 0.06678 /2 = 0.03339 moles
=> pOH = 5.91 + log (0.0666 + 0.01 / 0.03339 - 0.01)
=> pOH = 6.42
pH = 14 - pOH = 14 - 6.42 = 7.58
3)
To exhaust the remaining capacity of the buffer, all of Tris should react with HCl
Moles of Tris left 0.03339 - 0.01 = 0.02339
10 x V = 0.02339
=> V = 0.002339 L = 2.339 mL of HCl needed