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?   

Answer 1

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