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 pKb of 5.91. The hydrochloride of Tris is (HOCH2)3CNH3Cl, which can be abbreviated as TrisHCl. Part A 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? . CORRECT6.7 mL Part B This procedure represents only one of many ways to prepare a buffer. The exact method chosen depends on the reagents and glassware available as well as the accuracy required for the pH. The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0300 mol of hydrogen ions without changing the volume. What is the pH of the final solution? Express your answer numerically to two decimal places.CORRECT pH = 6.64 Notice that even though a significant amount of the strong acid was added to the buffer, the pH did not change drastically since the capacity of the buffer had not been exceeded. Part C 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? In other words, how much more of this HCl solution is required to consume the remaining Tris in the buffer? PLEASE HELP

Answer 1

Number of moles of Tris-HCl = 31.52/157.59 = 0.2 moles

Moles of NaOH = 10.0 M x 0.0067 L = 0.067 moles

The ICE table can be setup as,

                                     Tris-HCl    +    OH-         Tris    +    H₂O    +    Cl-

Initial                              0.2             0.067

Change                        -0.067          -0.067        +0.067

Final                             0.133            0                0.067

Since the buffer is diluted to 1L, number of moles will be reduced to half

Moles of Tris-HCl = 0.133/2 = 0.0665

Moles of Tris = 0.067/2 =0.0335

To this solution, 0.03 moles of H⁺ are added without changing the volume,

Therefore,

Moles of Tris-HCl = 0.0665 + 0.03 = 0.0965

Moles of Tris = 0.0335 -0.03 = 0.0035

The capacity of the buffer will be exhausted when the 0.0035 moles of Tris are consumed

Moles of HCl required = 0.0035 moles

Volume of HCl required = (0.0035 moles/10 M)= 0.00035 L = 0.35 mL