Preparation of a Tris Buffer Part A Express your answer in millilitersusing two
ID: 678900 • Letter: P
Question
Preparation of a Tris BufferPart A Express your answer in millilitersusing two significant figures Preparation of a Tris Buffer In the study of biochemicalprocesses, a common buffering agent is the weak basetrishydroxymethylaminomethane, (HOCH_2)_3CNH_2 , often abbreviated as Tris. At 25 ^degreeC, Tris has a pK_b of 5.91. The hydrochloride of Tris is (HOCH_2)_3CNH_3Cl, 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? Express your answer in millilitersusing two significant figures
Explanation / Answer
PartA: What volume of 10.0M NaOH is needed to prepare a bufferwith a pH of 7.79 using 31.52 g of TrisHCl? TrisHCl is a weak baseand the molecular weight is 157.67 g/mol. I found the solution, itis 6.67 milliliters. The main question is:Part B: "The buffer fromPart A is diluted to 1.00 L. To half of it (500.mL), you add 0.0150mol of hydrogen ions without changing the volume. What is theresulting pH? pKb for the base= 5.91" Here is how you came about the 6.67 mL. pH = pKa + log (Base/acid) If pKb = 5.91, then pKa = 8.09 7.79 = 8.09 + log(B/A) base/acid = 0.501 or base = 0.501(acid) If we call the trisHCl, TH^+ and the base tris, just H, then TH^+ + OH^- ==> H2O + T If we start with 31.52/157.67 = 0.2 mol TH^, and call the NaOHadded as y, then the final moles are TH^+ = 0.2-y T = y and we substitute into the above, base = 0.501*(acid) y moles base = 0.501(0.2-y) y = 0.0667 moles which is 0.00667 L of 10 M NaOH or 6.67 mL.