Chemistry
posted by Alexandra .
In the study of biochemical processes, a common buffering agent is the weak base trishydroxymethylaminomethane, often abbreviated as Tris. At 25 C, Tris has a pKb of 5.91.
The hydrochloride can be abbreviated as Tris HCl.
What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of Tris HCl?
I found this answer to be 6.6755 mL
This is what I can't figure out:
The buffer from Part A is diluted to 1.00L . To half of it (500 mL ), you add 0.0250 mol of hydrogen ions without changing the volume. What is the pH of the final solution?

I don't believe your answer to the first part is correct.
pKa = 14pKb = 8.09
7.79 = 8.09 + log[(base)/(acid)]
B/A = 0.501 or
base = 0.501*acid
If acid is 31.52 g then we start with 31.52/157.6 (check my molar mass) = 0.2 mole acid; therefore,
base = 0.2 mole x (0.501) = 0.1 mole for base.
With 10 M NaOH, we have M = moles/L
10M = 0.1/L or L = 0.1/10 = 0.01 L = 10 mL of the 10 M NaOH (and not the 6.6755 you calculated but check me out on that).
For part B, I would do it this way.
Diluting to 1.00 L and take 1/2 that means we have 0.05 moles acid and 0.05 mole base.
..........TRIS.HCl ==> B... +... H^+
initial:...0.05 mole....0.05 mole....0
add........0.............0......0.0250
change......+0.025....0.025....0.0250
final..... 0.075.......0.025.....0
Then use the HH equation to calculate the final pH. I get
pH = 8.09 + log(0.0250/0.0750) = 7.61
Check my work carefully.