Question
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?
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?
Answers
DrBob222
I don't believe your answer to the first part is correct.
pKa = 14-pKb = 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.
pKa = 14-pKb = 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.
X
wrong not 7.61
X
You may have used the ratio of Tris to TrisH+ from Part A. You need to calculate a new ratio based on the new conditions in Part B. Also, check your value for the pKa.