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 = 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.

wrong not 7.61

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.

To find the pH of the final solution, we need to consider the effect of adding 0.0250 mol of hydrogen ions to the buffer solution.

Here are the steps to calculate the pH of the final solution:

Step 1: Calculate the concentration of the Tris HCl in the solution.
First, convert the mass of Tris HCl given (31.52 g) to moles using its molar mass (1 mole = 121.14 g/mol).

Moles of Tris HCl = 31.52 g / 121.14 g/mol = 0.2604 mol

Next, divide the moles of Tris HCl by the total volume of the buffer solution (500 mL or 0.5 L) to find the concentration of Tris HCl.

Concentration of Tris HCl = 0.2604 mol / 0.5 L = 0.5208 M

Step 2: Determine the moles of Tris and Tris HCl in the solution.

Since Tris acts as a weak base, it reacts with the added hydrogen ions (H+) to form TrisH+.

Tris + H+ ⇌ TrisH+

The moles of Tris and Tris HCl are the same because Tris HCl dissociates in water to supply Tris.

For every mole of added hydrogen ions (H+), one mole of Tris reacts to form TrisH+.

Therefore, the moles of Tris in the solution after adding the hydrogen ions is:

Moles of Tris = Moles of Tris HCl = 0.2604 mol

Step 3: Calculate the concentration of Tris and TrisH+ in the final solution.

The final volume of the solution is 1.00 L, and we added 0.0250 mol of hydrogen ions. Therefore, the moles of Tris and TrisH+ in the final solution can be calculated as follows:

Moles of Tris = Moles of Tris HCl = 0.2604 mol
Moles of TrisH+ = Moles of added hydrogen ions = 0.0250 mol

We know that the moles of Tris are the same as the moles of Tris HCl because they react in a 1:1 ratio.

Therefore, the concentration of Tris and TrisH+ can be calculated as follows:

Concentration of Tris = Moles of Tris / Volume of solution = 0.2604 mol / 1.00 L = 0.2604 M
Concentration of TrisH+ = Moles of TrisH+ / Volume of solution = 0.0250 mol / 1.00 L = 0.0250 M

Step 4: Calculate the pOH using the Kb value.

The Kb value for Tris is given as pKb = 5.91. Kb can be calculated as 10^(-pKb).

Kb = 10^(-5.91) = 8.2285 x 10^(-6)

Next, use the concentration of TrisH+ and Kb to calculate the pOH:

pOH = -log10([TrisH+]) = -log10(0.0250) = 1.60

Step 5: Calculate the pH using the equation: pH + pOH = 14.

pH = 14 - pOH = 14 - 1.60 = 12.40

Therefore, the pH of the final solution is 12.40.

Note: The concentration of Tris in the buffer remains the same after adding hydrogen ions because Tris reacts with the added H+ to form TrisH+, which does not contribute to changes in the concentration of Tris.