PartA: What volume of 10.0M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl? TrisHCl is a weak base and the molecular weight is 157.67 g/mol. I found the solution, it is 6.67 milliliters. The main question is:Part B: "The buffer from Part A is diluted to 1.00 L. To half of it (500.mL), you add 0.0150 mol of hydrogen ions without changing the volume. What is the resulting pH? pKb for the base= 5.91"

0.84ml is the correct answer

10M = 0.084 mol/ X L
Xml= 0.084 x1000/10 = 0.84

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 NaOH added 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.

Now we do the same kind of thing for the part B.
We have a mixture of T and TH^+.
We are adding H^+ to it.
T + H^+ ==> TH^+
We have 0.06676 moles T in the diluted solution. We are adding 0.150 moles H^+, so what are the final moles?
T = 0.006676 - 0.0150 = ??
TH^+ = 0.1332 + 0.0150 = ??
pH = pKa + log (base/acid)
Plug and chug. I get an answer approximately 7.5 pH but you need to go through it, watch the significant figures, and don't estimate as I did. Check my thinking. Check my work.

it is the answer is not right...not the answer is not "write"

correct answer for C is 1.8

the answer is 2.34 for part C. i got it right to yeh.

A) What volume of 10.0 mol L−1 NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl?

Answer: 6.7 mL

B) The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0100 mol of hydrogen ions without changing the volume. What is the pH of the final solution?

Answer: pH = 7.58

C) What additional volume of 10.0 mol L−1 HCl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?

Answer: 2.3 mL

lol i m also looking for dis answer..

can u plz help me with
Methyl violet is an indicator that changes color over a range from PH = 0 to PH= 1.6. What is K(a) of methyl violet?

I think I worked this last night.

Halfway between 0 and 1.6 is about pKa 0.8 so you can get Ka from that. Watch the significant figures. By the way, the pKa I found on the Internet was pKa = 0.8

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???

it's 2.34 mL.

Find your number of moles from your ice chart, then use C = n/V, deviated to V = n/C x 1000mL = 2.34 mL