Asked by Raj

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"
Answered by julian
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?
Answered by DrBob222
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.
Answered by DrBob222
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
Answered by Angie
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???
Answered by Amy
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
Answered by spongebob
i don't understand how you got the 0.501.. how do u get the base and the acid
Answered by patrick star
the answer is 6.7 mL after using hints
Answered by rita
Part C
What additional volume of 10.0 would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?
Express your answer in milliliters using two significant figures.
Answered by rita
for part B i got
pH = 5.91 + log (0.1632/0.008324) =6.75
Answered by Linda
Can anyone help me with

Part C
What additional volume of 10.0 would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?
Express your answer in milliliters using two significant figures.

Thanks!
Answered by hope
answer for part C

0.34 ml
Answered by Dr. E
the answer for part C is not 0.34 mL by the way.
Answered by G
Yeah, it is not 0.34, and i cant seem to find the right answer, did anyone find an answer that actually works? help would be greatly appreciated
Answered by Dr. E
The answer is 1.34 mL
Answered by Anonymous
for partc 1.34 is right!
Answered by Anonymous
the answer is not write!!!
Answered by (:
for me I got 2.34 mL
Answered by spell check
it is the answer is not right...not the answer is not "write"
Answered by NICK
correct answer for C is 1.8
Answered by saizad
0.84ml is the correct answer
10M = 0.084 mol/ X L
Xml= 0.084 x1000/10 = 0.84
Answered by Lina
The correct answer is 1.8 mL
Answered by MasteringChemistry
The correct answer is 0.84mL.
Answered by April
In Mastering Chemistry none of those answers are right!
Answered by Um
Mastering Chem changes the individual numbers. Post work.
Answered by Rowena
It is 0.34 ml trust me
Answered by sweetblue
the answer is 2.34 for part C. i got it right to yeh.
Answered by Ruth
For Mastering Chem, its 0.34
Answered by mike
both 0.34 and 0.84 are wrong
Answered by ra
2.34 worked for my mastering chem assignment!!
Answered by :)
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
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