Chemistry

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"

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asked by Raj
  1. 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?

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    posted by julian
  2. 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.

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    posted by DrBob222
  3. 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

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    posted by DrBob222
  4. 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???

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    posted by Angie
  5. 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

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    posted by Amy
  6. i don't understand how you got the 0.501.. how do u get the base and the acid

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  7. the answer is 6.7 mL after using hints

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  8. 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.

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    posted by rita
  9. for part B i got
    pH = 5.91 + log (0.1632/0.008324) =6.75

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    posted by rita
  10. 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!

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    posted by Linda
  11. answer for part C

    0.34 ml

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    posted by hope
  12. the answer for part C is not 0.34 mL by the way.

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    posted by Dr. E
  13. 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

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    posted by G
  14. The answer is 1.34 mL

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    posted by Dr. E
  15. for partc 1.34 is right!

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  16. the answer is not write!!!

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  17. for me I got 2.34 mL

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    posted by (:
  18. it is the answer is not right...not the answer is not "write"

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  19. correct answer for C is 1.8

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    posted by NICK
  20. 0.84ml is the correct answer
    10M = 0.084 mol/ X L
    Xml= 0.084 x1000/10 = 0.84

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    posted by saizad
  21. The correct answer is 1.8 mL

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    posted by Lina
  22. The correct answer is 0.84mL.

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  23. In Mastering Chemistry none of those answers are right!

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    posted by April
  24. Mastering Chem changes the individual numbers. Post work.

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    posted by Um
  25. It is 0.34 ml trust me

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    posted by Rowena
  26. the answer is 2.34 for part C. i got it right to yeh.

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  27. For Mastering Chem, its 0.34

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    posted by Ruth
  28. both 0.34 and 0.84 are wrong

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    posted by mike

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