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'A buffer is prepared by mixing a 100.omL of a 0.100 M NH3 solution with a 0.200M solution of NH4CL solution and making the total volume up to 1.000L of water. What is the volume of NH4Cl solution required to achieve a buffer at ph=9.5? Ka of NH4=5.6x10^-10. The textbook says the answer is 28.1mL.

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4 answers
  1. mols NH3 = M x L = ?
    mols NH4Cl = M x L = ?
    pH = pKa + log(b/a)
    9.5-9.25 + log (10/0./2x)
    Solve for x and convert to mL.

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  2. thank you that definitely helps. i just have one question, line 5:
    9.5-9.25 + log (10/0./2x) is that supposed to read
    9.5-9.25 + log (10/0.1/2x)? just a bit confused.

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  3. thank you that definitely helps. i just have one question, line 5:
    9.5-9.25 + log (10/0./2x) is that supposed to read
    9.5-9.25 + log (10/0.1/2x)? just a bit confused., and also why is it divided by 2x? i have an exam for this on friday morning.

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  4. Sorry about that. I made several typos.
    First mols = M x L (which is right BUT I used millimoles in the equation that follows).
    100 mL x 0.100M = 10 millimols NH3.
    x mL x 0.2M = 0.2x millimols NH4^+. Using millimoles, it should read
    pH = pKa + log(base/acid)
    9.5 = 9.25 + log(10/0.2x)
    1.778 = 10/0.2x
    0.3557x = 10
    x = 10/0.3557 = 28.1 mL. Thank you for bringing this to my attention. Usually I check these fairly closely; I was in a hurry to leave today and didn't. I hope this clears things for you.

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