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PH of solution

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Calculate the pH of the solution

    a) 23.0 ml of 6.29 M HCl is added to 160.0 ml of distilled water
    2. Relevant equations

    pH = -log [H]

    Kw= [H][OH]

    Kw = 1 x 10^-14


    3. The attempt at a solution

    For part a) I found the no.of moles Of HCl as 0.14467mol and from this I found the no.of moles of Hydrogen ions as 0.14467 mol
    then I found the concentration of Hydrogen ions as 0.9041875
    then pH as 0.04374

    but the answer is coming worng .... and I also don't know the correct answer because this is an online hw.


    Please someone tell where I'm wrong .... any kind of guide or help.


    :tongue: :blushing:
     
    Last edited: Feb 20, 2007
  2. jcsd
  3. Feb 20, 2007 #2

    symbolipoint

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    Assume complete ionic dissociation of hydrogen chloride, since in water it is a strong acid.

    Hydronium ion molarity will be approximately:
    (0.023 liter)*(6.29 moles/liter)/(0.023 Liter + 0160 liter)

    Find pH of that expression's value.
     
  4. Feb 20, 2007 #3
    thanks a lot !!!
     
  5. Feb 20, 2007 #4
    Can you please help me with 1 more question .... please !!!

    Question:

    What is the pH of a 0.620 M solution of CH_3NH_3+Br- if the pKb of CH_3NH_2 is 10.62 ?

    Relevant Equation:

    pKa = - log Ka

    pKb = - log Kb

    Attempt:

    I tried finding Kb .... but I think I used the wrong formula ..... please guide me if you can. I'll be really thankful 2 you.
     
  6. Feb 20, 2007 #5

    symbolipoint

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    For your ammonium bromide question, I have no quick answer but I can discuss a few details about the problem:

    You seem to be representing monomethyl ammonium bromide salt, and methylamine. Negative logarithm of the unneutralized amine's Kb is 10.62, so find the antilog of 10.62 and find the negative of it (did I express this properly?). The bromide salt is like an acid, so you might actually want to use the Ka for this acidic salt, but not certain yet which will be more convenient.

    To help you any further, I would need maybe 2 or 3 days of review - been much time since last studied this stuff. A convenient expression for Ka or Kb can be constructed leading to a quadratic equation, but I would be mixed up if I do not restudy this carefully. Last time I restudied this was just for equilibria of weak acids.

    Anyway, -log(Kb) = 10.62, find Kb.
    Kw = Ka*Kb, find Ka if you need it.
    Maybe at this point you could treat the ammonium salt as a weak acid (not sure if this is the best approach) and use detailed expression for Ka and solve for pH.
     
  7. Feb 21, 2007 #6

    symbolipoint

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    (Excuse the slighly poor typesetting in this solution attempt)

    The solution being of just the neutralized amine, should be treated as
    an acid. Check the dissociation of the cation:

    MeNH3 + == MeNH2 + H+
    Methyl ammonium cation in equilibrium with Methylamine and hydrogen ion(or hydronium).

    Since -log(Kb)=10.62, Ka=4.169*10^(-4), so we should treat the salt
    as an acid.

    Ka = (H)*(MeNH2)/(F-H), for which F=formality of the methylammonium,
    and for which the expression (F - H) is the MOLARITY of the methylammonium.

    Note that H=MeNH2 in the above Ka formula, so you can simplify into
    Ka=(H)*(H)/(F-H)
    Notice QUADRATIC EQUATION, so just find H.

    H=molarity of hydronium, so you just need to find -log(H).
     
  8. Feb 21, 2007 #7

    I'll also try to find about this problem.
     
  9. Feb 21, 2007 #8

    chemisttree

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    Kw = Ka * Kb
    Kw = 1X10^-14
    You don't know the Kb yet... 10.62 is actually the pKa of methylamine (CRC Handbook). You need to calculate the Kb from that knowing that pKa + pKb = 14. You can do that yourself.

    The methylammonium bromide is a salt of a weak base and a strong acid. The strong acid would be HBr.
    Therefore a solution of MeNH3Br will be acidic because of the hydrolysis of the ammonium ion:

    MeNH3+ + HOH <-------> MeNH3OH + H+ (1)

    Ka = Kw/Kb = 1X10^-14/Kb(MeNH2) (2)

    You also know that the equilibrium constant of (1)

    Keq = [MeNH3OH][H+]/[MeNH3+] (3)

    this equilibrium constant is also equal to the Ka by definition.

    Try it yourself from here.

    One more equation....

    let x = [H+], [H+] = [MeNH3OH] (4)


    What does (3) become? Solve for x!
     
    Last edited: Feb 21, 2007
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