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Homework Help: Chemistry - pH of solution

  1. Apr 12, 2010 #1
    This isnt really homework, Its just a problem I'm trying to learn to solve. I'm not taking a chemistry course so I'm learning on my own.

    Calculate the pH of a solution that is 0.65 M NH4OH and 0.35 M NH4Cl.

    Well we can calulate the pOH = -log(OH-) and 14-pOH = pH

    I guess I need the equilibrium concentration of OH-.

    here is the chemical formula. I believe is necessary

    NH4OH -> NH4SUB]+/SUP] + OH-/SUP]
    the Kb/SUB] =1.5 x 10-3SUP]
  2. jcsd
  3. Apr 12, 2010 #2


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    Staff: Mentor

  4. Apr 12, 2010 #3
    Without using the Henderson–Hasselbalch equation, Can I do this?
    also different Kb value.

    The NN4CL will completely dissociate adding 0.35 NH4

    NH4OH -> NH4+ + OH-
    0.65-x. 0.35+x. . x

    H2O + NH3 -> NH4+ + OH-

    Kb =1.8 x 10^-5 = [NH4+][OH-]/[NH3]= (0.35+x)(x)/ 0.65-x

    x = [OH-]=0.000033 M

    pOH = 4.5
    pH = 14 - pOH = 14 - 4.5 =9.5
  5. Apr 12, 2010 #4


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    Staff: Mentor

    This is correct by accident.

    You look at the dissociation of NH4OH, but you ignore acidic dissociation of NH4+.

    NH4+ <-> NH3 + H+


    My bet is that if you will try to do calculations using your approach and based on this reaction, you will get almost the same result, again accidentally.

    For very small x (compared with 0.35 & 0.65), is (0.35+x)/(0.65-x) substantially different from 0.35/0.65?

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