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Homework Help: PH of Salts

  1. Mar 16, 2006 #1
    Which one of the following salts, when dissolved in water, produces the solution with the lowest pH?


    a.NaCl

    b.NH4Cl

    c.MgCl2

    d.AlCl3

    Ammonium chloride

    My answer lies with b or d since they are both acidic salts, but I think is it d since the K_a is greater for Al(3+) than NH4+.


    Thanks.
     
    Last edited: Mar 17, 2006
  2. jcsd
  3. Mar 18, 2006 #2

    siddharth

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    The general rule of thumb is that the salt of a
    • Strong Acid - Strong Base is neutral
    • Weak Acid - Strong Base is basic
    • Strong Acid - Weak Base is acidic
    • Weak acid - Weak base depends on ka and kb
     
  4. Mar 18, 2006 #3
    So the base for NH4Cl is NH4(OH).
    The base for AlCl3 is Al(OH)3.

    Both spawn from the strong acid HCl, so how do I rank them from here with this criteria?
     
  5. Mar 18, 2006 #4

    siddharth

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    Right

    In this case, look at the equilibrium which exists.

    R+ + H2O <====> ROH + H+
    where R+ can be NH4+ or Al3+.

    The challenge here is to relate the pH to Kb. You can do this mathematically and get a relationship from which you can get the answer. If you assume the initial concentration of R+ to be 'a' and the extent of dissociation to be 'x', can you come up with a equation relating the pH and Kb?
    (Hint: Try assuming that x is negligible when compared to 1 to simplify your calculations)
     
    Last edited: Mar 18, 2006
  6. Mar 18, 2006 #5
    pH = 14 + log (K_b*[A-]/[HA]), where K_b = [HA][OH-]/[A-]??

    Is this an expression for that relationship?
     
  7. Mar 18, 2006 #6

    siddharth

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    Not quite.
    First of all can you see that [tex] k_h = \frac{k_w}{k_b} [/tex]? This is because

    H2O <=======> H+ + OH- ---- kw

    ROH <=====> R+ + OH- ----- kb

    Now if you subtract the two equations above you get,
    R+ + H2O <====> ROH + H+
    which is your hydrolysis equilibrium. So K for the above equilibrium will be [tex] k_h = \frac{k_w}{k_b} [/tex].

    If the initial concentration of R+ is 'a' and the extent of dissociation is 'x', then at equilibrium, the concentration of ROH and H+ will be ax and the concentration of R+ will be a-ax. Is it clear till this?

    Now, you know that

    [tex] k_h = \frac{k_w}{k_b} = \frac{(ROH)(H^+)}{(R^+)} [/tex].

    So substitute the concentrations in the above equation. Remember, you need the concentration of H+ (ie, 'ax'). Can you take it from here? If any of the above isn't clear, don't hesitate to ask.
     
    Last edited: Mar 18, 2006
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