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Homework Help: Understanding Acids What is the relationship between Ka and Kw?

  1. Mar 27, 2009 #1
    I'm not sure if this belongs in the HW section but I'm not doing HW so I'll post it here.

    My book mentions the auto-ionization constant of water( Kw = [H+][OH-] = 1.0 E-7 ) and shows us how we can use it to find the pH if a strong base is added by solving for [H+].

    Later on in the book it introduces you to the acid-disassociation constant( Ka = ([H+][A-])/[HA] ). This makes some of the other information seem inconsistent to me.

    For example, pOH + pH = 14 is one equation that is given. But this equation was derived using the auto-ionization constant of water. Does this mean pOH = pH = 14 will only work with strong basic and acidic solutions then?
  2. jcsd
  3. Mar 27, 2009 #2


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    No wonder you're confused - the auto-ionization constant is Kw = [H+][OH-] = 10^-14 which is the same thing as pOH + pH = 14.

    Which also means that pOH = 14 - pH and pH = 14 - pOH. So pH=pOH only for pH=pOH=7, neutral.
  4. Mar 27, 2009 #3
    Thanks for the reply
    The 7 was a typo.

    That doesn't address the fragmentation in my understanding though.

    Can I use Kw equilibrium with weak acids/bases? Why?
  5. Mar 27, 2009 #4


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    For strong acids, calculating [H+] is easy because the acid fully dissociates. So, for a monoprotic acid like HCl, the concentration of protons is equal to the concentration for acid. For weak acids like HF, the acid does not fully dissociate so concentration of protons is not equal to the concentration of acid. Here, one uses the expression for Ka to solve for the concentration of protons when the weak acid is added to solution. One can then use the [H+] obtained from the equation for Ka to solve for the concentration of hydroxide ions in solution using the expression for Kw.

    For example, let's say we have weak acid HA with Ka = 0.01 at a concentration of 1M. When the acid is added to solution, approximately 1% of the acid will dissociate, giving a [H+] of approx. 0.01M. Using this value we can calculate [OH-] ~ 1.0x10-12M.
  6. Mar 27, 2009 #5
    Thanks, that sums it up nicely.

    What if an acid is being added to a basic solution? How does Kw play into that type of scenario?
  7. Mar 27, 2009 #6


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  8. Mar 28, 2009 #7
    Thanks, that looks comprehensive.
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