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Water solubility of benzoic acid

  • Thread starter phyzmatix
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[SOLVED] Water solubility of benzoic acid

Homework Statement


At [tex]25^{0}C[/tex] a saturated solution of benzoic acid with [tex]K_{a}=6.4 x 10^{-5}[/tex] (can't get this thing to display a simple multiplication sign in the equation) has a pH = 2.80. Calculate the water solubility of benzoic acid in moles per litre.


Homework Equations


We know [tex]pH = -log[H^{+}][/tex]


The Attempt at a Solution


I can't find examples of similar problems in any of my textbooks, but was thinking for the balanced reaction of benzoic acid, we have

[tex]C_{6}H_{5}OOH \rightarrow C_{6}H_{5}OO^{-} + H^{+}[/tex]

and calculating [tex][H^{+}][/tex] from the above equation gives [tex]1.584 x 10^{-3}mol/L[/tex]

Does this mean that (from the balanced equation) the water solubility is equal to this value? Or is there some intermediate steps I was supposed to follow?

Thanks peeps.
 
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Answers and Replies

  • #2
chemisttree
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Write the expression for Ka and plug in what you know.
 
  • #3
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Hi chemisttree!

I thought of that, but don't understand how that will help (I'm not saying you're wrong, I'd just like an explanation)...I thought that, if I substitute the values into the expression for Ka, I can calculate the concentration of [tex]C_{6}H_{5}OOH[/tex] at equilibrium, but that doesn't tell me how much of it was dissolved in the water. My understanding of the terminology is that the water solubility is the maximum amount of the substance that can be dissolved in water. Is that right? Which basically means that it's the difference between the initial concentration (whatever that is) minus what's left over at equilibrium, i.e. the change in concentration which is [tex]1.585x10^{-3}[/tex] (assuming that the initial concentration of [tex]H^{+}[/tex] is 0.

I hope you don't mind my asking, but please explain.
 
  • #4
chemisttree
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Hi chemisttree!

I thought of that, but don't understand how that will help (I'm not saying you're wrong, I'd just like an explanation)...I thought that, if I substitute the values into the expression for Ka, I can calculate the concentration of [tex]C_{6}H_{5}OOH[/tex] at equilibrium, but that doesn't tell me how much of it was dissolved in the water.
Assume that [CH3COOH] in the Ka expression is that amount which is dissolved... ignore the solid stuff. You know [H+] and you therefore know [CH3COO-].
 
  • #5
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Assume that [CH3COOH] in the Ka expression is that amount which is dissolved... ignore the solid stuff. You know [H+] and you therefore know [CH3COO-].
*PING!*

:smile:

Thank you very much for your time! Obviously my understanding of the chemistry underlying this problem was the limiting reagent here :biggrin:
 

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