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phyzmatix
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[SOLVED] Water solubility of benzoic acid
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.
We know [tex]pH = -log[H^{+}][/tex]
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.
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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