3
$\begingroup$

Benzoic acid is a weak acid, which means that it does not completely dissociate in water. I know that the dissociation equation of Benzoic acid is:

$$\ce{C6H5CO2H <=> H+ + C6H5CO2-}$$

The question i'm having trouble with is: "Would the dissociated form be more or less likely to dissolve in water?"

Is this basically asking if the ions on the right are more likely to dissolve when compared to $\ce{C6H5CO2H}$?

If this is what the question is asking, I think yes, the dissociated form will be more likely to dissolve as they are very polar when compared to $\ce{C6H5CO2H}$ and they can be surounded by polar water molecules. But I'm not sure.

$\endgroup$
3
$\begingroup$

Solubility of benzoic acid(mw 122.12):

  • 1.7 g/L (0 °C) - 0.00139 moles/100 ml
  • 2.7 g/L (18 °C)
  • 3.44 g/L (25 °C)
  • 5.51 g/L (40 °C)
  • 21.45 g/L (75 °C)
  • 56.31 g/L (100 °C) - 0.04611 moles/100 ml

Solubility of sodium benzoate (mw 144.10):

  • 62.69 g/100 mL (0 °C) - 0.4350 moles/100 ml
  • 62.78 g/100 mL (15 °C)
  • 62.87 g/100 mL (30 °C)
  • 71.11 g/100 mL (100 °C) - 0.4935 moles/100 ml

So yes, sodium benzoate is more soluble than benzoic acid, but consider calcium benzoate which is less soluble.

Calcium benzoate (mw 282.31):

  • 2.32 g/100 mL (0 °C) - 0.00822 moles/100 ml
  • 2.72 g/100 mL (20 °C)
  • 8.7 g/100 mL (100 °C) - 0.03081 moles/100 ml

So the "real" answer is - it depends...

$\endgroup$
1
$\begingroup$

The question i'm having trouble with is:

Would the dissociated form be more or less likely to dissolve in water?

Is this basically asking if the ions on the right are more likely to dissolve when compared to $\ce{C6H5CO2H}$?

Yes, that is the gist of that question.

It basically doesn’t only touch benzoic acid or medium-chain fatty acids which are both much better soluble than the corresponding aldehydes or alcohols. It is also leading the way to things like the structure of amino acids. Take glycine. In organic solvents, it is best described as $\ce{H2N-CH2-COOH}$ while in water it assumes the form $\ce{^{+}H3N-CH2-COO-}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.