# What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.84 and has a freezing point of -2.0 ∘C? (Assume complete dissociation of sodium benzoate and a density of 1.01 g/mL for the solution.)

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1. To be honest I've not seen a problem but here is what I would do. Check my thinking.
You need to know the reatio base/acid and you get that from the Henderson-Hasselbalch equation.
pH = pKa benzoic acid + log (base)/(acid). The base is sodium benzoate and the acid is benzoic acid.. I think the pKa for benzoic aic dis 4.20 but you should confirm that.
4.84 = 4.20 + log b/a
and solve for the ratio b/a. I get b = about 5a

From the freezing point data you get
2 = i*Kf*m
I would use 3 for i (2 for C6H5COONa) and 1 for C6H5COOH). 1.86 is Kf for water Solve for molality. But you still don't know grams each. I get approx 0.36 for molality.
Then with mm = molar mass i have
(b/mm base) + (a/mm acid) = 0.36 mols
(5a/mm b) + (a/mm a) = 0.36 where a = mols a and b = mols b.

That should give you mols acid and mols b (convert to grams).

If anyone else has an idea don't be shy.

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2. Your train of thought is right because thats how i was figuring it. using the buffer solution equation and then following it with the freezing point dep. There may be a shortcut but I cant remember what/how because you have a weak acid with the salt of its conjugate base.

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