cherry7girl
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A student dissolved some of the dicarboxylic acid from Q10.1 in water and made up the solution to 250cm3 in a volumetric flask.
In a titration, a 25.0 cm3 sample of the acid solution needed 21.60 cm3 of 0.109 mol dm-3 sodium hydroxide solution for neutralisation.

Calculate the mass in g of the dicarboxylic acid used.

Q10.1 answer: the molecular forumla is C4H6O4
and relative molecular mass is 118


i thought that since it's a dicarboxylic acid, the ratio would be 1:2 NaOH to the acid, but it isn't correct?
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golgiapparatus31
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(Original post by onedance1)
A student dissolved some of the dicarboxylic acid from Q10.1 in water and made up the solution to 250cm3 in a volumetric flask.
In a titration, a 25.0 cm3 sample of the acid solution needed 21.60 cm3 of 0.109 mol dm-3 sodium hydroxide solution for neutralisation.

Calculate the mass in g of the dicarboxylic acid used.

Q10.1 answer: the molecular forumla is C4H6O4
and relative molecular mass is 118


i thought that since it's a dicarboxylic acid, the ratio would be 1:2 NaOH to the acid, but it isn't correct?
Edit: Sorry, I misread your ratio. The dicarboxylic acid contains 2 acidic protons so needs 2 NaOH for 1 acid

(Original post by ColtTheWolf)
Remember that carboxylic acids do not fully dissociate, so you cannot assume that all the H+ ions in solution have evolved from the acid

Edit: This is only in response to the ratio of H+ to OH-
No, the equilibrium will shift continuously until all the acid has dissociated
Last edited by golgiapparatus31; 4 weeks ago
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charco
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(Original post by onedance1)
A student dissolved some of the dicarboxylic acid from Q10.1 in water and made up the solution to 250cm3 in a volumetric flask.
In a titration, a 25.0 cm3 sample of the acid solution needed 21.60 cm3 of 0.109 mol dm-3 sodium hydroxide solution for neutralisation.

Calculate the mass in g of the dicarboxylic acid used.

Q10.1 answer: the molecular forumla is C4H6O4
and relative molecular mass is 118


i thought that since it's a dicarboxylic acid, the ratio would be 1:2 NaOH to the acid, but it isn't correct?
other way round
NaOH to acid = 2:1
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ColtTheWolf
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(Original post by golgiapparatus31)
Edit: Sorry, I misread your ratio. The dicarboxylic acid contains 2 protons so needs 2 NaOH for 1 acid


No, the equilibrium will shift continuously until all the acid has dissociated
Yeah, my bad I totally misread what was going on in that question. Removed comment for clarity.
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