Question

Calculate the molarity of a sodium hydroxide solution if 25.00 mL 0.100 M maleic acid requires 22.10 mL of NaOH to reach the endpoint?

Answer 1

You have noted that maleic acid is a diacid.......

Explanation:

We always need a stoichiometric equation to assess the stoichiometry:

`"HO(O=)CCH=CHC(=O)OH(aq)" + "2NaOH(aq)"`

`rarr"Na"^+""^(-)"O(O=)CCH=CHC(=O)""O"^(-)""^+"Na" +"2H"_2"O"(l)`

`"Maleic acid"` is the cis isomer; whereas `"fumaric acid"` is the trans isomer.

`"Moles of acid"=25.00xx10^-3*Lxx0.100*mol*L^-1=2.500xx10^-3*mol.`

But because it is a diacid this is equivalent to `5.00xx10^-3*mol` with respect to `NaOH`........

And so...`[NaOH]=(5.00xx10^-3*mol)/(22.10xx10^-3*L)~=0.2*mol*L^-1`