Question

What volume of `2.138*mol*L^-1` `NaOH` is required to react with a `25.0*mL` volume of `0.3057*mol*L^-1` solution of `"potassium pthalate"`, `1,2-C_6H_4(CO_2H)(CO_2^(-)K^(+))`?

Answer 1

Approx. `"3.6"*"mL"`.

Explanation:

We need (i) a stoichiometrically balanced equation to represent the acid-base reaction (this should always be your priority in these types of questions!):

`"NaOH(aq)" + "1,2-C"_6"H"_4("CO"_2^(-)"K"^(+))"CO"_2"H"rarr"1,2-C"_6"H"_4("CO"_2^(-)"K"^(+))("CO"_2^(-)"Na"^(+)) +"H"_2"O"`

There is thus `"1:1 equivalence"` between `"moles of sodium hydroxide"`, and `"moles of KHP."`

And now (ii) we need to find the equivalent quantites of each reagent:

`"Moles of potassium bipthalate,"` `0.3057*mol*cancel(L^-1)xx25.0*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)`
`=7.643xx10^-3*mol`.

And thus volume of standardized `"NaOH(aq)"` required is:

`(7.643xx10^-3*cancel"mol")/(2.138*cancel"mol"*cancel("L"^-1))xx10^3*mL*cancel("L"^-1)=3.58*"mL"`

Do you think it would have been better to have used `0.2138*mol*L^-1` `"NaOH(aq)"` instead of `2.138*mol*L^-1` `"NaOH(aq)"` for this titration? Why, or why not?