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^(+))#?

1 Answer
Mar 2, 2017

Answer:

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?