What volume of #0.200*mol*L^-1# of #NaOH(aq)# would react with a mass of #0.5466*g# of #"KHP"#?

1 Answer
Nov 6, 2016

Answer:

By #"KHPh"# we assume you means #"potassium hydrogen phthalate"#.

Explanation:

#"KHP"# #=# #C_6H_4(CO_2H)(CO_2^(-)K^(+))#.

#"Moles of KHP"# #=# #(0.5466*g)/(204.22*g*mol^-1)=2.677xx10^-3*mol#.

Reaction:

#NaOH(aq) + "KHPh"(aq)rarr"KNaPh"(aq) + H_2O(aq)#

So there is #1:1# equivalence, and #"moles of KHP "-=" moles of sodium hydroxide"#.

And thus, if #0.200*mol*L^-1# #"sodium hydroxide solution"# is available, we need a volume of,

#(2.677xx10^-3*cancel(mol))/(0.200*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)~=13*mL.#