Question

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

Answer 1

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.`