Question

In standardization, 8.77 mL of `NaOH` neutralized 1.522 g of `KHP`. Given the molar mass of `KHP` is 204.22 g/mol, what is the molarity of the `NaOH` solution?

Answer 1

`"Molarity"=0.8500*mol*L^-1`

Explanation:

`KHP`, `"potassium hydrogen phthalate"`, `1,2-C_6H_4CO_2HCO_2^(-)K^(+)`, is the potassium salt of the diacid, `"pthalic acid"`, `1,2-C_6H_4(CO_2H)_2`.

It can be titrated as a base, or here as an acid. It is air stable, non-hygroscopic, and thus useful as a primary standard.

`1,2-C_6H_4CO_2HCO_2^(-)K^(+)+Na^(+)""^(-)OH rarr "1,2-"C_6H_4CO_2^(-)Na^(+)CO_2^(-)K^(+)+H_2O`

And thus moles of `NaOH` `-=` `"KHP"`, and thus............

`[NaOH]=((1.522*g)/(204.22*g*mol^-1))/(8.77xx10^3*L)=0.8500*mol*L^-1`.