Question

If 25.21 mL of `NaOH` solution is required to react completely with .550 g `KHP`, what is the molarity of the `NaOH` solution?

Answer 1

Rxn: `C_6H_4(CO_2H)(CO_2^(-)K^(+)) + NaOH(aq) rarr C_6H_4(CO_2^(-)Na^+)(CO_2^(-)K^(+)) + H_2O`

Explanation:

Given the equation, there is 1:1 equivalence between moles of KHP and moles of sodium hydroxide.

`"Moles of KHP"` `=` `(0.550*g)/(204.22*g*mol^-1)` `=` `2.69xx10^-3*mol`

`"Concentration of NaOH(aq) solution:"` `=` `(2.69xx10^-3*mol)/(25.21xx10^-3*L)` `~=` `0.100*mol*L^-1`, but do the calculation for an exact value.

`"KHP"` is an excellen primary standard, because we could also have used it as a base and not an acid, and used it to standardize a hydrochloirc acid solution.